1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ How did you get to your conclusion with the pV graphs? On an isothermal curve the pressure will decrease as volume increases, therefore there is less area under the curve for the isothermal case. Therefore, it seems like your math is correct, but your second section with the pV graphs is wrong. $\endgroup$ – Aaron Stevens May 14 '18 at 16:53
  • $\begingroup$ @Aaron Stevens I'v added the picture of the graphs that were used (similar to the graphs in my book). I don't understand why the value of initial p is less in the third graph unlike the first two. $\endgroup$ – suiz May 14 '18 at 17:07
  • $\begingroup$ Ah ok. So it all depends on the pressures involved. In your math you are assuming that you are starting at the same pressure in each case. You can get larger or smaller values for either scenario depending on your starting pressure. $\endgroup$ – Aaron Stevens May 14 '18 at 17:12
  • $\begingroup$ @Aaron Stevens But if we look at the question, it hasn't mentioned that the pressure are different. Do you think it was wrong of me to take same initial pressure? $\endgroup$ – suiz May 14 '18 at 17:27
  • $\begingroup$ No I do not think this is wrong. The question seems to assume the gases are starting at the same initial conditions and then expanding with two different processes. Therefore, I would trust your math. $\endgroup$ – Aaron Stevens May 14 '18 at 17:36
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

tl;dr- Your math's good, and the constant-pressure case does require more work. Just gotta remember that the $p$ in the constant-pressure case needs to be the maximum pressure that the system will experience, i.e. $\max{\left(p_1,\, p_2\right)}$.


We can transform the mechanical definition of work,$$ W~=~\int_{\vec{x}_{\text{start}}}^{\vec{x}_{\text{finish}}}\vec{F}_{\text{resistance}} \cdot \, \mathrm{d}\vec{x} \,,$$into the expression of work that you're using by just noting that the resistive force is equal to the resistive pressure over the interface area. So, we rewrite resistive force as resisitve pressure, then take the area factor and use it to transform the displacement distance into a displacement volume, yielding$$ W~=~\int_{V_{\text{start}}}^{V_{\text{finish}}}P_{\text{resistance}} \, \mathrm{d}V \,.$$ Your math in the question is correct: the work's greater in the case of constant pressure. The reason's that, in order for the constant pressure of the gas to be sufficient, it must be equal to the resistive pressure; this implies that the resistive pressure is at a maximum the entire time, forcing the gas to do more work. But when you allow the pressure to vary, the resistive pressure is assumed to respond in-kind, such that the gas ultimately gets resisted less.

If that seems a bit off, it's because it is. This equation assumes that the pressure asserted by the gas matches the resistive pressure, plus infinitely slightly more, to have a reversible process. The idea's that minimal work is done, and no entropy is created – reminding that the temperature is constant in this case.

In the constant-pressure case, more work is done as it's not a reversible process. The energy balance is satisfied by a change in entropy; this is, the temperature was increased (doubled, for an ideal gas) to double the volume.

Anyway, your math was correct. As for the graphs, just need to note that in the constant-pressure case, $p$ needs to be the larger of $p_1$ and $p_2$ from the other cases, showing that the constant-pressure case requires more work.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.