A gas expands in a cylinder, pushing a piston outward at a constant...

The rate at which work is done by the gas (mechanical power) is given by the product of force and velocity:

$\dfrac{dW}{dt} = F \times v = 400 \times 0.2 = 80\text{ W}$

Since the gas expands, work is done by the system, so $\dfrac{dW}{dt} = +80\text{ W}$.

The rate of heat supplied to the gas is $\dfrac{dQ}{dt} = 120\text{ W}$.

According to the first law of thermodynamics:

$\dfrac{dQ}{dt} = \dfrac{dU}{dt} + \dfrac{dW}{dt}$

Substituting the known values:

$120 = \dfrac{dU}{dt} + 80$

$\dfrac{dU}{dt} = 120 - 80 = 40\text{ W}$

The internal energy of the gas increases at a rate of $40\text{ W}$.

Answer: $40\text{ W}$