Question

One mole of an ideal gas is expanded
isothermally to four times its initial volume.
Calculate the entropy change in terms of R,
the gas constant.

Answer 1

The process in which there is no change in temperature is known as isothermal process. Entropy changes from S₁ to S₂ when gas absorbs heat during expansion. The heat taken by gas is given by area under the curve which represents work done during expansion.

$$Q=W$$

but,

$$Q=T({S_2}-{S_1})$$

and 

$$W={P_1}{V_1}\ln{(\frac{V_2}{V_1})}=RT\ln{(\frac{V_2}{V_1})}$$

so according to the question;

$${V_2}=4V , {V_1}=V$$

so we have a formula fro above equations;

$$T({S_1}-{S_2})=RT\ln{(\frac{V_2}{V_1})}$$

$$T({S_2}-{S_1})=RT\ln{(\frac{4V}{V})}$$

$$({S_2}-{S_1})=R\ln{4}$$

Answer:

$$({S_2}-{S_1})=1.38629R$$