• question_answer A diatomic gas initially at 18°C is compressed adiabatically to one-eighth of its original volume. The temperature after compression will be A) ${{10}^{o}}C$ B) ${{887}^{o}}C$ C) $668K$ D) ${{144}^{o}}C$

Solution :

 [c]$T{{V}^{\gamma -1}}=$constant $\Rightarrow {{T}_{2}}={{T}_{1}}{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}=(273+18)\ {{\left( \frac{V}{V/8} \right)}^{0.4}}=668K$

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