An ideal gas with an initial pressure P and volume V undergoes an isothermal and reversible expansion. If the change in entropy due to this expansion is ΔS, the magnitude of work done by the gas is

Correct option is C

In an isothermal expansion process, the temperature of the system remains constant throughout the expansion process. Since for an ideal gas, U depends only on temperature 

it follows that

Substituting the above expression in the first law of  thermodynamics (dU=dq+dw)

Hence in an isothermal expansion process, heat is converted into work and vice versa. Thus, if heat is supplied to the system (q positive), the equivalent amount of work is done by the system (w negative), and if some work is done on the system (w positive), the equivalent amount of heat is given out (q negative).

Reversible expansion process

The expression for the work involved in an isothermal expansion from volume V₁ to V₂ can be worked out as follows:

w=-q

w=-(-TΔS)=TΔS=(PV/nR)ΔS

[For an ideal gas, PV=nRT]