Question bank/NEET UG/Physics

NEET UGPhysicsModerateRe-NEET 2026

In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (γ=5/3\gamma = 5/3) decreases from 60 K to 50 K. The work done by the gas in the process is:
(Take the universal gas constant as
R=8.3 J mol−1 K−1R = 8.3~\text{J mol}^{-1}~\text{K}^{-1})

Show answer and solution
Sol. Work done in adiabatic process
W=nRΔT1−γW = \frac{nR\Delta T}{1-\gamma}
ΔT=−10K\Delta T = -10K
W=1×8.3×(−10)1−53W = \frac{1 \times 8.3 \times (-10)}{1-\frac{5}{3}}
W=83×32W = \frac{83 \times 3}{2}
W=124.5 JW = 124.5~\text{J}

Source: geekyrakshit/indian-exam-questions / reneet2026-code50

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