Starting at temperature 300 K,


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)


$T_{f}=2 T_{2}=300 \times 2 \times(16)^{0.4}=1818 \mathrm{~K}$

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