Figure shows an insulated cylinder of volume `V` containing monatomic gas in both the compartments. The pistone is diathermic. Initially the piston is kept fixed and the system is allowed to acquire a state of thermal equilibrium. The initial pressures and temperatures are as shown in the figure. Calculate

Final volume of the left compartment.

Options

A. `(V)/(2)((P_(2)T_(1))/(P_(2)T_(1)+P_(1)T_(2)))`
B. `V((P_(2)T_(1))/(P_(2)T_(1)-P_(1)T_(2)))`
C. `V((P_(2)T_(1))/(P_(2)T_(1)+P_(1)T_(2)))`
D. `V((2P_(2)T_(2))/(P_(2)T_(1)-P_(1)T_(2)))`

Answer

C

Explanation

`P'_(1)=(P_(1)T_(2)(P_(1)+P_(2)))/((P_(1)T_(2)+P_(2)T_(1)))` and `P'_(2)=(P_(2)T_(1)(P_(1)+P_(2)))/((P_(1)T_(2)+P_(2)T_(1)))`
As `P_(2)T_(1) gt P_(1) T_(2)` (given)
Hence, `P'_(2)gt P'_(1)`
Let `Delta V` be the change in volume of any compartments.
`(P((V//2)-DeltaV))/(T)=(P_(1)V)/(2T_(1))`
`(P(V//2)+DeltaV)/(T)=(P_(2)V)/(2T_(2))`
Dividing, we get
`(((V//2)+DeltaV))/(((V//2)-DeltaV))=(P_(2)T_(1))/(P_(1)T_(2))`
`DeltaV=(V)/( 2)((P_(2)T_(1)-P_(1)T_(2))/(P_(2)T_(1)+P_(1)T_(2)))`
Volume of left compartment
`V'_(1)=(V)/(2)-DeltaV=V((P_(1)T_(2))/(P_(2)T_(1)+P_(1)T_(2)))`
Volume of right compartment
`V'_(2)=(V)/(2)+DeltaV=V((P_(2)T_(1))/(P_(2)T_(1)+P_(1)T_(2)))`