7 mole of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is (Given R = 8 . 3 J K - 1 mol - 1 )

Step-by-step solution

For the given process pressure is constant therefore, it is an isobaric process. For a quasi-static process the change in internal energy of an ideal gas is independent of the nature of the process and is given by, Δ U = n C v Δ T = n × 3 R 2 × Δ T [molar heat capacity at constant volume for monatomic gas = 3 R 2 ] Δ U = 7 × 3 2 × 8 . 3 × 40 = 3486   J