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

Question

Quantrex Academy · Accepted Answer

Correct answer: D. 124.5 J

The work done by an ideal gas in an adiabatic process is given by the formula: W = nR(T_1 - T_2) - 1 Given values are: n = 1 mole R = 8.3 J mol^ -1 K^ -1 T_1 = 60 K T_2 = 50 K = 5 3 Substituting these values into the formula: W = 1 8.3 (60 - 50) 5 3 - 1 W = 8.3 10 2 3 W = 83 3 2 W = 249 2 = 124.5 J Answer: 124.5 J

NEET Physics Question (2026) — Solution

Question

Options

Answer

Step-by-step solution