Question

Equal volume of two gases which do not react together are enclosed in separate vessels. Their pressures are $10\text{mmHg}$ and $400\text{mmHg}$ respectively. If the two vessels are joined together, then what will be the pressure of the resulting mixture (temperature remaining constant)?

• A. $125\text{mm}Hg$
• B. $150\text{mm}Hg$
• C. $410\text{mm}Hg$
• D. $205\text{mm}Hg$

Answer 1

The correct option is D $205\text{mm}Hg$

Let, volume of the vessels be $V$ & temperature be $T$
$P_1=10\text{mm}{P}_2=400\text{mm}$

$\therefore n_1=\frac{P_{1}V}{RT}$ & $n_2=\frac{P_{2}V}{RT}$
$\therefore n_1+n_2=\frac{(P_1+P_2)\times V}{RT}$
After joining two containers final vol $=(V+V)=2V$ (for gases)
$P_{total}=\frac{(n_1+n_2)RT}{V_{total}}=\frac{(P_1+P_2)\times V}{RT}\times \frac{RT}{2V}=\frac{(P_1+P_2)}{2}=\frac{10+400\text{mmHg}}{2}=205\text{mmHg}$