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Question

# The pressure and density of a diatomic gas (γ=75) changes from (P,ρ) to (P′,ρ′) during an adiabatic change. If ρ′ρ=32, then the value of P′P is

A
36
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B
1128
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C
128
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D
136
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Solution

## The correct option is C 128Given, Initial state of gas (P,ρ) Final state of gas (P′,ρ′) We know that, mass density (ρ)∝1V Thus, V1V2=ρ2ρ1 .....(1) where V is the volume and ρ is mass density. Assuming no of moles of diatomic gas is constant. From the equation of state of gas during an adiabatic process P2Vγ2=P1Vγ1 we get, P2P1=(V1V2)γ, Using equation (1) and from the data given in the question, P′P=(ρ2ρ1)γ=(ρ′ρ)γ=(32)75 ⇒P′P=27=128 Thus, option (c) is the correct answer.

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