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Thermodynamics (Physics) · Question 2

RE-NEET (UG) 2026MCQEasy📈 Topic in 38% papers+4−1
1

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

  1. 166 J
  2. 41.5 J
  3. 83 J
  4. 124.5 J

Answer

(4) 124.5 J

Solution

Work done in an adiabatic process is:

W = nR(Ti - Tf) - 1

Given n = 1 mol, = 5/3, Ti = 60 K, Tf = 50 K:

W = 1 R (60 - 50) 53 - 1 = 10R2/3 = 15R

Substituting R = 8.3 J mol-1 K-1:

W = 15 8.3 = 124.5 J

Answer: (4)

RE-NEET (UG) 2026

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