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Question

Physics Question on Pressure

The pressure and density of a diatomic gas (γ=75)\left(\gamma=\frac{7}{5}\right) change adiabatically from (P,?)\left(P, ?\right) to \left(P? If \frac{\rho'}{\rho}=32,then, then \frac{P '}{P}$ is

A

1128\frac{1}{128}

B

3232

C

128128

D

256256

Answer

128128

Explanation

Solution

Here, γ=75,P1=P,ρ1=ρ;P2=P,ρ2=ρ\gamma=\frac{7}{5}, P_{1}=P, \rho_{1}=\rho; P_{2}=P', \rho_{2}=\rho' For an adiabatic process, PVγ=PV^{\gamma} = constant P1V1γ=P2V2γ\therefore P_{1}V_{1}^{\gamma }=P_{2}V_{2}^{\gamma} P2P1=(V1V2)γ=(ρ2ρ1)γ=(ρρ)7/5=(32)7/5\frac{P_{2}}{P_{1}}=\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=\left(\frac{\rho_{2}}{\rho_{1}}\right)^{\gamma}=\left(\frac{\rho'}{\rho}\right)^{7/5}=\left(32\right)^{7/5} PP=(25)7/5=27=128\frac{P'}{P}=\left(2^{5}\right)^{7/5}=2^{7}=128