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)
- 166 J
- 41.5 J
- 83 J
- 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)