A diatomic ideal gas is compressed adiabatically to $ \frac{1}{32} $ of its initial volume. If the initial temperature of the gas is $ T_{\mathrm{i}} $ (in kelvin) and the final temperature is $ T_{\mathrm{f}}=a T_{\mathrm{i}} $, then value of $ a $ is

Question

A diatomic ideal gas is compressed adiabatically to $ \frac{1}{32} $ of its initial volume. If the initial temperature of the gas is $ T_{\mathrm{i}} $ (in kelvin) and the final temperature is $ T_{\mathrm{f}}=a T_{\mathrm{i}} $, then value of $ a $ is, $ 4 $, $ 6 $, $ 5 $, $ 9 $

MARKS App · Accepted Answer

Let the initial volume of diatomic gas be V i , then its final volume will be V i 32 . In adiabatic process, T V γ - 1 = constant ∴ T i V i γ - 1 = T f V f γ - 1 . As γ = 1 . 4 for diatomic gas. ⇒ T i V i 0 . 4 = a T i V i 32 0 . 4 ⇒ a = 32 0 . 4 = 4 .

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