A diatomic gas initally at $18^{\circ} \mathrm{C}$ is compressed adiabatically to one eighth of its original…

A diatomic gas initally at $18^{\circ} \mathrm{C}$ is compressed adiabatically to one eighth of its original volume. The temperature after compression will be (take \(\left.8^{2 / 5}=2.3ight)\)
  1. $18^{\circ} \mathrm{C}$
  2. $887^{\circ} \mathrm{C}$
  3. $327^{\circ} \mathrm{C}$
  4. $395.5^{\circ} \mathrm{C}$

Solution

$\mathrm{T}_{1}=18^{\circ} \mathrm{C}=(273+18)=291 \mathrm{~K}$ and $\mathrm{V}_{2}=\mathrm{V}_{1} / 8$ We know that $\mathrm{TV}^{\gamma-1}=$ constant or, $\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}=\mathrm{T}_{1} \mathrm{~V}_{1}^{\gamma-1}$ $\therefore \quad \mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}ight)^{\gamma-1}=291 \times(8)^{1.4-1}$ $=668.5 \mathrm{~K}=395.5^{\circ} \mathrm{C}$ ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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