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JEE Main 202131 AugustEvening Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2021 · 31 August · Evening Shift · Physics · Thermodynamics

A sample of gas with γ = 1 . 5 is taken through an adiabatic...

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 .

Correct answer

0

Step-by-step solution

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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