The internal energy of an ideal gas depends only on its temperature. Do a first-law analysis of this process. A sample of an ideal gas is allowed to expand at constant temperature against atmospheric pressure.

Check Explanation

Explanation:

The Mathematical expression of the first law is given as

ΔU = Q - W

Note that depending on convention, the mathematical expression for the first law can be written as

ΔU = Q + W

a) Yes, the gas does work on its surroundings. Since it expands; indicating that there is a change in volume.

The work done is given as -PΔV if we are using the convention of (ΔU = Q + W). And this work is negative work done, since the system expands and does work on the surroundings.

But if we are using the convention of (ΔU = Q - W), the work done is given as PΔV and the workdone by the system on the environment is positive work during expansion.

b) Since this all takes place at constant temperature and against a constant atmospheric pressure, the change in internal energy for this system is 0. Change in internal energy depends on a change in temperature.

So, from the mathematical expression,

ΔU = Q + W or ΔU = Q - W

If ΔU = 0,

Q = - W or Q = W

But either ways,

Q = PΔV

(because, W = -PΔV for ΔU=Q+W and W = PΔV for ΔU=Q-W)

So, either ways, the heat transfer is the same and it is positive. This indicates heat is transferred from the surroundings to the system.

(c) What is ΔU for the gas for this process?

Since this all takes place at constant temperature and against a constant atmospheric pressure, the change in internal energy for this system is 0. Change in internal energy depends on a change in temperature.

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