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Q.

A diatomic ideal gas is compressed adiabatically to 132 of its initial volume. If the initial temperature of the gas is Ti (in Kelvin) and the final temperature is aTi, the value of a is ______.

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answer is 4.

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Detailed Solution

For an adiabatic process, TVγ−1=constant TiViγ−1=TfVfγ−1

Substituting the given values, we get  TiViγ−1=aTi(vi32)γ−1⇒a=32γ−1

For diatomic gas, γ=75, a=(32)75−1=(32)2/5=(2)2=4

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