2. The change in internal energy for the expansion of a gas sample is -4750 J. How much work is done if the gas sample loses 1125 J of heat to the surroundings? Is this work done by the gas or done by the surroundings?

The work done by the gas expansion is 5875 J,

Since the work done is positive, the work is done by the gas on the surroundings.

Explanation:

Given;

change in internal energy, ΔU = -4750 J

heat transferred to the system, Q = 1125 J

The change in internal energy is given by;

ΔU = Q - W

Where;

W is the work done by the system

The work done by the system is calculated as;

W = Q - ΔU

W =  1125 - (-4750)

W = 1125 + 4750

W = 5875 J

Since the work done is positive, the work is done by the gas on the surroundings (energy flows from the gas to the surroundings).

Therefore, the work done by the gas expansion is 5875 J

a. they are both space stations

happy to

pls mark as brainliest.

you need to give more information