Thermodynamics

JEE Main level, numerical answer, medium

109 of 190

Question 109 of 190

The pressure P1{P}_{1} and density d1{d}_{1} of diatomic gas (γ=75)\left(\gamma =\dfrac{7}{5}\right) changes suddenly to P2(>P1){P}_{2}\left(\gt {P}_{1}\right) and d2{d}_{2} respectively during an adiabatic process. The temperature of the gas increases and becomes____times of its initial temperature.
(given d2d1=32\dfrac{{d}_{2}}{{d}_{1}}=32)

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Answer: 4

For adiabatic process, PVγ=P{V}^{\gamma }= constant
We know, density d=massVolumed=\dfrac{mass}{Volume}.
Then, we have Pdγ=\dfrac{P}{{d}^{\gamma }}= constant
Given here, d2d1=32\dfrac{{d}_{2}}{{d}_{1}}=32 and γ=75\gamma =\dfrac{7}{5}.
The ratio P1P2=(d1d2)γ=(132)75=1128\dfrac{{P}_{1}}{{P}_{2}}={\left(\dfrac{{d}_{1}}{{d}_{2}}\right)}^{\gamma }={\left(\dfrac{1}{32}\right)}^{\dfrac{7}{5}}=\dfrac{1}{128}
The ratio of temperature using ideal gas equation is T1T2=P1V1P2V2=1128×32=14\dfrac{{T}_{1}}{{T}_{2}}=\dfrac{{P}_{1}{V}_{1}}{{P}_{2}{V}_{2}}=\dfrac{1}{128}\times 32=\dfrac{1}{4}
Thus, T2=4T1{T}_{2}=4{T}_{1}.

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