# Is work done in isothermal processes greater than work done in isobaric processes?

Now I am here to learn something from the experts of this site. It would be good if anyone could give me an explanation of my query. It would be helpful also if I got some links that are able to meet my need.

Thanks!

• Why don't you formulate the relevant equations and analyze the situation yourself? Jun 21, 2016 at 17:27
• I tried but I failed. Work done for isothermal process is, W=nRT ln(v2/v1), and for isobaric process it is W=P(v2-v1). From this two equations I cannot find out any clue for my question. Jun 21, 2016 at 17:30

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).

With the same starting state $(P_1, V_1, T_1)$, for gas expansion to volume $V_2$, the work done for an isobaric process ($P_2=P_1$) is as you wrote,

$$W=P_1(V_2-V_1)$$

For isothermal process, your result is good, but just need to move it further.

$$W=\int_1^2 PdV=P_1\int_1^2 \frac{P}{P_1}dV =P_1\int_1^2 \frac{V_1}{V}dV$$

We know $V \gt V_1$ because of expansion. So $$W=P_1\int_1^2 \frac{V_1}{V}dV \lt P_1\int_1^2 dV =P_1(V_2-V_1)$$

So $$W_{isothermal}<W_{isobaric}$$

Is work done in isothermal process greater than work done in isobaric process?

No, it isn't.

Counterexample:

Assume that we want to compare work done in isothermal and isobaric processes for expanding an ideal gas from same initial states ($P_1,V_1,T_1$) to final volume of $V_2=eV_1$ ($e$ is the Euler's number $\approx 2.7$). We have:

$$W_{\textrm {isothermal}}=P_1V_1\ln \frac{V_2}{V_1}=P_1V_1$$ And $$W_{\textrm{isobaric}}=P_1(V_2-V_1)=P_1V_1(e-1)$$ As $e-1\approx 1.7\gt 1$ then $W_{\textrm{isothermal}}\ngtr W_{\textrm{isobaric}}$

• Thanks to both lucas and user115350. But your explanation make me confused. I cannot decide which one is correct or which one is best. Jun 21, 2016 at 18:51
• @NazmulHassan I wanted to give a hint by a counterexample that you try to prove the question yourself. Jun 21, 2016 at 18:53