An ideal gas has specific heat capacity at constant pressure $\frac{11}{10} R$. If one mole of this ideal…

An ideal gas has specific heat capacity at constant pressure $\frac{11}{10} R$. If one mole of this ideal gas at $125^{\circ} \mathrm{C}$ does $83 \mathrm{~J}$ of work adiabatically, then the final temperature of the gas would be (Universal gas constant, $R=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )
  1. $25^{\circ} \mathrm{C}$
  2. $50^{\circ} \mathrm{C}$
  3. $75^{\circ} \mathrm{C}$
  4. $100^{\circ} \mathrm{C}$

Solution

Work done by gas in adiabatic process, $\Delta W=+83 \mathrm{~J}$ As in adiabatic process, $\Delta Q=0$, By first law of thermodynamics, $\Delta Q=\Delta U+\Delta W$ so change in internal energy of gas, $\Delta U=-83 \mathrm{~J}$ Also, $\Delta U=n C_V \Delta T=83$ Here, $n=1$ mole $\Rightarrow$ specific heat at constant volume $C_V=C_p-R$ $\Rightarrow \quad C_V=\frac{11}{10} R-R=\frac{1}{10} R$ $\begin{aligned} & \text { so, } \quad-83=1 \times \frac{1}{10} R \times\left(T_f-125\right) \\ & \Rightarrow T_f-125=\frac{-830}{8.3}=-100\end{aligned}$ or $\quad T_f=125-100=25^{\circ} \mathrm{C}$

Asked in: AP EAMCET 2022 (08 Jul Shift 2)

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