[Physics] Does the gas do more work on its surroundings if the expansion is at constant pressure or at constant temperature

thermodynamics

A quantity of ideal gas
undergoes an expansion that doubles its volume. Does the gas do more work on its surroundings
if the expansion is at constant pressure or at constant temperature?

The answer to this was based on the graph of the two cases and comparison of the two $pV$ graphs. Which gave a conclusion that $W$ at constant temperature gives a greater value for the given case.

But when I tried to solve it mathematically, I got greater value of $W$ at constant pressure which is plain wrong. I don't understand where did I go wrong.

For constant pressure my result was:

$$
W ~=~ p \left(2V-V\right) ~= ~ \frac{nRT}{V}\left(2V-V\right) ~=~nRT
\,.$$

For constant temperature my result was:$$
W~=~nRT \ln{\left(\frac{2V}{V}\right)}~=~nRT \ln{\left(2\right)}
\,.$$

So, I got lesser value for constant temperature than constant pressure.

Here are the graphs that were used to get the 1st conclusion:

enter image description here

Best Answer

A quantity of ideal gas undergoes an expansion that doubles its volume. Does the gas do more work on its surroundings if the expansion is at constant pressure or at constant temperature?

The correct answer: Not enough information. What constant pressure, what constant temperature? The pressure in the isothermal expansion obviously changes; in the case that the volume doubles, the pressure drops by a half. The isobaric expansion does less work than does the isothermal expansion if the isobaric expansion occurs at the final pressure of the isothermal expansion, but more work if the isobaric expansion occurs at the initial pressure of the isothermal expansion.

In your math, you implicitly assumed the two expansions start in the same state. Your interpretation of the graph did not assume this to be the case. To interpret the graph with this assumption, look at the leftmost graph. If you draw a horizontal line from P1,V1 to P1,V2, it's pretty clear that the area under this curve is greater than the yellow shaded area.

As for why the isobaric expansion results in more work than the isothermal expansion given the same initial conditions, think of the amount of heat that needs to be transferred to maintain the isobaric condition versus the isothermal condition. The isothermal expansion results in zero net change in the internal energy of the gas. By the first law of thermodynamics, this means the heat transfer needed to maintain the isothermal condition is exactly equal to the amount of work done by the gas. You've properly calculated that as $W_{\text{isothermal}} = NRT\ln 2$, and thus $Q_{\text{isothermal}} = NRT\ln 2$.

The heat transfer during a isobaric process is $Q_{\text{isobaric}} = nC_p\Delta T$. To maintain isobaric condition, the temperature must double when the volume doubles, so $Q_{\text{isobaric}} = nC_p T$. Since $C_p > R > R\ln 2$, a good amount more heat needs to be transferred to maintain the isobaric condition vs. the isothermal condition. Some of that extra heat results in a greater amount of work; the rest goes into raising the temperature.

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