Question

For an ideal gas its adiabatic exponent is $\gamma$. In some process its molar heat capacity varies as $C=\frac{\alpha }{T}$, where $\alpha$ is a constant. Work performed by one mole of gas, during its heating from $T_0$ to $n{T}_0$ will be ?

• A. $\alpha {\log }_{e}n$
• B. $\alpha {\log }_{e}n-(n-1)R{T}_0$
• C. $\alpha {\log }_{e}n-\frac{n-1}{\gamma -1}R{T}_0$
• D. $zero$

Answer 1

Answer: (C)

$\alpha {\log }_{e}n-\frac{n-1}{\gamma -1}R{T}_0$

$\text{Solution}:$

By the first law of thermodynamics

$A=Q-\Delta U$

or, $=CdT-C_{V}dT=(C-C_V)dT$ (for one mole)

Given $C=\frac{\alpha }{T}$

So, $A={\int }_{T_o}^{\eta T_o}\left(\frac{\alpha }{T}\right)-C_{V}dT=\alpha 1n\frac{\eta T_0}{T_0}-C_V(\eta T_0-T_0)=\alpha 1n\eta =C_V{T}_0(\eta -1)=\alpha 1n\eta +\frac{R{T}_o}{\gamma -1}(\eta -1)$ = $\alpha lo{g}_e\eta +\frac{R{T}_o}{\gamma -1}(\eta -1)$