### Which Is A Correct Statement Of Gay-Lussac’S Law Describing The Behavior Of A Fixed Amount Of Gas?

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• 8 Which is a correct statement of Gay-Lussac’s law describing the behaviour of a fixed amount of gas? Solution : According to Gay-Lussac’s law,at constant volume, the pressure of given mass of the gas is directly proportional to its absolute temperature.

### Which is a correct statement of Charles Law describing the behavior of a fixed amount of gas?

Charles’s law, a statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.

### What does Gay-Lussac law say?

Laws of Gas Properties – There are 4 general laws that relate the 4 basic characteristic properties of gases to each other. Each law is titled by its discoverer. While it is important to understand the relationships covered by each law, knowing the originator is not as important and will be rendered redundant once the combined gas law is introduced.

So concentrate on understanding the relationships rather than memorizing the names. Charles’ Law- gives the relationship between volume and temperature if the pressure and the amount of gas are held constant : 1) If the Kelvin temperature of a gas is increased, the volume of the gas increases. (P, n Constant) 2) If the Kelvin temperature of a gas is decreased, the volume of the gas decreases.

(P, n Constant) This means that the volume of a gas is directly proportional to its Kelvin temperature. Think of it this way, if you increase the volume of a gas and must keep the pressure constant the only way to achieve this is for the temperature of the gas to increase as well. Calculations using Charles’ Law involve the change in either temperature (T 2 ) or volume (V 2 ) from a known starting amount of each (V 1 and T 1 ): Boyle’s Law – states that the volume of a given amount of gas held at constant temperature varies inversely with the applied pressure when the temperature and mass are constant. The reduction in the volume of the gas means that the molecules are striking the walls more often increasing the pressure, and conversely if the volume increases the distance the molecules must travel to strike the walls increases and they hit the walls less often thus decreasing the pressure. Avagadro’s Law- Gives the relationship between volume and amount of gas in moles when pressure and temperature are held constant. If the amount of gas in a container is increased, the volume increases. If the amount of gas in a container is decreased, the volume decreases. Gay Lussac’s Law – states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. If you heat a gas you give the molecules more energy so they move faster. This means more impacts on the walls of the container and an increase in the pressure. Conversely if you cool the molecules down they will slow and the pressure will be decreased. To calculate a change in pressure or temperature using Gay Lussac’s Law the equation looks like this: To play around a bit with the relationships, try this simulation,

## Which of the following statements is true in describing the relationship between volume and temperature at constant pressure?

At constant temperature, pressure is inversely proportional to volume. At constant temperature, pressure is directly proportional to volume. No worries!

## Which of the following is the correct definition of Charles Law equation?

Definition of Charles Law Formula is, ‘When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and therefore the volume is going to be in direct proportion.’ The equation of the law is PV = k.

## Which of the following law explains the relationship between temperature and volume at constant pressure and fixed amount of an ideal gas?

Link to Learning – This video shows how cooling and heating a gas causes its volume to decrease or increase, respectively. These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Figure 9.12 The volume and temperature are linearly related for 1 mole of methane gas at a constant pressure of 1 atm. If the temperature is in kelvin, volume and temperature are directly proportional. The line stops at 111 K because methane liquefies at this temperature; when extrapolated, it intersects the graph’s origin, representing a temperature of absolute zero.

• The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles’s law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre César Charles.
• Charles’s law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant,

Mathematically, this can be written as: V α T or V = constant · T or V = k · T or V 1 / T 1 = V 2 / T 2 V α T or V = constant · T or V = k · T or V 1 / T 1 = V 2 / T 2 with k being a proportionality constant that depends on the amount and pressure of the gas.

## Which is true about the relationship between the volume and the temperature of a sample of a gas at constant pressure?

PV=nRT The ideal gas Law PV = nRT Where does this come from? Robert Boyle found PV = a constant That is, the product of the pressure of a gas times the volume of a gas is a constant for a given sample of gas. In Boyle’s experiments the Temperature (T) did not change, nor did the number of moles (n) of gas present.

So Boyle found PV = (nRT) but did not explore the effect the temperature, or the number of moles would have on pressure and volume. Jaques Charles found V = (a constant) T That is, the volume of a given sample of gas increases linearly with the temperature if the pressure (P) and the amount of the gas (n) is kept constant.

So Charles found V = (nR/P) T Avagadro’s Postulate At the same temperature and pressure equal volumes of all gasses contain the same number of molecules. V = n (a constant) V = n (RT/P) Guy Lussac found that 1 volume of Cl 2 combined with 1 volume of H 2 to make 2 volumes of HCl. With this example we can clearly see the relationship between the number of moles of a gas, and the volume of a gas. At constant temperature and pressure the volume of a gas is directly proportional to the number of moles of gas. Not so coincidentally if V is constant instead of P then P = n (RT/V) At constant temperature and volume the pressure of a gas is directly proportional to the number of moles of gas.

You could remember all the different gas laws, P 1 V 1 = P 2 V 2 P 1 /T 1 = P 2 /T 2 V 1 /T 1 = V 2 /T 2 and so on. Or you could think about the problem a bit and use PV=nRT. N 2 O is placed in a piston. Initially the volume of the piston is 3.0 L, and the pressure of the gas is 5.0 atm. The piston is used to compress the gas to a volume of 1.5 L; determine the pressure of the N 2 O.

well, before the compression P 1 V 1 = n 1 R 1 T 1 or after expansion P 2 V 2 = n 2 R 2 T 2 since n, R, and T do not change substituting P = 1.0 x 10 L

See, if you forget all those different relationships you can just use PV=nRT. A PV = nRT problem What is the volume of 1 mole of an ideal gas at STP (Standard Temperature and Pressure = 0 °C, 1 atm)?

PV = nRT (1) V = 1(0.08206)(273.15) V = 22.41 L So, the volume of an ideal gas is 22.41 L/mol at STP. This, 22.4 L, is probably the most remembered and least useful number in chemistry. Another example What is the volume of 5.0 g NH3 at 25 °C and 1 atm. of pressure? Well we just found that the volume of 1 mole of an ideal gas is 22.41 L so we can use this as a conversion factor.right? Everyone remembers that 1 mol of an ideal gas occupies a volume of 22.4 L, but this is probably the least useful number in chemistry. Alot of people forget that this relationship is only true at STP (0 °C and 1 atm.). So, use PV=nRT To use PV=nRT we need to have moles of NH 3, It is not practical to use PV=nRT as a conversion in a factor label problem so we will just solve for V. V = 7.18 = 7.2 L NH 3 Another Problem Seltzer water is made by dissolving CO 2 in water. Seltzer can be made at home using small containers of pressurized CO 2, If one of the cartridges contains 20.00 mL CO 2 at 55.00 atm at 23.0 °C and it expands into an empty seltzer bottle with a volume of 1.000 L and the resulting pressure is 1.000 atm what is the temperature of the gas. and after the gas expands. or since R has not changed (it is called the universal gas CONSTANT for a reason) and we have not changed the number of moles of CO 2, T = 274.61 K or. T = 274.61 – 273.15 = 1.5 °C continuing the previous problem What will the pressure be when the gas warms to 23.0 °C? before expansion and warming.

#### Which of the following statements explains the relationship between temperature and volume of a gas?

Answer and Explanation: The correct option is (c) As temperature increases, the volume of a gas also increases. Since according to Boyle’s Law, the volume of a gas is directly proportional to the temperature at constant pressure and constant amount of the gas.

### Which of the following is the correct mathematical equation of Charles Law Brainly?

V1T1=V2T2 (Pressure and mass constant)

## Which of the following is the correct mathematical relation of Charles Law at constant pressure?

According to Charles’ law V/T = constant.

## Which of the following is a correct statement about the gas law?

The correct option is A At a given temperature, pressure of a given mass of gas is inversely proportional to its volume.

## What is the theory used to explain the relationship between pressure temperature and volume in gas laws?

The Relationship between Pressure and Volume: Boyle’s Law – As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart.

• Weather balloons get larger as they rise through the atmosphere to regions of lower pressure because the volume of the gas has increased; that is, the atmospheric gas exerts less pressure on the surface of the balloon, so the interior gas expands until the internal and external pressures are equal.
• The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas.

Boyle used a J-shaped tube partially filled with mercury, as shown in Figure $$\PageIndex$$. In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature.

More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure plus the difference in the heights of the mercury columns, and the resulting volume is measured. This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately.

Data such as those from one of Boyle’s own experiments may be plotted in several ways (Figure $$\PageIndex$$). A simple plot of $$V$$ versus $$P$$ gives a curve called a hyperbola and reveals an inverse relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two. Figure $$\PageIndex$$: Boyle’s Experiment Using a J-Shaped Tube to Determine the Relationship between Gas Pressure and Volume. (a) Initially the gas is at a pressure of 1 atm = 760 mmHg (the mercury is at the same height in both the arm containing the sample and the arm open to the atmosphere); its volume is V,

B) If enough mercury is added to the right side to give a difference in height of 760 mmHg between the two arms, the pressure of the gas is 760 mmHg (atmospheric pressure) + 760 mmHg = 1520 mmHg and the volume is V /2. (c) If an additional 760 mmHg is added to the column on the right, the total pressure on the gas increases to 2280 mmHg, and the volume of the gas decreases to V /3.

Dividing both sides by $$P$$ gives an equation illustrating the inverse relationship between $$P$$ and $$V$$: \ or \ where the ∝ symbol is read “is proportional to.” A plot of V versus 1/ P is thus a straight line whose slope is equal to the constant in Equation 6.2.1 and Equation 6.2.3. Figure $$\PageIndex$$: Plots of Boyle’s Data. (a) Here are actual data from a typical experiment conducted by Boyle. Boyle used non-SI units to measure the volume (in.3 rather than cm 3 ) and the pressure (in. Hg rather than mmHg). (b) This plot of pressure versus volume is a hyperbola.

### Which of the simple gas laws describes the relationship between volume and pressure when temperature and moles are kept constant?

The Relationship between Pressure and Volume: Boyle’s Law – As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart. Figure $$\PageIndex$$: Boyle’s Experiment Using a J-Shaped Tube to Determine the Relationship between Gas Pressure and Volume. (a) Initially the gas is at a pressure of 1 atm = 760 mmHg (the mercury is at the same height in both the arm containing the sample and the arm open to the atmosphere); its volume is V,

1. B) If enough mercury is added to the right side to give a difference in height of 760 mmHg between the two arms, the pressure of the gas is 760 mmHg (atmospheric pressure) + 760 mmHg = 1520 mmHg and the volume is V /2.
2. C) If an additional 760 mmHg is added to the column on the right, the total pressure on the gas increases to 2280 mmHg, and the volume of the gas decreases to V /3 (CC BY-SA-NC; anonymous).

The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas. Boyle used a J-shaped tube partially filled with mercury, as shown in Figure $$\PageIndex$$.

In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature. More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure plus the difference in the heights of the mercury columns, and the resulting volume is measured.

This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately. Data such as those from one of Boyle’s own experiments may be plotted in several ways (Figure $$\PageIndex$$).

• A simple plot of $$V$$ versus $$P$$ gives a curve called a hyperbola and reveals an inverse relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two.
• This relationship between the two quantities is described as follows: \ Dividing both sides by $$P$$ gives an equation illustrating the inverse relationship between $$P$$ and $$V$$: \ or \ where the ∝ symbol is read “is proportional to.” A plot of V versus 1/ P is thus a straight line whose slope is equal to the constant in Equations $$\ref$$ and $$\ref$$.

Dividing both sides of Equation $$\ref$$ by V instead of P gives a similar relationship between P and 1/ V, The numerical value of the constant depends on the amount of gas used in the experiment and on the temperature at which the experiments are carried out. Figure $$\PageIndex$$ : Plots of Boyle’s Data. (a) Here are actual data from a typical experiment conducted by Boyle. Boyle used non-SI units to measure the volume (in.3 rather than cm 3 ) and the pressure (in. Hg rather than mmHg). (b) This plot of pressure versus volume is a hyperbola.

1. Because PV is a constant, decreasing the pressure by a factor of two results in a twofold increase in volume and vice versa.
2. C) A plot of volume versus 1/pressure for the same data shows the inverse linear relationship between the two quantities, as expressed by the equation V = constant/ P (CC BY-SA-NC; anonymous).

At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure

#### Which one of the following laws expresses the relationship between the temperature and pressure of a gas?

15 Ideal Gases – An ideal gas is defined as a gas in which the volume of the gas molecules is negligible compared to the volume occupied by the gas. Also, the attraction or repulsion between the individual gas molecules and the container are negligible.

1. Further, for an ideal gas, the molecules are considered to be perfectly elastic and there is no internal energy loss resulting from collision between the molecules.
2. Such ideal gases are said to obey several classical equations such as the Boyle’s law, Charles’s law and the ideal gas equation or the perfect gas equation.

We will first discuss the behavior of ideal gases and then follow it up with the behavior of real gases. If M represents the molecular weight of a gas and the mass of a certain quantity of gas is m, the number of moles is given by (3.42) n = m / M where n is the number that represents the number of moles in the given mass.

• As an example, the molecular weight of methane is 16.043.
• Therefore, 50 lb of methane will contain approximately 3 mol.
• The ideal gas law, sometimes referred to as the perfect gas equation simply states that the pressure, volume, and temperature of the gas are related to the number of moles by the following equation.

(3.43) PV = nRT where P – Absolute pressure, psia V – Gas volume, ft 3 n – Number of lb moles as defined in Equation (3.42) R – Universal gas constant T – Absolute temperature of gas, °R (°F + 460). The universal gas constant R has a value of 10.732 psia ft 3 /lb mole °R in USCS units.

• We can combine Eqn (3.42) with Eqn (3.43) and express the ideal gas equation as follows (3.44) PV = mRT / M where all symbols have been defined previously.
• It has been found that the ideal gas equation is correct only at low pressures close to the atmospheric pressure.
• Because gas pipelines generally operate at pressures higher than atmospheric pressures, we must modify Eqn (3.44) to take into account the effect of compressibility.

The latter is accounted for by using a term called the compressibility factor or gas deviation factor. We will discuss the compressibility factor later in this chapter. In the perfect gas Eqn (3.44), the pressures and temperatures must be in absolute units.

Absolute pressure is defined as the gauge pressure (as measured by a gauge) plus the local atmospheric pressure. Therefore (3.45) P abs = P gauge + P atm Thus if the gas pressure is 20 psig and the atmospheric pressure is 14.7 psia, we get the absolute pressure of the gas as P abs = 20 + 14.7 = 34.7 psia Absolute pressure is expressed as psia, whereas the gauge pressure is referred to as psig.

The adder to the gauge pressure, which is the local atmospheric pressure, is also called the base pressure. In SI units, 500 kPa gauge pressure is equal to 601 kPa absolute pressure if the base pressure is 101 kPa. The absolute temperature is measured above a certain datum.

In USCS units, the absolute scale of temperatures is designated as degree Rankin (°R) and is equal to the sum of the temperature in °F and the constant 460. In SI units, the absolute temperature scale is referred to as degree Kelvin (K). Absolute temperature in K is equal to °C + 273. Therefore, Absolute temperature, °R = Temp °F + 460.

Absolute temperature, K = Temp °C + 460. It is customary to drop the degree symbol for absolute temperature in Kelvin. Ideal gases also obey Boyle’s law and Charles’s law. Boyle’s law is used to relate the pressure and volume of a given quantity of gas when the temperature is kept constant.

• Constant temperature is also called isothermal condition.
• Boyle’s law is as follows P 1 / P 2 = V 2 / V 1 or (3.46) P 1 V 1 = P 2 V 2 where P 1 and V 1 are the pressure and volume at condition 1 and P 2 and V 2 are the corresponding value at some other condition 2 where the temperature is not changed.

Charles’s law states that for constant pressure, the gas volume is directly proportional to the gas temperature. Similarly, if volume is kept constant, the pressure varies directly as the temperature. Therefore we can state the following. (3.47) V 1 / V 2 = T 1 / T 2 at constant pressure (3.48) P 1 / P 2 = T 1 / T 2 at constant volume

## What are the relationships among the volume temperature pressure and amount of a given sample of gas?

Learning Objectives – By the end of this section, you will be able to:

Identify the mathematical relationships between the various properties of gases Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions

During the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly ( ), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas.

Although their measurements were not precise by today’s standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an ideal gas—a hypothetical construct that real gases approximate under certain conditions.

Eventually, these individual laws were combined into a single equation—the ideal gas law —that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law. Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in, We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then P and T are directly proportional (again, when volume and moles of gas are held constant ); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor. Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the P – T relationship for gases is known as either Amontons’s law or Gay-Lussac’s law,

• Under either name, it states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant,
• Mathematically, this can be written: $$P\propto T\phantom }\text \phantom }P=\text \phantom }×\phantom }T\phantom }\text \phantom }P=k\phantom }×\phantom }T$$ where ∝ means “is proportional to,” and k is a proportionality constant that depends on the identity, amount, and volume of the gas.

For a confined, constant volume of gas, the ratio $$\frac$$ is therefore constant (i.e., $$\frac \phantom }=k$$). If the gas is initially in “Condition 1” (with P = P 1 and T = T 1 ), and then changes to “Condition 2” (with P = P 2 and T = T 2 ), we have that $$\frac _ } _ }\phantom }=k$$ and $$\frac _ } _ }\phantom }=k,$$ which reduces to $$\frac _ } _ }\phantom }=\phantom }\frac _ } _ }.$$ This equation is useful for pressure-temperature calculations for a confined gas at constant volume.

1. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called absolute zero ).
2. Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.) Predicting Change in Pressure with Temperature A can of hair spray is used until it is empty except for the propellant, isobutane gas.

(a) On the can is the warning “Store only at temperatures below 120 °F (48.8 °C). Do not incinerate.” Why? (b) The gas in the can is initially at 24 °C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 °C on a hot day, what is the new pressure in the can? Solution (a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately.

High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.) (b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons’s/Gay-Lussac’s law. Taking P 1 and T 1 as the initial values, T 2 as the temperature where the pressure is unknown and P 2 as the unknown pressure, and converting °C to K, we have: $$\frac _ } _ }\phantom }=\phantom }\frac _ } _ }\phantom }\text \phantom }\frac }\text } }\text }\phantom }=\phantom }\frac _ } }\text }$$ Rearranging and solving gives: $$_ =\phantom }\frac }\text \phantom }×\phantom }323\phantom }\overline }} }\overline }}\phantom }=390\phantom }\text$$ Check Your Learning A sample of nitrogen, N 2, occupies 45.0 mL at 27 °C and 600 torr.

What pressure will it have if cooled to –73 °C while the volume remains constant? If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let’s say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). This video shows how cooling and heating a gas causes its volume to decrease or increase, respectively. These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles’s law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre César Charles. Charles’s law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant,

Mathematically, this can be written as: $$V\phantom }\text \phantom }T\phantom }\text \phantom }V=\text \text T\phantom }\text \phantom }V=k\text T\phantom }\text \phantom } _ \text _ = _ \text _$$ with k being a proportionality constant that depends on the amount and pressure of the gas. For a confined, constant pressure gas sample, $$\frac$$ is constant (i.e., the ratio = k ), and as seen with the P – T relationship, this leads to another form of Charles’s law: $$\frac _ } _ }\phantom }=\phantom }\frac _ } _ }.$$ Predicting Change in Volume with Temperature A sample of carbon dioxide, CO 2, occupies 0.300 L at 10 °C and 750 torr.

What volume will the gas have at 30 °C and 750 torr? Solution Because we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles’s law. Taking V 1 and T 1 as the initial values, T 2 as the temperature at which the volume is unknown and V 2 as the unknown volume, and converting °C into K we have: $$\frac _ } _ }\phantom }=\phantom }\frac _ } _ }\phantom }\text \phantom }\frac }\text } }\text }\phantom }=\phantom }\frac _ } }\text }$$ Rearranging and solving gives: $$_ =\phantom }\frac }\text \phantom }×\phantom }\text \phantom }\overline }} }\overline }}\phantom }=0.321\phantom }\text$$ This answer supports our expectation from Charles’s law, namely, that raising the gas temperature (from 283 K to 303 K) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L).

• Check Your Learning A sample of oxygen, O 2, occupies 32.2 mL at 30 °C and 452 torr.
• What volume will it occupy at –70 °C and the same pressure? Measuring Temperature with a Volume Change Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure.

The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm 3 when immersed in a mixture of ice and water (0.00 °C). When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm 3, Find the temperature of boiling ammonia on the kelvin and Celsius scales.

Solution A volume change caused by a temperature change at constant pressure means we should use Charles’s law. Taking V 1 and T 1 as the initial values, T 2 as the temperature at which the volume is unknown and V 2 as the unknown volume, and converting °C into K we have: $$\frac _ } _ }\phantom }=\phantom }\frac _ } _ }\phantom }\text \phantom }\frac } }^ } }\text }\phantom }=\phantom }\frac } }^ } _ }$$ Rearrangement gives $$_ =\phantom }\frac } }}^ \phantom }×\phantom }273.15\phantom }\text } } }}^ }\phantom }=239.8\phantom }\text$$ Subtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is –33.4 °C.

Check Your Learning What is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm? If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say 25 °C.

• If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases.
• This example of the effect of volume on the pressure of a given amount of a confined gas is true in general.

Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in, Unlike the P – T and V – T relationships, pressure and volume are not directly proportional to each other. Instead, P and V exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written: $$P\phantom }\text \phantom }1\text V\phantom }\text \phantom }P=k\text 1\text V\phantom }\text \phantom }P\text V=k\phantom }\text \phantom } _ _ = _ _$$ with k being a constant.

Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure $$\left(\frac \right)$$ versus the volume ( V ), or the inverse of volume $$\left(\frac \right)$$ versus the pressure ( P ). Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data.

For those reasons, scientists often try to find a way to “linearize” their data. If we plot P versus V, we obtain a hyperbola (see ). The relationship between pressure and volume is inversely proportional. (a) The graph of P vs. V is a hyperbola, whereas (b) the graph of $$\left(\frac \right)$$ vs. The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as Boyle’s law : The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.

Volume of a Gas Sample The sample of gas in has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL, using: (a) the P – V graph in (b) the $$\frac$$ vs. V graph in (c) the Boyle’s law equation Comment on the likely accuracy of each method. Solution (a) Estimating from the P – V graph gives a value for P somewhere around 27 psi.

(b) Estimating from the $$\frac$$ versus V graph give a value of about 26 psi. (c) From Boyle’s law, we know that the product of pressure and volume ( PV ) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have P 1 V 1 = k and P 2 V 2 = k which means that P 1 V 1 = P 2 V 2,

Using P 1 and V 1 as the known values 13.0 psi and 15.0 mL, P 2 as the pressure at which the volume is unknown, and V 2 as the unknown volume, we have: $$_ _ = _ _ \phantom }\text \phantom }13.0\phantom }\text \phantom }\phantom }×\phantom }15.0\phantom }\text = _ \phantom }×\phantom }7.5\phantom }\text$$ Solving: $$_ =\phantom }\frac }\text \phantom }\phantom }×\phantom }15.0\phantom }\overline }} }\overline }}\phantom }=26\phantom }\text$$ It was more difficult to estimate well from the P – V graph, so (a) is likely more inaccurate than (b) or (c).

The calculation will be as accurate as the equation and measurements allow. Check Your Learning The sample of gas in has a volume of 30.0 mL at a pressure of 6.5 psi. Determine the volume of the gas at a pressure of 11.0 psi, using: (a) the P – V graph in (b) the $$\frac$$ vs.

V graph in (c) the Boyle’s law equation Comment on the likely accuracy of each method. Answer: (a) about 17–18 mL; (b) ~18 mL; (c) 17.7 mL; it was more difficult to estimate well from the P – V graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow Breathing and Boyle’s Law What do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing.

How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger.

• The increase in volume leads to a decrease in pressure (Boyle’s law).
• This causes air to flow into the lungs (from high pressure to low pressure).
• When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle’s law again), and air flows out of the lungs (from high pressure to low pressure).

You then breathe in and out again, and again, repeating this Boyle’s law cycle for the rest of your life ( ). Breathing occurs because expanding and contracting lung volume creates small pressure differences between your lungs and your surroundings, causing air to be drawn into and forced out of your lungs. The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by Avogadro’s law : For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant, Visit this interactive PhET simulation to investigate the relationships between pressure, volume, temperature, and amount of gas. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws).

Boyle’s law: PV = constant at constant T and n Amontons’s law: $$\frac$$ = constant at constant V and n Charles’s law: $$\frac$$ = constant at constant P and n Avogadro’s law: $$\frac$$ = constant at constant P and T

Combining these four laws yields the ideal gas law, a relation between the pressure, volume, temperature, and number of moles of a gas: $$PV=nRT$$ where P is the pressure of a gas, V is its volume, n is the number of moles of the gas, T is its temperature on the kelvin scale, and R is a constant called the ideal gas constant or the universal gas constant.

The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L atm mol –1 K –1 and 8.314 kPa L mol –1 K –1, Gases whose properties of P, V, and T are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behavior or to approximate the traits of an ideal gas,

An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature.

In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behavior observed for many gases at relatively high pressures and low temperatures. The ideal gas equation contains five terms, the gas constant R and the variable properties P, V, n, and T,

Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises. Using the Ideal Gas Law Methane, CH 4, is being considered for use as an alternative automotive fuel to replace gasoline.

One gallon of gasoline could be replaced by 655 g of CH 4, What is the volume of this much methane at 25 °C and 745 torr? Solution We must rearrange PV = nRT to solve for V : $$V=\phantom }\frac$$ If we choose to use R = 0.08206 L atm mol –1 K –1, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.

Converting into the “right” units: $$n=6\phantom }55\phantom }\overline \phantom } }_ }\phantom }×\phantom }\frac }\text } } }}_ }\phantom }=40.8\phantom }\text$$ $$T=25\phantom }\text +273=298\phantom }\text$$ $$P=745\phantom }\overline }\phantom }×\phantom }\frac }\text } }\overline }}\phantom }=0.980\phantom }\text$$ $$V=\phantom }\frac \phantom }=\phantom }\frac }\overline }\right)\left(0.08206\phantom }\text \phantom }\overline }^ \phantom } }^ }}\right)\left(298\phantom }\overline }\right)} }\overline }}\phantom }=1.02\phantom }×\phantom } ^ \text$$ It would require 1020 L (269 gal) of gaseous methane at about 1 atm of pressure to replace 1 gal of gasoline.

1. It requires a large container to hold enough methane at 1 atm to replace several gallons of gasoline.
2. Check Your Learning Calculate the pressure in bar of 2520 moles of hydrogen gas stored at 27 °C in the 180-L storage tank of a modern hydrogen-powered car.
3. If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: $$\frac _ _ } _ }\phantom }=\phantom }\frac _ _ } _ }$$ using units of atm, L, and K.

Both sets of conditions are equal to the product of n $$×$$ R (where n = the number of moles of the gas and R is the ideal gas law constant). Using the Combined Gas Law When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm ( ). Letting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37 °C, we have: $$\frac _ _ } _ }\phantom }=\phantom }\frac _ _ } _ }\phantom }⟶\phantom }\frac }\text \right)\left(13.2\phantom }\text \right)} }\text \right)}\phantom }=\phantom }\frac }\text \right)\left( _ \right)} }\text \right)}$$ Solving for V 2 : $$_ =\phantom }\frac }\overline }\right)\left(13.2\phantom }\text \right)\left(310\phantom }\overline }\right)} }\overline }\right)\left(3.13\phantom }\overline }\right)}\phantom }=667\phantom }\text$$ (Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures.

1. Despite this limitation, the calculated volume can be viewed as a good “ballpark” estimate.) Check Your Learning A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm.
2. Find the volume of this sample at 0 °C and 1.00 atm.
3. The Interdependence between Ocean Depth and Pressure in Scuba Diving Whether scuba diving at the Great Barrier Reef in Australia (shown in ) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety.

Scuba divers, whether at the Great Barrier Reef or in the Caribbean, must be aware of buoyancy, pressure equalization, and the amount of time they spend underwater, to avoid the risks associated with pressurized gases in the body. (credit: Kyle Taylor) Pressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as “atmospheres absolute” or ATA in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level.

1. As a diver descends, the increase in pressure causes the body’s air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs.
2. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization.

Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in his BCD expands because of lower pressure according to Boyle’s law (decreasing the pressure of gases increases the volume).

The expanding air increases the buoyancy of the diver, and she or he begins to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor.

The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume.

The diver uses up available air twice as fast as at the surface. We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 kPa).

At STP, an ideal gas has a volume of about 22.4 L—this is referred to as the standard molar volume ( ). Since the number of moles in a given volume of gas varies with pressure and temperature changes, chemists use standard temperature and pressure (273.15 K and 1 atm or 101.325 kPa) to report properties of gases. The behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons’s law).

• The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles’s law).
• The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle’s law).
• Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro’s law).

The equations describing these laws are special cases of the ideal gas law, PV = nRT, where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, T is its kelvin temperature, and R is the ideal (universal) gas constant.

Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why? Explain how the volume of the bubbles exhausted by a scuba diver ( ) change as they rise to the surface, assuming that they remain intact. As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle’s law.

One way to state Boyle’s law is “All other things being equal, the pressure of a gas is inversely proportional to its volume.” (a) What is the meaning of the term “inversely proportional?” (b) What are the “other things” that must be equal? An alternate way to state Avogadro’s law is “All other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas.” (a) What is the meaning of the term “directly proportional?” (b) What are the “other things” that must be equal? (a) The number of particles in the gas increases as the volume increases.

(b) temperature, pressure How would the graph in change if the number of moles of gas in the sample used to determine the curve were doubled? How would the graph in change if the number of moles of gas in the sample used to determine the curve were doubled? The curve would be farther to the right and higher up, but the same basic shape.

In addition to the data found in, what other information do we need to find the mass of the sample of air used to determine the graph? Determine the volume of 1 mol of CH 4 gas at 150 K and 1 atm, using, About 12.2 L Determine the pressure of the gas in the syringe shown in when its volume is 12.5 mL, using: (a) the appropriate graph (b) Boyle’s law A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 °C.

If the can is thrown into a fire (T = 475 °C), what will be the pressure in the hot can? 3.40 $$×$$ 10 3 torr What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 13.3 L at 55 °C and 744 torr? A 2.50-L volume of hydrogen measured at –196 °C is warmed to 100 °C.

Calculate the volume of the gas at the higher temperature, assuming no change in pressure.12.1 L A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon? A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 °C at ground level. 217 L The volume of an automobile air bag was 66.8 L when inflated at 25 °C with 77.8 g of nitrogen gas. What was the pressure in the bag in kPa? How many moles of gaseous boron trifluoride, BF 3, are contained in a 4.3410-L bulb at 788.0 K if the pressure is 1.220 atm? How many grams of BF 3 ? 8.190 $$×$$ 10 –2 mol; 5.553 g Iodine, I 2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed.

What is the temperature in a 73.3-mL bulb that contains 0.292 g of I 2 vapor at a pressure of 0.462 atm? How many grams of gas are present in each of the following cases? (a) 0.100 L of CO 2 at 307 torr and 26 °C (b) 8.75 L of C 2 H 4, at 378.3 kPa and 483 K (c) 221 mL of Ar at 0.23 torr and –54 °C (a) 7.24 $$×$$ 10 –2 g; (b) 23.1 g; (c) 1.5 $$×$$ 10 –4 g A high altitude balloon is filled with 1.41 $$×$$ 10 4 L of hydrogen at a temperature of 21 °C and a pressure of 745 torr.

What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr? A cylinder of medical oxygen has a volume of 35.4 L, and contains O 2 at a pressure of 151 atm and a temperature of 25 °C. What volume of O 2 does this correspond to at normal body conditions, that is, 1 atm and 37 °C? 5561 L A large scuba tank ( ) with a volume of 18 L is rated for a pressure of 220 bar.

The tank is filled at 20 °C and contains enough air to supply 1860 L of air to a diver at a pressure of 2.37 atm (a depth of 45 feet). Was the tank filled to capacity at 20 °C? A 20.0-L cylinder containing 11.34 kg of butane, C 4 H 10, was opened to the atmosphere. Calculate the mass of the gas remaining in the cylinder if it were opened and the gas escaped until the pressure in the cylinder was equal to the atmospheric pressure, 0.983 atm, and a temperature of 27 °C.46.4 g While resting, the average 70-kg human male consumes 14 L of pure O 2 per hour at 25 °C and 100 kPa.

How many moles of O 2 are consumed by a 70 kg man while resting for 1.0 h? For a given amount of gas showing ideal behavior, draw labeled graphs of: (a) the variation of P with V (b) the variation of V with T (c) the variation of P with T (d) the variation of $$\frac$$ with V For a gas exhibiting ideal behavior: A liter of methane gas, CH 4, at STP contains more atoms of hydrogen than does a liter of pure hydrogen gas, H 2, at STP. Using Avogadro’s law as a starting point, explain why. The effect of chlorofluorocarbons (such as CCl 2 F 2 ) on the depletion of the ozone layer is well known.

The use of substitutes, such as CH 3 CH 2 F( g ), for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by 10.0 g of each of these compounds at STP: (a) CCl 2 F 2 ( g ) (b) CH 3 CH 2 F( g ) (a) 1.85 L CCl 2 F 2 ; (b) 4.66 L CH 3 CH 2 F As 1 g of the radioactive element radium decays over 1 year, it produces 1.16 $$×$$ 10 18 alpha particles (helium nuclei).

Each alpha particle becomes an atom of helium gas. What is the pressure in pascal of the helium gas produced if it occupies a volume of 125 mL at a temperature of 25 °C? A balloon that is 100.21 L at 21 °C and 0.981 atm is released and just barely clears the top of Mount Crumpet in British Columbia.

#### Which of the following best describes the relationship between the pressure and the temperature of a gas?

The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas.

## How do you explain the relationship of volume and pressure and volume and temperature using the kinetic molecular theory?

The Kinetic Molecular Theory The Kinetic Molecular Theory Postulates The experimental observations about the behavior of gases discussed so far can be explained with a simple theoretical model known as the kinetic molecular theory, This theory is based on the following postulates, or assumptions.

Gases are composed of a large number of particles that behave like hard, spherical objects in a state of constant, random motion. These particles move in a straight line until they collide with another particle or the walls of the container. These particles are much smaller than the distance between particles. Most of the volume of a gas is therefore empty space. There is no force of attraction between gas particles or between the particles and the walls of the container. Collisions between gas particles or collisions with the walls of the container are perfectly elastic. None of the energy of a gas particle is lost when it collides with another particle or with the walls of the container. The average kinetic energy of a collection of gas particles depends on the temperature of the gas and nothing else.

The assumptions behind the kinetic molecular theory can be illustrated with the apparatus shown in the figure below, which consists of a glass plate surrounded by walls mounted on top of three vibrating motors. A handful of steel ball bearings are placed on top of the glass plate to represent the gas particles. When the motors are turned on, the glass plate vibrates, which makes the ball bearings move in a constant, random fashion (postulate 1). Each ball moves in a straight line until it collides with another ball or with the walls of the container (postulate 2).

• Although collisions are frequent, the average distance between the ball bearings is much larger than the diameter of the balls (postulate 3).
• There is no force of attraction between the individual ball bearings or between the ball bearings and the walls of the container (postulate 4).
• The collisions that occur in this apparatus are very different from those that occur when a rubber ball is dropped on the floor.

Collisions between the rubber ball and the floor are inelastic, as shown in the figure below. A portion of the energy of the ball is lost each time it hits the floor, until it eventually rolls to a stop. In this apparatus, the collisions are perfectly elastic, Any object in motion has a kinetic energy that is defined as one-half of the product of its mass times its velocity squared. KE = 1 / 2 mv 2 At any time, some of the ball bearings on this apparatus are moving faster than others, but the system can be described by an average kinetic energy, How the Kinetic Molecular Theory Explains the Gas Laws The kinetic molecular theory can be used to explain each of the experimentally determined gas laws. The Link Between P and n The pressure of a gas results from collisions between the gas particles and the walls of the container. T ) The last postulate of the kinetic molecular theory states that the average kinetic energy of a gas particle depends only on the temperature of the gas. Thus, the average kinetic energy of the gas particles increases as the gas becomes warmer. Because the mass of these particles is constant, their kinetic energy can only increase if the average velocity of the particles increases.

The faster these particles are moving when they hit the wall, the greater the force they exert on the wall. Since the force per collision becomes larger as the temperature increases, the pressure of the gas must increase as well. Boyle’s Law ( P = 1/ v ) Gases can be compressed because most of the volume of a gas is empty space.

If we compress a gas without changing its temperature, the average kinetic energy of the gas particles stays the same. There is no change in the speed with which the particles move, but the container is smaller. Thus, the particles travel from one end of the container to the other in a shorter period of time. T ) The average kinetic energy of the particles in a gas is proportional to the temperature of the gas. Because the mass of these particles is constant, the particles must move faster as the gas becomes warmer. If they move faster, the particles will exert a greater force on the container each time they hit the walls, which leads to an increase in the pressure of the gas. N ) As the number of gas particles increases, the frequency of collisions with the walls of the container must increase. This, in turn, leads to an increase in the pressure of the gas. Flexible containers, such as a balloon, will expand until the pressure of the gas inside the balloon once again balances the pressure of the gas outside.

Thus, the volume of the gas is proportional to the number of gas particles. Dalton’s Law of Partial Pressures ( P t = P 1 + P 2 + P 3 +,) Imagine what would happen if six ball bearings of a different size were added to the molecular dynamics simulator, The total pressure would increase because there would be more collisions with the walls of the container.

But the pressure due to the collisions between the original ball bearings and the walls of the container would remain the same. There is so much empty space in the container that each type of ball bearing hits the walls of the container as often in the mixture as it did when there was only one kind of ball bearing on the glass plate. Graham’s Laws of Diffusion and Effusion A few of the physical properties of gases depend on the identity of the gas. One of these physical properties can be seen when the movement of gases is studied. In 1829 Thomas Graham used an apparatus similar to the one shown in the figure below to study the diffusion of gases the rate at which two gases mix. This apparatus consists of a glass tube sealed at one end with plaster that has holes large enough to allow a gas to enter or leave the tube. When the tube is filled with H 2 gas, the level of water in the tube slowly rises because the H 2 molecules inside the tube escape through the holes in the plaster more rapidly than the molecules in air can enter the tube. Graham found that the rates at which gases diffuse is inversely proportional to the square root of their densities. This relationship eventually became known as Graham’s law of diffusion, To understand the importance of this discovery we have to remember that equal volumes of different gases contain the same number of particles. As a result, the number of moles of gas per liter at a given temperature and pressure is constant, which means that the density of a gas is directly proportional to its molecular weight. Similar results were obtained when Graham studied the rate of effusion of a gas, which is the rate at which the gas escapes through a pinhole into a vacuum. The rate of effusion of a gas is also inversely proportional to the square root of either the density or the molecular weight of the gas. Graham’s law of effusion can be demonstrated with the apparatus in the figure below. A thick-walled filter flask is evacuated with a vacuum pump. A syringe is filled with 25 mL of gas and the time required for the gas to escape through the syringe needle into the evacuated filter flask is measured with a stop watch. As we can see when data obtained in this experiment are graphed in the figure below, the time required for 25-mL samples of different gases to escape into a vacuum is proportional to the square root of the molecular weight of the gas. The rate at which the gases effuse is therefore inversely proportional to the square root of the molecular weight. The Kinetic Molecular Theory and Graham’s Laws The kinetic molecular theory can be used to explain the results Graham obtained when he studied the diffusion and effusion of gases. The key to this explanation is the last postulate of the kinetic theory, which assumes that the temperature of a system is proportional to the average kinetic energy of its particles and nothing else. This equation can be simplified by multiplying both sides by two. It can then be rearranged to give the following. Taking the square root of both sides of this equation gives a relationship between the ratio of the velocities at which the two gases move and the square root of the ratio of their molecular weights. This equation is a modified form of Graham’s law. It suggests that the velocity (or rate) at which gas molecules move is inversely proportional to the square root of their molecular weights.

#### What is the statement of Charles’s law?

Charle’s Law – Definition, Formula, Derivation, Application Charles law states that the volume of an ideal gas is directly proportional to the absolute temperature at constant pressure. The law also states that the Kelvin temperature and the volume will be in direct proportion when the pressure exerted on a sample of a dry gas is held constant.

## What is the behavior of Charles Law?

– Charles’s Law – Charles’s law was discovered in the 1700s by a French physicist named Jacques Charles. According to Charles’s law, if the pressure of a gas is held constant, increasing the temperature of the gas increases its volume. What happens when a gas is heated? Its particles gain energy. As the temperature of a gas increases, its volume also increases. Roger had a latex balloon full of air inside his air-conditioned house. When he took the balloon outside in the hot sun, it got bigger and bigger until it popped. Boyle’s law explains why. As the gas in the balloon warmed in the sun, its volume increased. It stretched and expanded the latex of the balloon until the balloon burst.

### Which of the following statements is true about Charles’s law?

Which of the following statements correctly describes Charles’s Law? The volume of a gas is directly proportional to the temperature of the gas, if the pressure and amount of gas remain constant.

## What are the following describe the Charles Law?

Charles’s Law – French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. Charles’s Law states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. Figure $$\PageIndex$$: As a container of confined gas is heated, its molecules increase in kinetic energy and push the movable piston outward, resulting in an increase in volume. (CC BY-NC; CK-12) Mathematically, the direct relationship of Charles’s Law can be represented by the following equation: \ As with Boyle’s Law, $$k$$ is constant only for a given gas sample.

Table $$\PageIndex$$: Temperature-Volume Data

Temperature $$\left( \text \right)$$ Volume $$\left( \text \right)$$ $$\frac = k$$ $$\left( \frac } } \right)$$
50 20 0.40
100 40 0.40
150 60 0.40
200 80 0.40
300 120 0.40
500 200 0.40
1000 400 0.40

When this data is graphed, the result is a straight line, indicative of a direct relationship, shown in the figure below. Figure $$\PageIndex$$: The volume of a gas increases as the Kelvin temperature increases. Notice that the line goes exactly toward the origin, meaning that as the absolute temperature of the gas approaches zero, its volume approaches zero. However, when a gas is brought to extremely cold temperatures, its molecules would eventually condense into the liquid state before reaching absolute zero.

• The temperature at which this change into the liquid state occurs varies for different gases.
• Charles’s Law can also be used to compare changing conditions for a gas.
• Now we use $$V_1$$ and $$T_1$$ to stand for the initial volume and temperature of a gas, while $$V_2$$ and $$T_2$$ stand for the final volume and temperature.

The mathematical relationship of Charles’s Law becomes: \ This equation can be used to calculate any one of the four quantities if the other three are known. The direct relationship will only hold if the temperatures are expressed in Kelvin. Temperatures in Celsius will not work.