Question

For one mole of an ideal gas which of the following relation is correct ?

• A. $\left( \frac{\partial\mathbf{P}}{\partial T} \right)_{V}$$\left( \frac{\partial P}{\partial V} \right)_{T}$= –1
• B. $\left( \frac{\partial\mathbf{P}}{\partial T} \right)_{V}$$\left( \frac{\partial V}{\partial P} \right)_{T}$= 1
• C. $\left( \frac{\partial\mathbf{P}}{\partial T} \right)_{V}$ $\left( \frac { \partial \mathrm { T } } { \partial \mathrm { V } } \right) _ { \mathrm { P } }$ $\left( \frac{\partial V}{\partial P} \right)_{T}$ = –1
• D. None of these

Answer 1

Answer: (C)

$\left( \frac{\partial\mathbf{P}}{\partial T} \right)_{V}$ $\left( \frac { \partial \mathrm { T } } { \partial \mathrm { V } } \right) _ { \mathrm { P } }$ $\left( \frac{\partial V}{\partial P} \right)_{T}$ = –1

Answer 2

For one mole of an ideal gas PV = RT

or PdV + VdP = RdT

Dividing the equation by dT and introducing the condition of constant volume, we get

V $\left( \frac{\partial\mathbf{P}}{\partial T} \right)_{V}$ = R

or $\left( \frac{\partial\mathbf{P}}{\partial T} \right)_{V}$ = $\frac{R}{V}$

$\left( \frac{\partial T}{\partial V} \right)_{P}$ = $\frac{P}{R}$ and $\left( \frac{\partial V}{\partial P} \right)_{T}$= –$\frac{V}{P}$

\$\left( \frac{\partial\mathbf{P}}{\partial T} \right)_{V}$ $\left( \frac { \partial \mathrm { T } } { \partial \mathrm { V } } \right) _ { \mathrm { P } }$ $\left( \frac{\partial V}{\partial P} \right)_{T}$ = –$\frac{R}{V}$ × $\frac{P}{R}$× $\frac{V}{P}$

= –1