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Today in our chemistry class we derived the pressure-volume work done on an ideal gas. Our foremost assumption was that $$P_{ext}=P_{int}+dP$$ so that all the time the system remains (approximately) in equilibrium with the surrounding and the process occurs very slowly (its a reversible process). Now $$W_{ext}=\int P_{ext}dV$$ $$\Rightarrow W_{ext}=\int (P_{int}+dP) dV$$ $$W_{ext}=\int P_{int}dV$$ (Since $ dPdV$ is very small $\Rightarrow \int dPdV =0$, though it is an approximation I guess?)

Now, the question is :

  • In the case of (say) pushing a book the force on the book and that on the pusher form action reaction pair hence their work shows the same energy transfer but such isn't the case here and hence their work done does not represent the same energy transfer. So what does it represent? As in non-approximate case $W_{ext}-W_{int}=\int dPdV$. What does $ \int dPdV$ mean physically ?

[Note that I ain't equalizing the case of book with that of gas but giving (a kind of analogy or something) with respect to which I want the answerer to compare/contrast the compressing situation]


EDIT

I posted a similar on Maths SE to realize the mathematical significance of the term $ \int dPdV$. I got this answer over there. Though it mostly satisfies what I wanted to know but states that

The last term ( I believe is referring to $ \int \Delta PdV$ ) is then the energy “lost” e.g. by friction, that is, it is not reversible.

Now I'm wondering how does this external pressure term incorporates the frictional force in it?

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    $\begingroup$ The product of two differentials is considered insignificant. $\endgroup$ – Chet Miller Nov 18 at 12:14
  • $\begingroup$ @Aaron What I was trying to( say )is that when we apply some force say $2N$ on a book to move it some distance (say) $3m$ then work done by the push force on book is $6J$ whereas the push force on us by the book (via $3^{rd}$ law) is $-6J$. Here both represent the same energy transfer (i.e., $6J$ from the pusher to the book). But in case of ideal gas this isn't so, so what does that represent? $\endgroup$ – The Last Airbender Nov 18 at 16:44
  • $\begingroup$ A book is a solid. $\endgroup$ – VK_fan Nov 19 at 3:17
  • $\begingroup$ @Shreyansh I wasn't equalizing both the case but giving (a kind of analogy or something) with respect to which I want the answerer to compare/contrast the compressing situation. $\endgroup$ – The Last Airbender Nov 19 at 6:14
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Let me try to convince you that $ \int dPdv$ is almost negligible. As you have said, $P_{ext} = P_{int} + dP$ but what $dP$ really is? Well I think it is better to assume $dP$ as very small number and hence just adding it to $P_{int}$ will give a value bigger than $P_{int}$ at any moment no matter whatever $P_{int}$ is. So, in this sense $dP$ is just acting as constant. Let's see what this angle of thinking about $dP$ can lead to $$ W_{ext} = \int_{V_i}^{V_f} (P_{int}+dP)dV$$ $$ W_{ext} = \int_{V_i}^{V_f} P_{int}dV + \int_{V_i}^{V_f}dPdV$$ Now, let's just focus on the $dP$ part $$ X= dP\int_{V_i}^{V_f}dV$$ as $dP$ is constant. $$X= dP (V_i - V_f)$$ We agreed that $dP$ is a very small number and hence if we multiply it with any other thing no matter what the result will be very very small and therefore $X$ will be a very small number. $$ W_{ext} = \int_{V_i}^{V_f} P_{int}dV + X$$ Now, we can neglect $X$ and hence write $$ W_{ext} = \int_{V_i}^{V_f} P_{int}dV = W_{int}$$. Your argument that $ \int dPdV$ is negligible is quite sloppy as the integral adds many many pieces of small things ($f(x)dx$ is a very small number as $dx$ is very very small but adding many many of them would produce a different result).
Even in mechanics, when calculate gravitational potential energy we take the working force to be just a little more than $mg$ and hence calculate the work done just plugging the work with $mg$, however, the actual force is more than that.
I said that your argument was sloppy because it’s a matter of hyperreal numbers that when and when we cannot consider something negligible, your argument is vey all right if we just accept the rules of differentials.

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Your exact question is what is the physical interpretation of $$\int_{V_i}^{V_f}dPdV$$

I will try to explain this without using mathematics. Suppose you have a cylinder with piston at its one end which is free to move and the cylinder is filled with compressed gas at pressure $P_{int}$.

Your task here is to keep the piston stationary. You will have to apply exactly same pressure at its other end to keep it stationary and hence maintaining the thermodynamic state of gas at its initial condition.

$dP$, work done by you on the gas and work done by the gas on you are all zero in this case. This is the equilibrium state.

However, if your task was to slowly push the piston inwards further compressing the gas, you will have to increase the pressure applied by you on piston. The piston's acceleration will be dictated by how much you increased external pressure. Let's assume external pressure is increased by an amount $\delta$.

One assumption in your derivation is that the process occurs very slowly meaning the piston's acceleration is almost zero. Even if we assume that piston is not accelerating at all, we still need to increase the external pressure. Why? Because there is friction between piston and cylinder wall in real life scenario and additional pressure $\delta$ is used to overcome this friction.

And, the work done by $\delta$ is $$\int_{V_i}^{V_f}\delta.dV$$ And $\delta$ is your $dP$ .

Therefore, $\int_{V_i}^{V_f}dPdV$ represents nothing energy lost to over any dissipative force present in the system due to irreversibility of the process.

Since one of the assumption in your derivation is process is irreversible hence there is no friction and $\int_{V_i}^{V_f}\delta.dV$ is Zero.

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