Change in entropy per mole for an isothermal process

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

$dS = \frac { dQ_{rev} } { T }$

Assuming that isothermal process is a reversible processes,

$dU = dQ – pdV$

For isothermal process, dU = 0.

$dQ = pdV$

$pV = nRT$, where n is number of moles.

For one mole,

$p = \frac { RT} { V }$

So, $\Delta Q = RT \int_V ^ {2V} \frac { RT} { V } dV$

$= RT \ln2$

So, $\Delta S = \frac { \Delta Q } { T } = R \ln2$Hence, the correct option is (c).

 

[Pushoam]

Thanks Delta².