JEE Main 2019 Physics question
Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20°C to 90°C. Work done by gas is close to: (Gas constant R = 8.31 J/mol.K)
Answer: (D)
Solution
W = nRΔT = 0.5 × 8.31 × 70 ≈ 291 J.
Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20°C to 90°C. Work done by gas is close to: (Gas constant R = 8.31 J/mol.K)
W = nRΔT = 0.5 × 8.31 × 70 ≈ 291 J.