A cylindrical chamber `A` of uniform cross section is divided into two parts `X` and `Y` by a movalbe piston `P` which can slide without friction inside the chamber. Initially part `X` contains `1 "mol"` of a monochromatic gas and `Y` contains `2 mol` of a diatomic gas, and the volumes of `X` and `Y` are in the ratio `1:2` with both parts `X` and `Y` being at the same temperature `T`. Assuming the gases to be ideal, the work `W` that will be done in moving the piston slowly to the position where the ratio of the volumes of `X` and `Y` is `2:1` will be

Options

A. `-5.8T`
B. `8.3 T`
C. `12.3 T`
D. zero

Answer

A

Explanation

a. Since the piston is moved slowly we assume isothermal condition for both the gases as thermal equilibrium is maintained throughout.
Let the total of chamber be. `V`. Then volume of gas in `X` increase (expand) from `V//3` to `2V//3`. The work done is positve and given (for 1 mol of monatomic gas by
`W_(x) = + RT log_(e) .(2 V//3)/(V//3) = RT log_(e) 2`
The volume of gas in `Y` decrease (compressed) from `2V//3` to `V//3`.
The work done (isothermally) is negative and given (for 2 mol of diatomic gas) by
`W_(x) = 2 RT log_(e) (V//3)/(2V//3) = - RT log_(e) 2`
Hence the total work done on the system is
`W = W_(x) + W_(Y) = - RT log_(e) 2`
Substituting `R = 8.3, log_(2) 2 = 0.6996` and simplifiying `W = - 5.8 T`.