Answer to Question #88753 in Molecular Physics | Thermodynamics for Shivam Nishad

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:

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