Question

The molar heat capacity C for an ideal gas going through a given process is given by $C=\frac{a}{T}$, where $a$ is a constant, and $T$ is the temperature. If $\gamma =\frac{C_P}{C_V},$ the work done by one mole of gas during heating from $T_0$ to $\eta T_0$ through the given process will be

• A. $\frac{1}{a}\ln \eta$
• B. $a\ln \eta -\left(\frac{\eta -1}{\gamma -1}\right)R{T}_0$
• C. $a\ln \eta -(\gamma -1)R{T}_0$
• D. None of the above

Answer 1

The correct option is B $a\ln \eta -\left(\frac{\eta -1}{\gamma -1}\right)R{T}_0$

For 1 mole of gas the heat transfer can be written as,
$Q={\int }_{T_0}^{\eta T_0}nCdT={\int }_{T_0}^{\eta T_0}CdT(\because n=1)=a\ln \frac{\eta T_0}{T_0}=a\ln \eta$Internal energy for the gas is given by, $\Delta U=n{C}_{V}dT=C_{V}dT$ $(\because n=1)$
Molat heat capacity for 1 mole of ideal gas, $C_V=\frac{R}{\gamma -1}$
Thus change in internal energy, $\Delta U=\frac{R}{\gamma -1}(\eta T_o-T_o)=\frac{R{T}_o}{\gamma -1}(\eta -1)$
From first law of thermodynamics,
$Q=\Delta U+W$
Thus work done by gas will be $W=Q-\Delta U$
$\Rightarrow W=aln\eta -\frac{R{T}_o}{\gamma -1}(\eta -1)$