First law of thermodynamics confusion

Question

The first law of thermodynamics tells us that: $dQ=dU+dW$

We know that for real gas, $dU=f(V,T)$ and $dW=f(P,V)$

Therefore, $dQ=f(V,T)+f(P,V)$

We know, Internal energy $(dU)$ = Kinetic energy+Potential energy

Therefore $dQ=K.E(T)+P.E(V)+dW$

Here doesn't the work done while heating the gas stored as the potential energy? If that is true, why can't we just write $dQ=K.E+P.E$, as $P.E$ and $dW$ are same thing.

What am I missing here? Please help in simple words as possible

Answer 1

Internal energy consists of many other forms of energies not just KE or PE.

Dont confuse it with mechanical energy conservation.

When you heat the gas you dont know for sure where does the heat energy goes. But the energy is always conserved.

Answer 2

I think i get your question, its perfectly a valid question.

Look at this formula, dU = dQ + dW.

dQ is the heat change and dW is the work done by or on the system.

First lets get this formula clear.

If heat is added to this then dq is +ve and if heat is given out by the system then dq is -ve.

The work done is a little different case.

The formula for work done here is dW = -PdV (note the -ve sign here).

Now if work is done by the system then dw is -ve (as dV is +ve, system expands).

Now if work is done on the system then dw is +ve (as dV is -ve, system is compressed).

The system is the key here not the surrounding. The signs of the heat and work done are in reference to the system. That explains the -ve sign the formula for the work done.

Now think for yourself what are the things that can bring a change in the internal energy of a system by keeping the energy is the universe constant. (We cannot determine the absolute internal energy, only the change in it). The first law's formula gives us the only two things that can change the internal energy.

Let me give you a case, Consider an adiabatic process, in which dQ = 0.

Now the formula becomes dU = dW. According to you this formula becomes dU = 0 as you are considering dU and dW as one term.

Say, an external agent perform some work on the system dW = +ve which means that dU will also be +ve. The work that we have done must go somewhere and thus it becomes the internal energy of the system. This work INCREASES the internal energy.

But according to your formula, dU = 0. Now this formula implies that even if some work is performed on the system the internal energy wont change. As dQ=0, this energy has to go somewhere because energy is always conserved. So Your formula fails to explain this.

This argument should answer your question. Understand this explanation.