Question

Consider the following diagram and mark the correct answer.

• (A) Pressure in container-I is 3 atm before opeing the valve.
• (B) Pressure after opening the valve is 3.57 atm
• (C) Moles in each compartment are same after opening the valve.
• (D) Pressure in each compartment is the same after opening the valve.

Answer 1

The Correct Option is B, D

Explanation:
$P_1\times 16.42=0.0821\times 300\times 2=3\mathrm{atm}$$P_2\times 8.2=0.0821\times 400\times 1=4\mathrm{atm}$$P_{1\text{ initial }}=3$ atm$P_{2\text{ initial }}=4$ atmAfter opening the valve (to reach equilibrium), pressure and temperature in compartment must be same. \(3$\mathrm{PV}=\mathrm{nRT}$After opening value, Pfinal, Tfinal will be same in both containers $P_f(16.42+8.21)=(2+1)(0.0821)T_f$${\mathrm{T}}_{\mathrm{f}}={\mathrm{X}}_{\mathrm{I}}{\mathrm{T}}_{\mathrm{I}}+{\mathrm{X}}_{\mathrm{II}}{\mathrm{T}}_{\Vert }=\frac{2}{3}\times 300+\frac{1}{3}\times 400=333.33\mathrm{K}$$P_f(16.42+8.2)=(2+1)(0.0821)\times 333.33$$P_f=3.334$ atm