4.0 Dalton’s Law of Partial Pressure

This law states that ' The total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases' . In a mixture of gases, the pressure exerted by the individual gas is called partial pressure. Mathematically, $${P_{total}} = {P_1} + {P_2} + {P_3} + ...... + {P_n}$$

There are few assumptions in this law.

Let us consider $n$ gases with number of moles ${n_1},{n_2},{n_3},.....{n_n}$ and partial pressure ${P_1},{P_2},{P_3},.....{P_n}$ respectively. Also, the temperature and pressure be $T \ and\ V $ respectively. Now, partial pressure of ${i^{th}}$ gas can be given as $${P_i} = {n_i}RT/V$$.

From Dalton’s Law of Partial Pressures, we know that $$\begin{equation} \begin{aligned} {P_{total}} = {P_1} + {P_2} + {P_3} + ...... + {P_n} \\ {P_{total}} = {n_1}RT/{V_1} + {n_2}RT/{V_2}+......+ {n_n}RT/{V_n} \\\end{aligned} \end{equation} $$ Now, $$\begin{equation} \begin{aligned} {P_i}/{P_{total}} = {n_i}/({n_1} + {n_2} + ...{n_n}) \\ {P_i}/{P_{total}} = {X_i} \\ {P_i} = {X_i}{P_{total}} \\\end{aligned} \end{equation} $$Therefore, partial pressure of a gas is equal to the product of its mole fraction and total pressure.

Question 3. When $2$g of a gaseous substance $A$ is introduced into an initially evacuated flask kept at ${25^ \circ }C$, the pressure is found to be $1$ atm. The flask is evacuated and $3$g of other gas $B$ is introduced. The pressure is found to be $0.5$ atm at same temperature. Calculate the ratio ${M_A} = {M_B}$.

Solution : $$\begin{equation} \begin{aligned} {P_A} = 1atm \\ {P_B} = 0.5atm \\ \frac{{{P_A}}}{{{P_B}}} = \frac{{{X_A} \times {P_{total}}}}{{{X_B} \times {P_{total}}}} \\ \frac{{{P_A}}}{{{P_B}}} = \frac{{\frac{{{n_A}}}{{{n_A} + {n_B}}}}}{{\frac{{{n_B}}}{{{n_A} + {n_B}}}}} \\ \frac{{{P_A}}}{{{P_B}}} = \frac{{\frac{{{w_A}}}{{{M_A}}}}}{{\frac{{{w_B}}}{{{M_B}}}}} \\ \frac{{{M_A}}}{{{M_B}}} = \frac{{{P_B}}}{{{P_A}}} \times \frac{{{w_A}}}{{{w_B}}} = \frac{2}{3} \times \frac{{0.5}}{1} = \frac{1}{3} \\\end{aligned} \end{equation} $$

Question 4. A neon-dioxygen mixture contains $70.6$ g dioxygen and $167.5$ g neon. If pressure of the mixture of gases in the cylinder is $25$ bar. What is the partial pressure of dioxygen and neon in the mixture ?

Solution: $$\begin{equation} \begin{aligned} {n_{{o_2}}} = \frac{{70.6}}{{32}} = 2.2065 \\ {n_{Ne}} = \frac{{167.5}}{{20}} = 8.375 \\ {X_{{o_2}}} = \frac{{2.20625}}{{2.20625 + 8.375}} = 0.2085 \\ {X_{Ne}} = 1 - {X_{{o_2}}} = 0.791 \\ {P_{{o_2}}} = {X_{{o_2}}} \times {P_{total}} = 0.2085 \times 25 = 5.2125\ bar \\ {P_{Ne}} = {X_{_{Ne}}} \times {P_{total}} = 0.791 \times 25 = 19.7875\ bar \\\end{aligned} \end{equation} $$

Question 5. A mixture of dihydrogen and dioxygen at one bar pressure contains $20\%$ by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

Solution: ${P_{total}} = 1bar$

Let, the total weight of mixture be $w$.

$$\begin{equation} \begin{aligned} {w_{{H_2}}} = \frac{{20}}{{100}}w = 0.2w \\ {w_{{O_2}}} = \frac{{80}}{{100}}w = 0.8w \\ {n_{{H_2}}} = \frac{{0.2w}}{2} = 0.1w \\ {n_{{O_2}}} = \frac{{0.8w}}{{32}} = 0.025w \\ {X_{{H_2}}} = \frac{{0.1w}}{{0.1w + 0.025w}} = 0.8 \\ {P_{{H_2}}} = {X_{{H_2}}} \times {P_{total}} = 0.8 \times 1 = 0.8bar \\\end{aligned} \end{equation} $$