Question

The Vanderwaal equation is given as

$(P+\frac{a}{v^2}(v-b\left)\right)=RT$

if the system executes some process from state 1 to state 2 the change in entropy is given as
$s_2-s_1=C_v\text{log}\left(\frac{T_2}{T_1}\right)+k$
The value of k is

• A. $R{\text{log}}_e\left(\frac{v_2-b}{v_1-b}\right)$
• B. $R{\text{log}}_e\left(\frac{P_2-b}{P_1-b}\right)$
• C. $R{\text{log}}_e\left(\frac{v_1-b}{v_2-b}\right)$
• D. $R{\text{log}}_e\left(\frac{P_1-b}{P_2-b}\right)$

Answer 1

The correct option is A $R{\text{log}}_e\left(\frac{v_2-b}{v_1-b}\right)$

The change in entropy is given as
$ds=cv\frac{dT}{T}+{\left(\frac{\partial P}{\partial T}\right)}_{v}dv$ for Vanderwaal's equation.

${\left(\frac{\partial P}{\partial T}\right)}_v=\frac{R}{v-b}$

$ds=c_v\frac{\partial T}{T}+\frac{R}{v-b}dv$

$={\int }_1^2{c}_v\frac{\partial T}{T}+R{\int }_1^2\frac{dv}{v-b}$

$s_2-s_1=cv{\text{log}}_e\frac{T_2}{T_1}+R{\text{log}}_e\left(\frac{v_2-b}{v_1-b}\right)$

Comparing it with given equation we have

$k=Rlo{g}_e\left(\frac{v_2-b}{v_1-b}\right)$