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.
[Physics] Work done in isothermal vs adiabatic process
adiabatichomework-and-exercisesthermodynamics
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Yes quite! The ideal gas law (and in general any state equation) holds only on equillibrium while the 1st law (and all the rest) hold in general. So, your anslysis mr James Hoyland is inacurrate.
Mr Steven, the post includes the word "expansion" so volume changes by assumption.
And mr or ms PhysC, the first two cases are correct. About the third, expansion would occur by pulling the piston thus removing energy from the system and causing decrease on temperature
I'm guessing that you want to compare the work done in an expansion from volume $V_1$ to volume $V_2$ following an isothermal and an isobaric curve respectively.
First, let's look at the $PV$ plane:
(The curve b is an isotherm, which we know is an hyperbola in the $PV$ plane)
Since work is the area under the curve, it is clear that sometimes the work done in the isobaric process will be more and sometimes it will be less. For example, the work done in process a (isobaric) is clearly more than the work done in process b (isothermal), but the latter is more than the work done in process c (isobaric).
Now, some math: let's find the condition under which the isothermal work is less than the isobaric work.
The work performed during the expansion is
$$W=\int_{V_1}^{V_2} P dV$$
So, for an isobaric process:
$$W_{ib}=P \int_{V_1}^{V_2} dV = P (V_2-V_1)$$
While for an isothermal process:
$$W_{it}=\int_{V_1}^{V_2} P dV=nRT\int_{V_1}^{V_2} \frac{dV}{V} = nRT \log \left(\frac{V_2}{V_1}\right) $$
So, if we impose
$$W_{it}<W_{ib}$$
we obtain
$$nRT \log \left(\frac{V_2}{V_1}\right)<P (V_2-V_1)$$
that is to say
$$\frac{\log(V_2)-\log(V_1)}{V_2-V_1}<\frac P {nRT} $$
This equation can be solved numerically.
For example, if $P/nRT=1$ (in the correct units), we get the folowing inequality plot:
where $x=V_2$ and $y=V_1$ (you have of course to consider only the $x>0,y>0$ quadrant).
Best Answer
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:
Clearly the area under the curve for isothermal processes is greater, so isothermal processes require more work.
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).