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So the equation for reversible work is $xRT \sim \ln\left(v_2/v_1\right)$ where $x = \textrm{no. of moles of gas}$. This is calculated on the basis of internal pressure. This equation is also given as the intermediate step in the carnot engine cycle, where heat from the hot reservoir is converted into this work. My question is, for a reversible process, this expansion should happen infinitely slowly, with the external pressure almost equalling the internal pressure at every point. So you need to do work to change the external pressure at every point as well which would be equal to the work done by the gas in expansion. So why do we say that the energy from the hot reservoir gets converted into the work done by the gas? I get that the reservoir is the reason delta u is equal to 0 and thus the heat it puts in should be equal to the work done by the system, but the work done by the system clearly wouldn't be reversible if the external pressure wasn't changing in such a way so as to keep the system in equilibrium the whole time. And you'd need to do an equivalent amount of work to do that. So you need to do work as well as put the same amount of heat? I think I'm missing something.

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    $\begingroup$ This should help: math.meta.stackexchange.com/q/5020 $\endgroup$ – bemjanim Feb 9 at 8:40
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    $\begingroup$ Please note that the quantity “amount of substance” shall not be called “number of moles”, just as the quantity “mass” shall not be called “number of kilograms”. $\endgroup$ – user59991 Feb 9 at 9:55
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So you need to do work to change the external pressure at every point as well which would be equal to the work done by the gas in expansion.

It is in this statement where you are going wrong.

Consider the following thought experiment. (See diagrams below). Let's say you have the gas in a vertically oriented frictionless cylinder and piston. On top of the piston you have a number of weights which, in addition to the external air pressure, provides the external pressure on the system. You now slide a little weight horizontally off of a platform connected to the shaft of the piston and onto another platform along side the cylinder. That reduces the external pressure allowing the gas to expand and perform work in raising the remaining weight.

Sliding the weight horizontally off the piston would theoretically require very little effort (work) compared to the work done by the gas in raising the remaining weight. In order to carry this process out extremely slowly (reversibly) we can imagine the weight as a pile of sand. We remove the sand one grain at a time causing an infinitely small reduction in external pressure. That results in an infinitely small expansion of the gas, $Adh$. The expansion causes an infinitely small decrease in the gas temperature $dT$ and infinitely small transfer of heat $dQ$ into the gas to bring its temperature and pressure back into equilibrium with surroundings, awaiting the next removal of a weight.

Finally, for the process to be reversible, we must return both the system and the surroundings exactly to their original state. Note that after the last weight is removed (top of figure to the right), the platform is above the last weight that was removed. The requires us to take an additional weight from somewhere in the surroundings and place it on our platform to begin the reverse process. The obvious choice in order to return the system to its original state (pressure and volume) is to take the first weight that was removed and raise it to the top of the platform. At the completion of the reversed process the system (gas) has been returned to its original state but the surroundings has been altered as it had to do work to raise the first weight. This demonstrates that in order for the process to be reversible the weights must be infinitesimally small.

Hope this helps.

enter image description here

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  • $\begingroup$ This is the scenario which led me to ask this question. You say sliding the weight off horizontally requires less work, but the increase in height of the piston is infinitesimal as well, right? $\endgroup$ – kimi Feb 9 at 9:58
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    $\begingroup$ @kimi each step is infinitesimal but the accumulation for the entire process is not. For example after half the sand has been removed the other half has been elevated the entire height of the expansion to that point. And don’t forget the gas expands against constant atmospheric pressure knee the entire process $\endgroup$ – Bob D Feb 9 at 12:19
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    $\begingroup$ And BTW that’s why the work is proportional to the log of the ratio of the final to initial volumes and not simply the difference as in the case of a constant pressure process where more work is done $\endgroup$ – Bob D Feb 9 at 12:24
  • $\begingroup$ @kimi I have added more detail on the thought experiment you might find helpful. $\endgroup$ – Bob D Feb 9 at 21:01
  • $\begingroup$ Yes, thank you so much. $\endgroup$ – kimi Feb 10 at 0:55

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