Thermodynamics

JEE Main level, numerical answer, medium

120 of 190

Question 120 of 190

A sample of gas with γ=1.5\gamma =1.5 is taken through an adiabatic process in which the volume is compressed from 1200cm31200{cm}^{3} to 300cm3.300{cm}^{3}. If the initial pressure is 200kPa.200kPa. The absolute value of the workdone by the gas in the process =________J.

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Answer: 480

Vi=1200×10−6m3,Pi=200KPa{V}_{i}=1200\times {10}^{-6}{m}^{3},{P}_{i}=200KPa
Vf=300×10−6m3,Pf=?{V}_{f}=300\times {10}^{-6}{m}^{3},{P}_{f}=?
γ=1.5\gamma =1.5
PiViγ=PfVfγ{P}_{i}{V}_{i}^{\gamma }={P}_{f}{V}_{f}^{\gamma }
200×103(1200×10−6300×10−6)γ=Pf200\times {10}^{3}{\left(\dfrac{1200\times {10}^{-6}}{300\times {10}^{-6}}\right)}^{\gamma }={P}_{f}
Pf=200×103×43/2{P}_{f}=200\times {10}^{3}\times {4}^{3/2}
=200×103×8=200\times {10}^{3}\times 8
=1600×103=16×105Pa=1600\times {10}^{3}=16\times {10}^{5}Pa
W=PiVi−PfVfγ−1W=\dfrac{{P}_{i}{V}_{i}-{P}_{f}{V}_{f}}{\gamma -1}
W=2×105×1200×10−6−16×105×300×10−61.5−1=240−4800.5=−240×2=−480JW=\dfrac{2\times {10}^{5}\times 1200\times {10}^{-6}-16\times {10}^{5}\times 300\times {10}^{-6}}{1.5-1}=\dfrac{240-480}{0.5}=-240\times 2=-480J∣W∣=480J|W|=480J

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