# Doubt regarding isothermal reversible process- Thermodynamics

Since its a reversible process, I can say

$$P_{ext}=P_{int}\pm dP$$ so now

$$W_{int}=\int_{V_1}^{V_2} P_{int}dV$$

My question is why can't I assume $$P_{ext}=P_{int}$$ here and take it as a constant outside the integral then finally get $$W = P_{ext}(V_2-V_1)$$?

(I am aware of the equation $$W = -nRTln({V_2}/{V_1})$$ but I really want to know what was done above doesn't work

• For isothermal reversible process. p=f(V), (for ideal gas p=nRT/V), so integral p.dV is not p . ΔV. Nov 9, 2023 at 15:30
• I'm not sure that your first statement is correct. Why is the external pressure so close in value to the internal pressure?
– Zhe
Nov 9, 2023 at 21:15
• @Zhe if it was not, it would be irreversible process, out of scope of the question. Nov 10, 2023 at 5:42
• Oops. Thanks. I missed that.
– Zhe
Nov 11, 2023 at 13:46

For a reversible process, you are right in saying that $$P_\text{int}= P_\text{ext}$$; however, this pressure is not constant. As the gas compresses or expands, the pressure changes, and as a result it is variable and cannot be taken out of the integration as a constant.

When the reversible process is also isothermal, Boyle's law applies and $$P_\text{int} = P_\text{ext} = \dfrac{n\text{R}T}{V}$$ where the numerator is constant and can be taken out of the integration:

\begin{align} \int_{V_\text{i}}^{V_\text{f}}P_\text{int}dV &= \int_{V_\text{i}}^{V_\text{f}}P_\text{ext}dV \\ &= \int_{V_\text{i}}^{V_\text{f}}\dfrac{n\text{R}T}{V}dV\\ &= n\text{R}T\int_{V_\text{i}}^{V_\text{f}}\dfrac{1}{V}dV\\ &= n\text{R}T\ln{\dfrac{V_\text{f}}{V_\text{i}}} \end{align}

Your approach works for a reversible isobaric process, the pressures are constant and can be taken out of the integration. Say $$P_\text{int}=P_\text{int}=P$$:

\begin{align} \int_{V_\text{i}}^{V_\text{f}}P_\text{int}dV &= \int_{V_\text{i}}^{V_\text{f}}P_\text{ext}dV \\ &= \int_{V_\text{i}}^{V_\text{f}}PdV\\ &= P\int_{V_\text{i}}^{V_\text{f}}dV\\ &= P(V_\text{f}-V_\text{i}) \end{align}

For a massless-frictionless piston, applying Newton's 2nd law gives $$F_g=P_{ext}A$$where $$F_g$$ is the force that the gas exerts on the piston. So the work the gas does on the piston is $$W=\int{F_gds}=\int{P_{ext}Ads}=\int{P_{ext}dV}$$

Do you think that, in a rapid irreversible expansion or compression, the average pressure of the gas on the piston is given by the ideal gas law? Or does the ideal gas law apply only in a reversible process.

• If I'm not wrong then ideal gas equation is applicable throughout a reversible process but only applicable on the initial and final states for an irreversible one Nov 10, 2023 at 15:57

Welcome to Chemistry SE Ainesh.

This is a common doubt that students all over the world have. In order to dispel it, it pays to have a glimpse of exactly what process each equation is describing. The irreversible expansion can be illustrated by the following example.

In this example, the external pressure is the same through out and is lower than the internal pressure. The only thing preventing the achievement of mechanical equilibrium between the two sides is the grey lock. Once the lock is removed, in the second step, the internal pressure is allowed to do work and the gas expands. Notice that from the second to the third step, the external pressure is constant at all times, and it only becomes equal to the internal pressure when equilibrium is reach. hence, when we compute the work exerted by the gas throughout the entire process, like so:

$$W = - \int_{V_i}^{V_f} P_{ext} \mathrm{d} V$$

The external pressure, as we have just established, is constant and can leave the integral, leaving us with the familiar expression for irreversible expansion:

$$W = - P_{ext} \int_{V_i}^{V_f} \mathrm{d} V = - P_{ext} \Delta V$$

In the reversible expansion we have more like the following situation:

Now here, things are different. The gas is not allowed to expand with its total power against a constant external pressure. Rather, there is an external gas, whose pressure is allowed to lower slowly, allowing the gast to undergo several differential steps of expansion against a constantly dwindling external pressure. At each step in the expansion, the internal and external pressure are in equilibrium, so at each differential step of expansion we have that $$P_{ext} = P_{int}$$.

If the expanding gas is an ideal gas, then $$P_{int} = \frac{n R T}{V}$$, which means that $$P_{ext} = \frac{n R T}{V}$$. Under these conditions we have that:

$$W = - \int_{V_i}^{V_f} P_{ext} \mathrm{d} V = - \int_{V_i}^{V_f} \frac{n R T}{V} \mathrm{d} V$$

Now, during this expansion we are assuming that there is no leakage in the internal gas. If this expansion can be made slowly enough and the temperature of the gas is maintained at a constant value throughout, then we can simplify the above expression to:

$$W = - n R T \int_{V_i}^{V_f}\frac{1}{V}\mathrm{d} V = - n R T \ln \left( \frac{V_f}{V_i} \right)$$

• If I do a force balance on a massless frictionless piston (or partition), I find, using Newton's 2nd law, that what you call $P_{Int}$ is equal to $P_{Ext}$ throughout the irreversible expansion. How do you reconcile that with your diagram showing that they are not equal? Do you think that, in an irreversible expansion, an ideal gas continues to exhibit gas behavior throughout the expansion, or is some other feature of gas behavior coming into play? Nov 10, 2023 at 12:02
• This clarifies it, thanks alot. I see now that $P_{ext}$ isn't necessarily constant. Also if P was constant then V would have to be too right? Then it would also imply work done is $0$ as $ln1=0$. Nov 10, 2023 at 16:04