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


2 Answers 2


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.

  • $\begingroup$ This still doesn't answer my question.While it may consist of other forms of energy but why don't we just write dQ=dU rather than dQ=dU+PdV because PdV is just work done which is stored as potential energy which is compensated by dU=K.E+P.E+.....? $\endgroup$ Jan 19, 2017 at 2:22
  • $\begingroup$ @SurazBasnet $U$ is a function of thermodynamic parameters (such as $P,V$), while P.E. and K.E. in general are not. So if you wish to do thermodynamics then you must separate $U$ from P.E. and K.E. $\endgroup$
    – Deep
    Jan 19, 2017 at 5:29
  • $\begingroup$ The general formula for First law of thermodynamics is dU=dQ + dW and not dQ=dU+dW $\endgroup$
    – Mitchell
    Jan 19, 2017 at 7:00
  • $\begingroup$ @BhavyaSharma The formula is absolutely correct . The sign of W is taken opposite its the work done by the system ! $\endgroup$
    – Shashaank
    Feb 21, 2017 at 7:24
  • $\begingroup$ @SurazBasnet Wo don't write dU=dQ because dU considers only the internal microscopic energies whether potential or kinetic or electrostatic (potential) not the macroscopic energy of the system. That is if the system is moving with a velocity v or is in a gravitational or electrostatic field , then those energies are not accounted in the internal energy. Internal energy consists of the microscopic energies of the atoms. I guess you can make it up from here . $\endgroup$
    – Shashaank
    Feb 21, 2017 at 7:28

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.


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