Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container ([link]). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton’s law of partial pressures: *The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases*: $$P_{Total}=P_A+P_B+P_C+…=Σ_iP_i$$

In the equation *P _{Total}* is the total pressure of a mixture of gases,

*P*is the partial pressure of gas A;

_{A}*P*is the partial pressure of gas B;

_{B}*P*is the partial pressure of gas C; and so on.

_{C}The partial pressure of gas A is related to the total pressure of the gas mixture via its **mole fraction ( X)**, a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components:

$$P_A=X_A×P_{Total}\qquad where\qquad X_A=\frac{n_A}{n_{Total}}$$

where *P _{A}*,

*X*, and

_{A}*n*are the partial pressure, mole fraction, and number of moles of gas A, respectively, and

_{A}*n*is the number of moles of all components in the mixture.

_{Total}**The Pressure of a Mixture of Gases**

A 10.0-L vessel contains 2.50 ×10^{−3} mol of H_{2}, 1.00 ×10^{−3} mol of He, and 3.00 ×10^{−4} mol of Ne at 35 °C.

(a) What are the partial pressures of each of the gases?

(b) What is the total pressure in atmospheres?

**Solution**

The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using $P=\frac{nRT}{V}$:

$$P_{H_2}=\frac{(2.50×10^{−3}\;\require{enclose}\enclose{horizontalstrike}{mol})(0.08206\; \enclose{horizontalstrike}{L}\;atm\; \enclose{horizontalstrike}{mol^{-1}K^{−1}})(308\; \enclose{horizontalstrike}{K})}{10.0\; \enclose{horizontalstrike} {L}}=6.32×10^{−3}\;atm$$

$$P_{He}= \frac{(1.00×10^{−3}\;\enclose{horizontalstrike}{mol})(0.08206\; \enclose{horizontalstrike}{L}\;atm\; \enclose{horizontalstrike}{mol^{-1}K^{−1}})(308\; \enclose{horizontalstrike}{K})}{10.0\; \enclose{horizontalstrike} {L}}=2.53×10^{−3}\;atm$$

$$P_{Ne}=\frac{(3.00×10^{−4}\;\enclose{horizontalstrike}{mol})(0.08206\; \enclose{horizontalstrike}{L}\;atm\; \enclose{horizontalstrike}{mol^{-1}K^{−1}})(308\; \enclose{horizontalstrike}{K})}{10.0\; \enclose{horizontalstrike} {L}}=7.58×10^{−4}\;atm$$

The total pressure is given by the sum of the partial pressures:

$$P_T=P_{H_2}+P_{He}+P_{Ne}=(0.00632+0.00253+0.00076)\;atm=9.61×10^{−3}\;atm$$

**Check Your Learning**

A 5.73-L flask at 25 °C contains 0.0388 mol of N_{2}, 0.147 mol of CO, and 0.0803 mol of H_{2}. What is the total pressure in the flask in atmospheres?

**Answer:**

1.137 atm

Here is another example of this concept, but dealing with mole fraction calculations.

**The Pressure of a Mixture of Gases**

A gas mixture used for anesthesia contains 2.83 mol oxygen, O_{2}, and 8.41 mol nitrous oxide, N_{2}O. The total pressure of the mixture is 192 kPa.

(a) What are the mole fractions of O_{2} and N_{2}O?

(b) What are the partial pressures of O_{2} and N_{2}O?

**Solution**

The mole fraction is given by $X_A=\frac{n_A}{n_{Total}}$ and the partial pressure is *P _{A}* =

*X*×

_{A}*P*.

_{Total}For O_{2},

$$X_{O_2}=\frac{n_{O_2}}{n_{Total}}=\frac{2.83\;mol}{(2.83+8.41)\;mol}=0.252$$

and

$$P_{O_2}=X_{O_2}\times P_{Total}=0.252\times 192\;kPa=48.4\;kPa$$

For N_{2}O,

$$X_{N_2O}=\frac{n_{N_2O}}{n_{Total}}=\frac{8.41\;mol}{(2.83+8.41)\;mol}=0.748$$

and

$$P_{N_2O}=X _{N_2O} \times P_{Total}=0.748\times 192\;kPa=143.6\;kPa$$

**Check Your Learning**

What is the pressure of a mixture of 0.200 g of H_{2}, 1.00 g of N_{2}, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C? Answer:

1.87 atm