Which of the following partial derivatives are incorrect for one mole of an ideal gas?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

${\frac{\partial S}{\partial T})}_{V} = \frac{C_{v, m}}{T}$

${\frac{\partial S}{\partial V})}_{T} = \frac{n R}{P}$

${\frac{\partial G}{\partial P})}_{T} = V$

${\frac{\partial H}{\partial P})}_{T} = 0$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \frac{d q_{rev}}{T} = d S = \frac{1}{T} d U + \frac{p}{T} d V \dots (i); S = f (U, V) \\ d S = {(\frac{\partial S}{\partial U})}_{V} \cdot d U + {(\frac{\partial S}{\partial V})}_{V} \cdot d V . . . (ii) \\ \Rightarrow {(\frac{\partial S}{\partial U})}_{V} = \frac{1}{T} and {(\frac{\partial S}{\partial V})}_{U} = \frac{P}{T} \end{array}$

Also, $d U = T d S - P d V$

Since, $d U$ is an exact differential, we can apply Euler's condition of reciprocity to this equation. This yields

${(\frac{\partial T}{\partial V})}_{S} = - {(\frac{\partial T}{\partial V})}_{V}$ ......(iii)

Similarly,

$S = f (T, V)$

$d S = {(\frac{\partial S}{\partial T})}_{V} \cdot d T + {(\frac{\partial S}{\partial V})}_{T} \cdot d V$ ...(iv)

$d U = {(\frac{\partial U}{\partial T})}_{V} \cdot d T + {(\frac{\partial U}{\partial V})}_{T} \cdot d V$ ...(v)

Substituting (v) in (i), we get,

$Z = \frac{1}{T} {(\frac{\partial U}{\partial T})}_{V} d T + \frac{1}{T} [P + {(\frac{\partial U}{\partial V})}_{T}] d V$ ...(vi)

Comparing, we get, ${(\frac{\partial S}{\partial T})}_{V} = \frac{1}{T} {(\frac{\partial U}{\partial T})}_{V} = \frac{n C_{v}}{T}$

Also, ${(\frac{\partial S}{\partial V})}_{T} = \frac{1}{T} [P + {(\frac{\partial U}{\partial V})}_{T}]; {(\frac{\partial U}{\partial V})}_{T} = 0 \forall$ idealgas