The four variables used when discussing the physical behavior of any gas regardless of its identity are: pressure (P), volume (V), temperature (T), and amount in moles (n). In this lesson, we will use the gas laws to predict the influence of these four variables for any gas.

Pressure and Volume: Boyle’s Law

To relate pressure and volume at a constant temperature and for a fixed amount of gas, we use Boyle’s Law. Boyle’s law states that the gas pressure is inversely proportional to the volume of that gas at constant temperature and for a fixed amount of gas. For two quantities to be inversely related it means that while one quantity increases the other will decrease and vice versa (see figure below). This means that according to Boyle’s law, if the volume is increased, the pressure will decrease. If the pressure of a gas is increased, the volume will decrease.

In the figure below, a sample of gas is shown in a cylinder that has a moveable plunger. Keeping the temperature constant, the plunger is pushed half way down, decreasing the volume by one half. The gas particles now have less volume to move around in resulting in an increase of the number of collisions with the walls of the cylinder and therefore, an increase in pressure.

The relationship between pressure and volume for a fixed amount of gas at constant temperature can be expressed as:

P \(\propto\;\)1/V
or
PV = k where k is a constant value

Because the product of the pressure and the volume is constant for a fixed amount of gas at a constant temperature, the product of the initial pressure (P1) and initial volume (V1) is equal to the product of the final pressure (P2) and the final volume (P2).

P₁V₁ = k and P₂V₂ = k

P₁V₁ = P₂V₂ at constant temperature for a fixed amount of gas.

Boyle’s law can be used to determine the final pressure or volume if a change is made to the initial pressure or volume of a fixed amount of gas at constant temperature.

A person takes a breath and the lungs expand from 4.48 L to 5.59 L. If the initial pressure, before the air was taken in, was 754 mm Hg, what is the pressure after the expansion? Assume constant temperature.

This is a Boyle’s Law problem, and P₁V₁ = P₂V₂.

P₁ = 754 mm Hg, V₁ = 4.48 L, and V₂ = 5.59 L. We need to find P₂. Solve the equation for P₂.

\(\displaystyle P_2\;=\;\frac{P_1V_1}{V_2}\;=\;\frac{754\;mm\;Hg\times\;4.48\;L}{5.59\;L}\;=\;604\;mm\;Hg\)

The answer makes sense as the volume was increased, therefore, we see a decrease in the pressure.

Volume and Temperature: Charles’s Law

What happens if you take a helium filled balloon outside where it is below zero degrees? It begins to deflate. This occurs because the volume of a gas is dependent on the temperature. If the deflated balloon is taken back into the warm house, it slowly begins to inflate again. Again, the gas volume is affected by temperature. Assuming constant pressure, the volume decreases as the temperature is decreased and the volume increases back to the original volume as the temperature is increased.

Kinetic energy is proportional to the temperature in Kelvin.

E_k \(\propto\;\) T (K)

When the temperature is increased, the kinetic energy of the gas particles also increases resulting in an increase in the volume. If temperature is decreased, the kinetic energy of the gas particles decreases which is why a balloon will deflate when taken outside in zero degree weather. The motion of the particles slows down and the forces of collision with the walls of the balloon decrease resulting in a decrease in volume.

Charles’s kaw states for a fixed amount of gas at constant pressure, the volume of a gas is directly proportional to the Kelvin temperature. A direct proportion means that as one quantity increases, the other quantity also increases. The change for both quantities is in the same direction. At higher temperatures the gas particles move with increased kinetic energy and will collide with the container walls with greater force. The rising of a hot air balloon also depends on Charles’s law. In order to maintain constant pressure, the volume of the gas must increase when the temperature is increased. We can state Charles’s law as

V \(\propto\;\) T at constant T for a fixed amount of gas

and

\(\displaystyle\frac{V}{T}\;=\;k\)

where k is a constant.

Because dividing the gas volume by the temperature, at constant pressure, gives a constant, and if the volume and temperature are known under an initial set of conditions, we can calculate the temperature or volume under a different set of conditions when either the volume or temperature are changed.

V₁/T₁ = k and V₂/T₂ = k
\(\displaystyle \frac{V_1}{T_1}\;=\;\frac{V_2}{T_2}\) at constant pressure for a fixed amount of gas.

Charles’s law can be used to determine the final temperature or volume (T₂,V₂) if a change is made to the initial temperature or volume (T₁,V₁) of a fixed amount of gas at constant pressure. In calculations that involve the gas laws, Kelvin temperature must be used. If temperatures are given in °F or °C they must first be converted to Kelvin temperature before carrying out the calculation. In any calculations involving gas laws the temperature must be converted to Kelvin.

Below is a plot of volume vs temperature. Notice the lines can all be extrapolated to zero volume at -273.15°C (0 K).

The coldest absolute temperature is 0 K or -273.15°C. At this temperature the volume of the gas is zero, but gases will condense to either liquid or solid well before reaching 0 K. Once the sample of gas becomes a liquid or a solid, Charles’s Law is no longer valid.

Pressure and Temperature: Gay-Lussac’s Law

On a cold winter day, the air pressure in a tire is lower than it would be on a hot summer day. This is because the gas pressure is temperature dependent. According to Gay-Lussac’s law, the pressure of a fixed amount of gas at constant volume is directly proportional to its Kelvin temperature.

P \(\propto\;\) T at constant P for a fixed amount of gas

and

\(\displaystyle \frac{P}{T}\;=\;k\)

where k is a constant.

Because dividing the gas pressure by the temperature, at constant volume, gives a constant, and if the pressure and temperature are known under an initial set of conditions, we can calculate the pressure or volume under a different set of conditions when either the volume or temperature are changed.

P₁/T₁ = k and P₂/T₂ = k
\(\displaystyle \frac{P_1}{T_1}\;=\;\frac{P_2}{T_2}\) at constant volume for a fixed amount of gas.

Gay-Lussac’s law can be used to determine the final pressure or temperature (P₂, T₂) if a change is made to the initial pressure or temperature (P₁, T₁) of a fixed amount of gas at constant volume. Recall that in calculations that involve the gas laws, Kelvin temperature must be used.

Volume and Moles: Avogadro’s Law

At a constant pressure and temperature, the volume of a gas is directly proportional to the amount of gas, in moles. This relationship is known as Avogadro’s Law.

V \(\propto\;\) n at constant T and P

and

\(\displaystyle \frac{V}{n}\;=\;k\)

where k is a constant.

If the initial volume and number of moles (V₁ and n₁) is known, we can calculate the number of moles or volume when one of the initial quantities is changed.

V₁/n₁ = k and V₂/n₂ = k
\(\displaystyle \frac{V_1}{n_1}\;=\;\frac{V_2}{n_2}\) at constant temperature, T, and Pressure, P, for a fixed amount of gas.

According to Avogadro’s law, as the number of moles of gas increase, the volume also increases. Recall from kinetic molecular theory the particles of gas are very small compared to the empty volume that surrounds them. Because of this there are no attractions between gas particles. This means that the identity of the gas is not important. The value of k (the constant) is the same for all gases, regardless of identity. This allows us to compare the molar quantities of any two gases by just comparing their volumes as long as the temperature and pressure are the same for each gas. Chemists compare the amounts of gas at a set of standard conditions called standard temperature and pressure abbreviated STP. Under standard conditions the pressure is 1 atm and the temperature is 273.15 K (0°C). At STP one mole of any gas occupies a volume of 22.4 L. This is called the standard molar volume. One mole of any gas at STP contains 6.02 x 10²³ molecules of gas and occupy a volume of 22.4 L.

The Combined Gas Law

Recall that PV, V/T, and P/T have constant values for a fixed amount of gas.

\(\displaystyle PV\;=\;k,\;\;\;\;\; \frac{V}{T}\;=\;k,\;\;and\;\;\frac{P}{T}\;=\;k\)

We can combine Boyle’s, Charles’s, and Gay-Lussac’s law into the combined gas law which relates pressure, volume and temperature.

\(\displaystyle\frac{PV}{T}\;=\;k\;and\;\frac{P_1V_1}{T_1}\;=\;\frac{P_2V_2}{T_2}\)

If five of the six quantities are known, we can calculate the sixth quantity. The combined gas law is the only equation you need to recall for a fixed amount of gas. If there is a variable amount of gas, we can use the following:

\(\displaystyle\frac{P_1V_1}{T_1n_1}\;=\;\frac{P_2V_2}{T_2n_2}\)

A 10.5 L sample of carbon dioxide gas that is stored at 22.0°C and 1.0 atm pressure is transferred to a 1.5 L tank. What temperature, in °C, is required to maintain the pressure at 3.2 atm?

Convert 22.0°C to K

K = 22.0°C + 273 = 295 K.

Determine both the known and unknown quantities:

P₁ = 1.0 atm, V₁ = 10.5 L, T₁ = 294 K, P₂ = 3.2 atm, V₂ = 1.5 L, and T_{2 = ?}

Use the combined gas law

\(\displaystyle\frac{P_1V_1}{T_1}\;=\;\frac{P_2V_2}{T_2}\)

Solve for T₂

\(\displaystyle T_2\;=\;\frac{P_2V_2T_1}{P_1V_1}\;=\;\frac{3.2\;atm\times\;1.5\;L\times\;295\;K}{1.0\;atm\times\;10.5\;L}\;=\;135\;K\)

Convert 135 K to °C.

°C = K – 273 = 135 K – 273 = -138°C.

Worksheet: Gas Laws: Part 1
Worksheet: Gas Laws: Part 2

Watch the following problem help videos.

Calculate the Density of a Gas at STP
Use Charles’s Law to Solve for Volume
Use the Combined Gas Law to Solve for Temperature

Exercises

Exercise 1. A gas has a pressure of 675 mm Hg in a volume of 455 mL. What is the pressure of the gas if the volume is changed to 250. mL? Assume a constant temperature. Which law does this follow?

Check Answer/Solution to Exercise 1

Exercise 2. If gas at 75.0°C has a volume of 255 mL. The temperature is increased to 125.2°C, what is the volume? Assume the pressure is constant. Which law does this follow?

Check Answer/Solution to Exercise 2

Exercise 3. One mole of gas at STP occupies a volume of 22.41 L. What is the volume at 25.0°C and 760 mm Hg?

Check Answer/Solution to Exercise 3

Exercise 4. A sample of nitrogen gas at 19.0°C and 760 mm Hg has a volume of 2.83 mL. What is the volume at STP?

Check Answer/Solution to Exercise 4

Exercise 5. A sample of a B vitamin, pantothenic acid, has a mass of 72.52 mg and gives 3.87 mL of nitrogen gas at 22.7°C at 784 mm Hg. What is the volume of nitrogen at STP?

Check Answer/Solution to Exercise 5

Exercise 6. Exercise 6. A sample of helium has a volume of 475 mL at 45.3°C and 742 mm Hg. The temperature is decreased to 21.0°C and the volume is increased to 555 mL. What is the new pressure?

Check Answer/Solution to Exercise 6

Exercise 7. An experiment called for 5.75 L of sulfur dioxide, SO₂, at 0.00°C and 1.00 atm of pressure. What is the volume at 25.00°C and 3.62 atm?

Check Answer/Solution to Exercise 7

Exercise 8. A hot air balloon has a volume of 136 L on a 22.0°C day. What would the volume be if the temperature plummeted to -3.6°C?

Check Answer/Solution to Exercise 8

Exercise 9. The pressure of a gas is reduced from 1205.0 mm Hg to 795.0 mm Hg as the volume is increased by the movement of a piston from 83.5 to 365.0 mL. The original temperature was 86.2°C. What is the final temperature?

Check Answer/Solution to Exercise 9

Exercise 10. A scuba tank with a volume of 13.0 L has a pressure of 156 atm. If the water temperature is 28.0°C, how many liters of air will the tank provide to the diver’s lungs at a depth of 75 feet in the ocean where the pressure is 3.25 atm?

Check Answer/Solution to Exercise 10

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