A gas expands in a cylinder, pushing a piston outward at a constant speed of 0.2 m/s against a constant external force of 400 N . If heat is being supplied to the gas at a rate of 120 W , what is the rate of increase of the internal energy of the gas?

Step-by-step solution

The rate at which work is done by the gas (mechanical power) is given by the product of force and velocity: dW dt = F v = 400 0.2 = 80 W Since the gas expands, work is done by the system, so dW dt = +80 W . The rate of heat supplied to the gas is dQ dt = 120 W . According to the first law of thermodynamics: dQ dt = dU dt + dW dt Substituting the known values: 120 = dU dt + 80 dU dt = 120 - 80 = 40 W The internal energy of the gas increases at a rate of 40 W . Answer: 40 W