JEE Main 2024 Physics question
A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2°C. The work done by the gas is: (Given, R = 8.3 J K⁻¹mol⁻¹.)
Answer: (D)
Solution
ΔU = (3/2)RΔT = 24.9 J, so W = Q − ΔU = 48 − 24.9 = 23.1 J.