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In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (γ=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⁻¹)

Asked in Re-NEET 2026 · Adiabatic Work and Internal Energy

• (1) 41.5 J
• (2) 83 J
• (3) 124.5 J
• (4) 166 J

Step-by-step solution

Why the other options are wrong

• (1) No single slip gives 41.5 J; it is only 1/2RΔ T, a third of the fall in internal energy 3/2RΔ T, and in an adiabatic process all of that fall is work.
• (2) This is RΔ T, the work in an isobaric process with the same change of temperature, not an adiabatic one.
• (4) No single slip gives 166 J; it is 2RΔ T, more than the whole fall in internal energy, 3/2RΔ T=124.5 J.

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