Thermodynamics

JEE Main level, single correct, medium

113 of 190

Question 113 of 190

A certain amount of gas of volume VV at 27∘C27^{\circ}C temperature and pressure 2×107Nm−22\times {10}^{7}N{m}^{-2} expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use γ=1.5\gamma =1.5)

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Answer: (B) 3.536×106Pa3.536\times {10}^{6}Pa

Given here, P1=2×107Nm−2=2×107Pa{P}_{1}=2\times {10}^{7}N{m}^{-2}=2\times {10}^{7}Pa
Using Boyle's law,
P1V1=P2V2{P}_{1}{V}_{1}={P}_{2}{V}_{2}
Since V2=2V1{V}_{2}=2{V}_{1}.
Hence, P2=P12{P}_{2}=\dfrac{{P}_{1}}{2} (isothermal expansion)
P2=1×107Pa{P}_{2}=1\times {10}^{7}Pa
Now, during adiabatic expansion,
P2(V2)γ=P3(2V2)γ{P}_{2}{\left({V}_{2}\right)}^{\gamma }={P}_{3}{\left(2{V}_{2}\right)}^{\gamma }
P3=1×10721.5=3.536×106Pa{P}_{3}=\dfrac{1\times {10}^{7}}{{2}^{1.5}}=3.536\times {10}^{6}Pa

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