Question

Two gases occupy two containers A and B. The gas in A, of volume $0.10{\text{m}}^3$, exerts a pressure of $1.40\text{MPa}$ and that in B, of volume $0.15{\text{m}}^3$, exerts a pressure of $0.7\text{MPa}$. The two containers are joined by a tube of negligible volume and the gases are allowed to intermingle. Then, if the temperature remains constant, the final pressure in the containers will be $\left(\text{in MPa}\right)$:

• A. $0.70$
• B. $0.98$
• C. $1.40$
• D. $2.10$

Answer 1

The correct option is B $0.98$

Given,
Volume of gas in container A $\left(V_A\right)=0.10{\text{m}}^3$
Volume of gas in container B $\left(V_B\right)=0.15{\text{m}}^3$
Pressure exerted by gas in container A $\left(P_A\right)=1.40\text{MPa}$
Pressure exerted by gas in container B $\left(P_B\right)=0.7\text{MPa}$
Let $P$ and $V$ be the pressure and volume of the gas after the two containers are joined together.
As the quantity of gas remains constant, $n=n_A+n_B$
Since the temperature is kept constant, by using ideal gas equation, we can write that,
$\frac{P(V_A+V_B)}{RT}=\frac{P_A{V}_A}{RT}+\frac{P_B{V}_B}{RT}(\because V=V_A+V_B)$
$\Rightarrow P=\frac{P_A{V}_A+P_B{V}_B}{V_A+V_B}=\frac{1.40\times 0.1+0.7\times 0.15}{0.1+0.15}$
$\Rightarrow P=0.98\text{MPa}$
Thus, option (b) is the correct answer.