Question

30 J of heat are removed from a gas sample while it is being compressed by a piston that does 20 J

of work. What is the change in the internal energy of the gas sample?

Answer 1

The change in internal energy of the gas sample when 30 J of heat are removed and 20 J of work is done on it is -50 J, according to the first law of thermodynamics.

Step-by-step explanation:

The change in the internal energy of a gas sample when 30 J of heat are removed and 20 J of work is done on it can be found using the first law of thermodynamics. This law states that the change in internal energy (ΔU) is equal to the heat (Q) added to the system minus the work (W) done by the system.

Thus, if heat is removed, Q is negative, and if work is done on the system, W is negative. In this scenario, the heat Q is -30 J since it is removed, and the work done on the system is +20 J.

Using the formula ΔU = Q - W, we have:

ΔU = (-30 J) - (+20 J)

= -30 J - 20 J

= -50 J

Therefore, the change in internal energy of the gas sample is -50 J.