Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures `T_(1)` and `T_(2)(T_(1)ltT_(2))`. The correct graphical depiction of the dependence of work done (w) on the final volume (V) is:

Options

A.
B.
C.
D.

Answer

C

Explanation

`w = - nRT ln V_2/V_1`
`abs(omega) = nRT ln V/V_1` (given final valume = V)
`abs(omega) = (nRT) ln V + (- nRT ln V_1)`
On comparing this graph with `y = mx + c , c = - nRT` ln `V_1`
If V = 1litre, then `0 lt V_1 lt 1`,
so at temperature `T_1 and T_2`,
`abs(omega)_(T_1) = - nRT_1` ln `V_1` , (n, R and T are positive and ln `V_1` is negative)
`abs(omega)_(T_2) = - nRT_2 ln V_1`
As `T_2 gt T_1 rArr abs(omega)_(T_2) gt abs(omega)_(T_1)`
Hence intercept on y-axis will be positive at
ln = V and the magnitude will be more for `T_2`. Only given in option (c) is possible.