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JEE Main 20262 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2026 · 2 April · Morning Shift · Physics · Thermodynamics

A vessel contains 0.15 m ^3 of a gas at pressure 8 bar and...

A vessel contains 0.150.15 m3^3 of a gas at pressure 88 bar and temperature 140°140°C with cp=3Rc_p = 3R and cv=2Rc_v = 2R. It is expanded adiabatically till pressure falls to 11 bar. The work done during this process is _______ kJ. (RR is gas constant)

Correct answer

0

Step-by-step solution

Given P1=8P_1 = 8 bar =8×105= 8 \times 10^5 Pa, V1=0.15V_1 = 0.15 m3^3, and P2=1P_2 = 1 bar =105= 10^5 Pa.

The adiabatic index γ\gamma is given by:

γ=cpcv=3R2R=1.5\gamma = \dfrac{c_p}{c_v} = \dfrac{3R}{2R} = 1.5

For an adiabatic process, P1V1γ=P2V2γP_1 V_1^\gamma = P_2 V_2^\gamma.

V2=V1(P1P2)1γV_2 = V_1 \left( \dfrac{P_1}{P_2} \right)^{\dfrac{1}{\gamma}}

V2=0.15×(81)11.5=0.15×823=0.15×4=0.6V_2 = 0.15 \times \left( \dfrac{8}{1} \right)^{\dfrac{1}{1.5}} = 0.15 \times 8^{\dfrac{2}{3}} = 0.15 \times 4 = 0.6 m3^3

The work done during the adiabatic expansion is:

W=P1V1−P2V2γ−1W = \dfrac{P_1 V_1 - P_2 V_2}{\gamma - 1}

W=8×105×0.15−1×105×0.61.5−1W = \dfrac{8 \times 10^5 \times 0.15 - 1 \times 10^5 \times 0.6}{1.5 - 1}

W=1.2×105−0.6×1050.5W = \dfrac{1.2 \times 10^5 - 0.6 \times 10^5}{0.5}

W=0.6×1050.5=1.2×105W = \dfrac{0.6 \times 10^5}{0.5} = 1.2 \times 10^5 J

W=120W = 120 kJ

Answer: 120120
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