Answer to Question #120767 in General Chemistry for Aakash

Question

1 mole of non ideal gas at 25*C is subjected to expand reversibly ten times of the initial volume. The change in entropy of expansion is,

Answer 1

We have the state relation -

$TdS=dU+PdV$

R=8.314Jmol−1

​$​\frac{V2}{V1}$ =10

Since expansion is isothermal, there will be no change in temperature and consequently no change in internal energy ($dU = 0$ ).

It reduces to -

$ds=\frac{PdV}{T}$

From ideal gas relation, $Pv= nRT$

$dS=\frac{nRdV}{V} ​$

​$S 2−S1 =nRln( \frac{V1}{V2} ​ ​ ​ )$

n = 1 (1 mole of gas), R=8.314Jmol^-1K,

S2−S1=8.314×ln(10)

⟹S2−S1=19.15K^-1mole^-1