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If we include the sign then work done in adiabatic expansion as well as contraction is greater than the work done in isothermal process (as although area under $pV$ curve for isothermal process is greater than that for adiabatic process for expansion...work is negative area under curve ($\Delta V$ is positive) and for contraction work done in adiabatic process is anyhow greater)...then why do we say work done in isothermal expansion is greater? Does sign not matter? please help as soon as possible...my test is coming in a week.

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  • $\begingroup$ An adiabatic expansion and an isothermal expansion can't be between the same two end points (since, in an adiabatic expansion, the temperature changes and, in an isothermal expansion, the temperature does not change. So how can you compare them on a common basis? $\endgroup$ – Chet Miller Dec 2 '18 at 15:44
  • $\begingroup$ For the same volume change, if the initial pressures match, the isothermal expansion work will be greater than the adiabatic expansion work. But, if the final pressures match, the adiabatic expansion work will be greater than the isothermal expansion work. $\endgroup$ – Chet Miller Dec 3 '18 at 1:45
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If we include the sign then work done in adiabatic expansion as well as contraction is greater than the work done in isothermal process

This is true for compression, not expansion. I'll get to this soon.

Isothermal processes follow $PV = constant$ while adiabatic processes follow $PV^{\gamma} = constant$ with $\gamma > 1$. We can therefore easily compare the two processes:

enter image description here

Clearly the area under the curve for isothermal processes is greater, so isothermal processes require more work.

Does sign not matter?

It does matter, but we compare absolute values when making claims like the "work done in isothermal expansion is greater."

For expansion, volume starts at $V_1$ and ends at some greater volume $V_2$. If you integrate the curves in the figure, you'll get positive work for both cases, meaning that work is performed on the surroundings. Clearly, $W_{isothermal} > W_{adiabatic}$ for expansion, meaning that an isothermal expansion does more work on the surroundings.

For compression, integrate the $PV$ curve from a larger volume $V_2$ to a smaller volume $V_1$. You'll have the same magnitudes of work as we did for expansion, but they are now negative. This means that work is input into the system. I think you're confused because $|W_{isothermal}| > |W_{adiabatic}|$ here (which is always true), but $W_{isothermal} < W_{adiabatic}$ since the adiabatic work is less negative. However, the isothermal compression requires more work to complete the process. When we say that isothermal compression requires more work, we mean that more work is input into the system (it is more negative).

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Does sign not matter?

The sign simply tells us whether the system (say ideal gas) is performing work on the surroundings (positive work, where energy is transferred out of the system), or the surroundings are performing work on the system (negative work, where energy is transferred into the system). In each case, the amount of work equals the magnitude of the energy transferred, regardless of whether it is positive or negative. In all cases the transfers are governed by the first law: $\Delta U=Q-W$.

If we include the sign then work done in adiabatic expansion as well as contraction is greater than the work done in isothermal process

From the graphs provided by @drew, if the system starts at the same equilibrium state and expands to the same final volume, you can clearly see that the positive isothermal work is greater than the positive adiabatic work since the area under the $pV$ curve is greater. The reason is the isothermal expansion process uses the heat transferred from the surroundings to do its work, whereas for the adiabatic expansion $Q=0$ and the process uses the system’s internal energy to perform its work. This results in a greater drop in pressure for the adiabatic expansion to reach the final volume than that for the isothermal expansion, and thus less area under the $pV$ curve and less work.

When the processes are reversed, the pressure rises at a faster rate for the adiabatic process (because all of the energy of the work done on the system increases its internal energy) than the isothermal process (because all of the energy of the work done on the system transfers out as heat). As a result, more negative work has to be done by the isothermal process to return to the same initial pressure as the adiabatic process. The adiabatic work may be less negative, but as previously stated, the amount of work depends only on its magnitude.

Bottom line: The magnitude of the work for the isothermal process for both expansion and compression is greater than the magnitude of the work for the adiabatic process. Although the adiabatic compression work is less negative than the isothermal compression work, the amount of work depends only on its magnitude.

Hope this helps.

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  • $\begingroup$ @ Bob D and Chester Miller...thanks a lot...so that means work done being negative or positive just denotes the direction and hence we can only compare with magnitude...that is area under curve?Great help...I will keep asking physics doubts from you all $\endgroup$ – Simran girdhar batra Dec 6 '18 at 13:18

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