Question

For an ideal gas equation of a process for which the heat capacity of the gas varies with temperature as C = $\frac{\alpha }{T}$ ($\alpha$ is a constant) is given by

• A. VlnT = constant
• B. $V{T}^{\frac{1}{(\gamma -1)}}{e}^{\frac{\alpha }{RT}}$ = constant
• C. $V^{\frac{1}{(\gamma -1)}}{T}^{\frac{\alpha }{RT}}$ = constant
• D. $V^{\gamma -1}T$ = constant

Answer 1

The correct option is B $V{T}^{\frac{1}{(\gamma -1)}}{e}^{\frac{\alpha }{RT}}$ = constant

For any process

$dQ=dV+dW$

$dQ=n{C}_{v}dT+PdV$

$dQ=nCvdT+\frac{nRT}{v}.dv$........(1) $[\because PV=nRT]$

also, we know that

molar that capacity of the process

$C=\frac{dQ}{ndT}\Rightarrow C=\frac{1}{ndT}[n{C}_{v}dT+\frac{nRT}{v}dV]$

$c=C_v+\frac{RT}{v}.\frac{dV}{dT}$

$\frac{\alpha }{T}=C_v+\frac{RT}{v}.\frac{dV}{dT}$ $[\because c=\frac{\alpha }{T}]$

$\frac{\alpha }{T}=\frac{R}{\gamma -1}+\frac{RT}{v}.\frac{dV}{dT}$

$\frac{\alpha }{T}.\left(\frac{dT}{RT}\right)=\left(\frac{R}{\gamma -1}\right).\frac{dT}{RT}+\frac{RT}{v}.\frac{dV}{dT}\times \left(\frac{dT}{RT}\right)$

$\frac{\alpha dT}{R{T}^2}=\frac{R}{(\gamma -1)T}.dT+\frac{dV}{v}\Rightarrow \frac{dv}{v}=dT[\frac{\alpha }{R{T}^2}$

On Integrating

$\int (\gamma -1).\frac{dV}{V}-=\int dT[\frac{\alpha (\gamma -1)}{T^{2}R}-\frac{1}{T}]$

$(\gamma -1).lnV=-\frac{\alpha (\gamma -1)}{R}.\frac{1}{T}-lnT+lnK$

Integrating

$ln{V}^{\gamma -1}=-\frac{\alpha (\gamma -1)}{RT}-lnT+lnK$

$ln{V}^{(\gamma -1)}+lnT-lnK=-\frac{\alpha (\gamma -1)}{RT}$

$ln\left[\frac{v^{(\gamma -1)T}}{k}\right]=-\frac{\alpha (\gamma -1)}{RT}$

$\frac{v^{(\gamma -1)T}}{k}=e^{-\frac{\alpha (\gamma -1)}{RT}}$

$v^{(\gamma -1)T}=k.e^{-\frac{\alpha (\gamma -1)}{RT}}$

$v^{(\gamma -1)T}=e^{-\frac{\alpha (\gamma -1)}{RT}}$
$V{T}^{\frac{1}{(\gamma -1)}}{e}^{\frac{\alpha }{RT}}$ = Constant