# Lecture 7.003- Work and Heat in Thermodynamics

Work

The first thing that strike is Work is product of force applied and displacement produced.

So, not going to much different we can take the simple piston example and prove the pressure volume work,

Now the work done here, W= Force * displacement

= Pressure* area * displacement

= pressure *(volume change)

That is generalize version of pressure volume work

Wpressure-volume = Psurrounding * V

• The volume change is V, its always V final –V initial, it might be negative or positive but the value is calculated as such, the sign have a very detailed discussion later.
• Psurrounding  is the pressure applied by the surrounding on the piston to compress it. It has nothing to do with internal pressure of the system.

Sign convention of work done

Here is where we have confusion, which is positive and which is negative work.

Now the problem is that convention are taken different in physics and chemistry but we will use only one of them and learn how to distinguish,

Are results for a thermodynamic system different in physics and chemistry?

Obviously not, it’s just the convention we follow, not the nature and it gives same result irrespective of whether you are using physics or chemistry.

W = Pext x △V               … (in physics)

Let’s take compression work, and then the change in volume is negative.

So the work done is a negative, here work is done by the surrounding on the system

In physics this is negative work done, but in chemistry we assume this as positive so,

W = — Pext x △V               … (in chemistry)

The consequences of this change in convention can be seen at later stage, but we will use all the conventions in chemistry but now you know the all difference, so try everything with both yourself.

Now the total work done when the pressure is not constant,

Not there is something very interesting about the dw, the term in integral makes it a

Non-exact integral the bar over the function represent that only.

The meaning of a non-exact integral is that the integral depend on the path taken and is not a fixed value for given upper and lower limits. The meaning is explained with the example given-

Taking a thermodynamic process for example,

Calculation of work

So if I just state that the work done can be calculated by integral dw then there are infinite possible ways in which we can go from (P1,V1) to (P2,V2) therefore we will have infinite solution of the integral in other words that integral don’t give a “Exact” or is inexact integral.Now on one thing,

The area highlighted is the work done.That gives rise to

1) Why does the area represents the work done

2) What work does it represent on the system or by the system or how can we differentiate between positive and negative work.

For those whose know meaning of integrals it’s easy

(Tool’s in mathematics T.01 cover a very short to explain go look up that video first)

But luckily here the example is simple enough to be solved by

W = — Pext x △V

Path 1 :- Now for the process (P1, V1) to (P2, V1) although we said formula is valid only for a constant pressure process but it doesn’t matter because the other term is zero so the net result is zero as, we can say that the term △V= Vfinal – Vinitial

Is zero as there is no change in process analogically the area under the line is also zero.

Now for the other half of the process (P2,V1) to (P2,V2) the pressure remains constant so we have W = — P2 x △V

Now here we are using a chemistry formula but as we concluded before it doesn’t matter therefore it △V= Vfinal – Vinitial  and since the volume is decreasing the term will be negative i.e. the Work done is positive.

Abhishek kumar jha

(Chemistry at Utkarshini)

7-003-work-in-thermodynamics