A system containing an ideal gas at a constant pressure of 1.42 × 10⁵ Pa gains 2320 J of heat. During the process, the internal energy of the system increases by 2540 J. What is the change in volume of the gas?

Question

Quizplus · Accepted Answer

The work done by the system is given by W = Q - ΔU, where Q is the heat added and ΔU is the change in internal energy. Here, W = 2320 J - 2540 J = -220 J. The work done by the system is also given by W = PΔV, where P is the pressure and ΔV is the change in volume. Solving for ΔV gives ΔV = W/P = -220 J / 1.42 × 10^5 Pa = -1.55 × 10^-3 m^3.

A System Containing an Ideal Gas at a Constant Pressure