Question

Two flasks of equal volume are connected by a narrow tube (of negligible volume) all at ${27}^0$C and contain 0.70 mole of $H_2$ at 0.5 atm. One of the flask is then immersed into a bath kept at ${127}^0$C, while the other remains at ${27}^0$C. The final pressure in each flask is :

• A. Final pressure = 0.5714 atm
• B. Final pressure = 1.5714 atm
• C. Final pressure = 0.5824 atm
• D. None of these

Answer 1

The correct option is A Final pressure = 0.5714 atm

Let us consider $0.35$ is the number of moles would be equal in two flasks.

Let in one number of moles would decrease and other would increase but pressure remains constant.

$P_2=P_1$

Hence,

$\frac{n_{2}RT}{V}=\frac{n_{1}RT}{V}$

Implies that,

$\left(0.35+x\right)300=\left(0.35-x\right)400$

Hence the value of $x$ is $0.05$.

In the vessel A,

$P_1=n_{1}RT$

$P_2=n_{2}RT$

$\frac{P_1}{P_2}=\frac{n_1}{n_2}$

Putting all given values :

$\frac{0.05}{P_2}=\frac{0.35}{0.4}$

$P_2=\frac{4}{7}atm$

Hence the final pressure is $0.5714$ atm

Therefore, the correct option is A.