Starting at temperature $300 \mathrm{~K}$, one mole of an ideal diatomic gas $(\gamma=1.4)$ is first compressed adiabatically from volume
$\mathrm{V}_{1}$ to $\mathrm{V}_{2}=\frac{\mathrm{V}_{1}}{16}$. It is then allowed to expand isobarically
to volume $2 \mathrm{~V}_{2}$. If all the processes are the quasi-static then the final temperature of the gas (in ${ }^{\circ} \mathrm{K}$ ) is (to the nearest integer)__________
(1818) For an adiabatic process,
$\mathrm{TV} \gamma^{-1}=$ constant
$\therefore T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}$
$\Rightarrow T_{2}=(300) \times\left(\frac{V_{1}}{\frac{V_{1}}{16}}\right)^{1.4-1}$
$\Rightarrow T_{2}=300 \times(16)^{0.4}$
Ideal gas equation, $P V=n R T$
$\therefore \quad V=\frac{n R T}{P}$
$\Rightarrow V=k T$ (since pressure is constant for isobaric process)
So, during isobaric process
$V_{2}=k T_{2}$ ....(1)
$2 \mathrm{~V}_{2}=k T_{f}$ .....(2)
Dividing (i) by (ii)
$\frac{1}{2}=\frac{T_{2}}{T_{f}}$
$T_{f}=2 T_{2}=300 \times 2 \times(16)^{0.4}=1818 \mathrm{~K}$
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