An ideal gas having initial pressure P volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66V, while its temperature falls to T/2. Find the number of degrees of freedom of the gas molecule.
(Take (5.66)0.4=2)
A
2
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B
3
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C
5
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D
6
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Solution
The correct option is C5 For an adiabatic process, TVγ−1 = constant
i.e. T1Vγ−11=T2Vγ−12 TVγ−1=T2(5.66V)γ−1 ⇒(5.66)γ−1=2 ⇒γ=1.4
We know that, γ=1+2f ∴ we get f=2γ−1=20.4=5