The constant mass volume of an ideal monatomic gas increases by 0.03 m3

The constant mass volume of an ideal monatomic gas increases by 0.03 m3 at a constant pressure of 500 kPa. How much has the internal energy of the gas increased?

Initial data: P (constant pressure of an ideal monatomic gas) = 500 kPa (500 * 10 ^ 3 Pa); ΔV (change in the volume of a monatomic gas) = 0.03 m3; the mass of the gas is constant.

Reference values: i (the number of degrees of freedom of molecules of a monatomic gas) = 3.

The increase in the internal energy of a given gas during the isobaric process is calculated by the formula: ΔU = 0.5 * i * P * ΔV.

Let’s make the calculation: ΔU = 0.5 * 3 * 500 * 10 ^ 3 * 0.03 = 22500 J (22.5 kJ).

Answer: The internal energy of the gas has increased by 22.5 kJ.