Answer:
Option B
Explanation:
Given p1=5 bar, V1 =1 m3, T1=400K
So, $n_{1}=\frac{5}{400R}$ (from pV=nRT)
Similarly , p2= 1 bar, V2 = 3 m3, T2 =300K
$n_{2}=\frac{3}{300R}$
Let at equilibrium the new volume of A will be (1+x)
So, the new volume of B will be (3-x)
Now, from the ideal gas equation
$\frac{p_{1}V_{1}}{n_{1}RT_{1}}=\frac{P_{2}V_{2}}{n_{2}RT_{2}}$
and at equilibrium (due to conduction of heat )
$\frac{p_{1}}{T_{1}}=\frac{p_{2}}{T_{2}}$
So $\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}$ or V1n2 = V2n1
After putting the values
$(1+x)\times \frac{3}{300R}=(3-x)\times \frac{5}{400R}$
or $(1+x)= \frac{(3-x)5}{4} or 4(1+x)= 15-5x$
or $4+4x=15-5x$ or $x=\frac{11}{9}$
Hence, new volume of A i.e, (1+x) will comes as $1+\frac{11}{9}=\frac{20}{9} or 2.22$