Book-

"5.2.1 Work

First of all, let us concentrate on the nature of work a system can do. We will consider only mechanical work i.e., pressure-volume work. For understanding pressure-volume work, let us consider a cylinder which contains one mole of an ideal gas fitted with a frictionless piston. Total volume of the gas is Vi and pressure of the gas inside is p. If external pressure is pex which is greater than p, piston is moved inward till the pressure inside becomes equal to ex. Let this change be achieved in a single step and the final volume be V f . During this compression, suppose piston moves a distance, l and is cross-sectional area of the piston is A [Fig. 5.5(a)].

then, volume change = l × A = ∆V = (V f – Vi ) We also know, pressure = Therefore, force on the piston = pex . A If w is the work done on the system by movement of the piston then w = force × distance = pex . A .l = p ex . (–∆V) = – pex ∆V = – pex (Vf – V i ) (5.2)

The negative sign of this expression is required to obtain conventional sign for w, which will be positive. It indicates that in case of compression work is done on the system. Here (V f – V i ) will be negative and negative multiplied by negative will be positive. Hence the sign obtained for the work will be positive. If the pressure is not constant at every stage of compression, but changes in number of finite steps, work done on the gas will be summed over all the steps and will be equal to – Σ р ∆V [Fig. 5.5 (b)]"

Edited September 2Sep 2 by HbWhi5F

28 minutes ago, HbWhi5F said:

Book-

"5.2.1 Work

First of all, let us concentrate on the nature of work a system can do. We will consider only mechanical work i.e., pressure-volume work. For understanding pressure-volume work, let us consider a cylinder which contains one mole of an ideal gas fitted with a frictionless piston. Total volume of the gas is Vi and pressure of the gas inside is p. If external pressure is pex which is greater than p, piston is moved inward till the pressure inside becomes equal to ex. Let this change be achieved in a single step and the final volume be V f . During this compression, suppose piston moves a distance, l and is cross-sectional area of the piston is A [Fig. 5.5(a)].

then, volume change = l × A = ∆V = (V f – Vi ) We also know, pressure = Therefore, force on the piston = pex . A If w is the work done on the system by movement of the piston then w = force × distance = pex . A .l = p ex . (–∆V) = – pex ∆V = – pex (Vf – V i ) (5.2)

The negative sign of this expression is required to obtain conventional sign for w, which will be positive. It indicates that in case of compression work is done on the system. Here (V f – V i ) will be negative and negative multiplied by negative will be positive. Hence the sign obtained for the work will be positive. If the pressure is not constant at every stage of compression, but changes in number of finite steps, work done on the gas will be summed over all the steps and will be equal to – Σ р ∆V [Fig. 5.5 (b)]"

As it explains (admittedly a bit confusingly badly😀), the convention is for work done on the system, i.e. when the system gains energy, to be +ve. Conversely, work done by the system, losing energy , is -ve.

In this case the volume decreases, pushing the piston in. So ΔV is -ve. pΔV is therefore also -ve. But we need to express the work as +ve, in line with the convention. So we need to say that W = - pΔV.

Here is a better explanation, again from the Libretext source: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/19%3A_The_First_Law_of_Thermodynamics/19.02%3A_Pressure-Volume_Work

This uses an integral which is better since, as V decreases, p will generally not stay constant but will increase. That means p is a function of V, so you actually need to integrate PdV with respect to V.

What your text is trying to express, not very well, is the principle of integration as the limit of a sum of lots of little steps, each with a slightly bigger p than the one before. (Personally I think it's a lousy idea to try to teach the mathematical principle of integration as a sidebar to a discussion on the gas laws. In my view integration should be taught on its own, as a piece of the mathematical toolkit you need to do this kind of science. It's just confusing to cover it here.)

You never answered my previous question about what text you are quoting from. Can you reply, please?

Edited September 2Sep 2 by exchemist

It’s bookkeeping. Energy is conserved, so work being done means the energy has to come from some source, and go somewhere else. The sign is telling you whether energy was deposited in the system (+), or spent by the system (-)

You are very welcome here.

Your questions are good technical questions.

Your written English is quite good.

But can you read English, or are you using a translator ?

As already said, bu others it is a matter of sign conventions.

Sign conventions are very important - Have you heard of them ?

They are important partly because Physicists and Engineers use a dfferent sign convention compared to Chemists.

So help us to help you by answering some simple questions about your circumstance.

We don't care what they are.

But we can then understand how best to help.

Why do all your questions seem like homework questions ?
They are not general interest but relate to specific problems such as would be found in homework assignments.

@exchemist why is it more when done in finite steps ? The same process takes more energy ?

59 minutes ago, HbWhi5F said:

@exchemist why is it more when done in finite steps ? The same process takes more energy ?

What book are you using?

3 hours ago, HbWhi5F said:

@exchemist why is it more when done in finite steps ? The same process takes more energy ?

I'm not sure why you think it is saying it takes more energy when done in finite steps. I don't think the text is saying that. What is giving you that impression?

This whole business about finite steps is because the moment you start to compress the gas in the cylinder, the pressure starts to go up. It is to show you that you can't just multiply the pressure at the start by the volume change, or you will get the wrong answer for the work done (too low). The pressure is a function of volume, i.e. p is p(V). So as I said before you need to calculate this as an integral: W = - ∫p(V) dV. If you have studied calculus, you will know that integration is the limiting case of adding up a number of steps, each with a different value of the quantity.

(Again as I said previously, I personally think it is mistake on the part of the writer of this text to try to teach integral calculus as a sidebar to a discussion of the gas laws. It's just confusing - and, lo and behold, you have got confused! Some simple calculus is something you need to know to do physics at this level anyway.)

But luckily for us chemists, a lot of the time we only need to consider PV work in the context of lab experiments done under ambient atmospheric pressure. Under these conditions the volume changes during a chemical reaction occur under constant pressure, so we can calculate PV work, e.g. to work out enthalpy changes, just on the basis of volume change, without needing to do an integration, i.e. we can say ΔH = ΔU + PΔV. (That would not be true if we were reacting gases, in a vacuum line or something.)

Edited September 8Sep 8 by exchemist

12 hours ago, HbWhi5F said:

@exchemist why is it more when done in finite steps ? The same process takes more energy ?

The single step compression process described in the section leading up to Fig. 5.5b ignores the pressure of the gas inside the piston acting in opposition to P_ex. It therefore greatly overestimates the work done.

Also, bear in mind that for an isothermal compression process, there's a significant amount of heat of compression being removed from the system.

On 9/2/2025 at 12:34 PM, exchemist said:

As it explains (admittedly a bit confusingly badly😀)...

You're being very generous to the author here. I found the quoted text to be utterly appalling. If anything was ever designed to bewilder young minds...

8 minutes ago, sethoflagos said:

The single step compression process described in the section leading up to Fig. 5.5b ignores the pressure of the gas inside the piston acting in opposition to P_ex. It therefore greatly overestimates the work done.

Also, bear in mind that for an isothermal compression process, there's a significant amount of heat of compression being removed from the system.

You're being very generous to the author here. I found the quoted text to be utterly appalling. If anything was ever designed to bewilder young minds...

I don't know how this NCERT material is compiled, but I suspect it will be by various authors contributing different sections on a pro bono basis. What may happen, I imagine, is an author may not know exactly what has been taught elsewhere in other modules and hence what level of knowledge to assume in the student. That might explain this rather ham-fisted attempt to teach integral calculus, sitting in the middle of a piece of physics. It does make it hard for @HbWhi5F , so I sympathise with his or her difficulties with some of this.

I don't think these questions are homework, by the way. They read to me like attempts by a student to understand the (not always clearly explained) text.