The air was heated at 30 K at a constant pressure of 10 ^ 5 Pa. At the same time, its volume changed from 1.5 to 1.67 m3.

The air was heated at 30 K at a constant pressure of 10 ^ 5 Pa. At the same time, its volume changed from 1.5 to 1.67 m3. Determine the work done by the air during expansion and the increment of internal energy.

To find the values of the work of air and the increment of its internal energy, we will use the formulas: A = P * ΔV = P * (Vк – Vн) and ΔU = i * P * ΔV / 2 = i * P * (Vк – Vн) / 2.

Constants and variables: P – constant air pressure (P = 10 ^ 5 Pa); Vк – air volume after heating (Vк = 1.67 m3); Vн – initial volume (Vн = 1.5 m3); i is the number of degrees of freedom of air molecules (i = 5, since air mostly consists of diatomic molecules).

Calculation: a) Work: A = 10 ^ 5 * (1.67 – 1.5) = 17 * 10 ^ 3 J = 17 kJ;

b) Increase in internal energy: ΔU = 5 * 10 ^ 5 * (1.67 – 1.5) / 2 = 42.5 * 10 ^ 3 J = 42.5 kJ.