Thermodynamics

JEE Main level, single correct, medium

133 of 190

Question 133 of 190

In an adiabatic process, the density of a diatomic gas becomes 3232nn times its initial value. The final pressure of the gas is found to be nn times the initial pressure. The value of nn is:

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Answer: (C) 128128

Given,
Final density of diatomic gas is ρf=32nρi{\rho }_{f}=32n{\rho }_{i}
Final pressure is Pf=nPi{P}_{f}=n{P}_{i}
In adiabatic process, equation of state is given by,
PVγ=constant...(1)P{V}^{\gamma }=\text{constant}...\left(1\right)
Where, VV is the volume, PP is the pressure and γ=75\gamma =\dfrac{7}{5} is the ratio of specific heat of diatomic gas
P(mρ)γ=constant...(2)P{\left(\dfrac{m}{\rho }\right)}^{\gamma }=\text{constant}...\left(2\right), As V=mρV=\dfrac{m}{\rho }
and mass is constant, then the pressure is given by
∴P∝ργ∴P\propto {\rho }^{\gamma }
⇒PfPi=(ρfρi)γ\Rightarrow \dfrac{{P}_{f}}{{P}_{i}}={\left(\dfrac{{\rho }_{f}}{{\rho }_{i}}\right)}^{\gamma }
⇒nPiPi=(32ρiρi)7/5=(32)7/5\Rightarrow \dfrac{n{P}_{i}}{{P}_{i}}={\left(\dfrac{32{\rho }_{i}}{{\rho }_{i}}\right)}^{{}^{7/5}}=(32{)}^{7/5}
∴n=27=128∴n={2}^{7}=128

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