Thermodynamics and Thermochemistry - Result Question 30

Answer:

Correct Answer: 30. (a)

Solution:

Given, $\dfrac{p}{V}=1 \Rightarrow p=V \hspace{6mm}…(i)$

Also from first law : $d q=C_V d T+p d V$

For one mole of an ideal gas : $p V = n R T$

$\Rightarrow \hspace{5mm} p d V+V d p =R d T \hspace{6mm}…(ii) $

From (i) p dV = V dp

Substituting in Eq. (ii) gives

$2 p d V =R d T \quad \Rightarrow \quad p d V=\dfrac{R}{2} d T $

$\Rightarrow \quad d q =C _V d T+\dfrac{R}{2} d T + \dfrac{R}{2} d T$

$\Rightarrow \quad \int \dfrac{d q}{d T} =C _V+\dfrac{R}{2}=\dfrac{3}{2} R+\dfrac{R}{2}=2 R$