Comprehension:
Solutions
Let us consider a binary solution of two volatile liquids and denote the two components as 1 and 2. When taken in a closed vessel, both the components would evaporate and eventually an equilibrium would be established between vapour phase and the liquid phase. Let the total vapour pressure at this stage be ptotal and p1 and p2 be the partial vapour pressures of the two components 1 and 2 respectively. These partial pressures are related to the mole fractions x1 and x2 of the two components 1 and 2 respectively. The French chemist, Francois Marie Raoult (1886) gave the quantitative relationship between them. The relationship is known as the Raoult's law which states that for a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction (x1) present in solution.
At constant temperature, 250 mL of nitrogen under 720 mm pressure and 380 mL of oxygen under 650 mm pressure were put into a one liter flask. The final pressure of the mixture will be;
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 427 mm Hg.
Two laws are needed. By Boyle’s law, when a gas expands at constant temperature its pressure falls in proportion to the increase in volume. By Dalton’s law of partial pressures, the total pressure of a mixture of non-reacting gases is the sum of the partial pressures each gas would exert alone in the same volume:
\( \displaystyle p_1V_1=p_2V_2 \)
\( \displaystyle p_{total}=p_{\ce{N2}}+p_{\ce{O2}} \)
Each gas expands into the 1 L (1000 mL) flask. Its partial pressure there is:
\( \displaystyle p_{\ce{N2}}=\frac{720\times250}{1000}=180\ \mathrm{mm\ Hg} \)
\( \displaystyle p_{\ce{O2}}=\frac{650\times380}{1000}=247\ \mathrm{mm\ Hg} \)
\( \displaystyle p_{total}=180+247=427\ \mathrm{mm\ Hg} \)
So the final pressure of the mixture is 427 mm Hg.