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Let's consider this example enter image description here

We have an isolated cylinder, inside of it is an ideal gas. The pressure is constant (note that the thing where the constant force is acting upon can move up and down). We bring some work to the gas. The first law of thermodynamics states:

Q1-2=U2-U1+W1-2

(U2-U1 is change in internal energy). Considering that the system is isolated we can re-write this as:

W1-2=U1-U2

Let's suppose we bring 1 joule of work to the system. Using the formula above, that work will cause a change in internal energy. But if internal energy increases, that means the temperature also increases. And since:

pV=mRT

is a formula of ideal gas, if temperature rises, the volume will also rise (since other numbers are constant). Positive change in volume means that the gas is doing work.

To summarize: we bring work to the gas, it's temperature rises, it's volume increases. I don't understand where is that work done by the gas visible in the formula for the first law of thermodynamics? What am I doing wrong? All I see is that the work I brought increased it's internal energy.

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  • $\begingroup$ "since other numbers are constant" That's your error. It's not true. $\endgroup$
    – garyp
    Mar 13 '17 at 19:11
  • $\begingroup$ hmm... which ones are not constant? Mass... there is not reason for mass to change. R is a gas property which is a constant. Pressure is also constant if we consider this to be a balanced system, which (among others things) means that there is a mechanical balance. And since the force F on the picture is constant, so is the pressure of the gas. $\endgroup$
    – M. Wother
    Mar 13 '17 at 19:34
  • $\begingroup$ "I don't understand where is that work done by the gas visible in the formula for the first law of thermodynamics..." I'm confused by this. What exactly do you mean? Note that $W_{12}$ is made up of both the positive work done by the paddle and the negative work done by the piston (in rising a distance $h$, the piston does work $-Fh$ on the system, assuming that the piston is massless). $\endgroup$
    – march
    Mar 13 '17 at 20:05
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There are two types of work being done: work done to stir the gas using the paddle and work done by the gas to cause the piston to rise against the constant force F. So the total work done by the gas is $$W=-W_p+P\Delta V$$ where $P=F/A$, $\Delta V=A\Delta x$, $\Delta x$ is the upward displacement of the piston, A is the cross sectional area of the cylinder, and $W_P$ is the work done by the paddle on the gas. If the system is insulated (adiabatic), we have $$\Delta U=nC_v\Delta T=W_p-P\Delta V$$ or equivalently, in terms of enthalpy,$$\Delta H=nC_p\Delta T=W_p$$ So the temperature rises.

If the system is not adiabatic, and heat transfer is also involved, then $$\Delta U=nC_v\Delta T=Q+W_p-P\Delta V$$or, in terms of enthalpy,$$\Delta H=nC_p\Delta T=Q+W_p$$

It's as simple as that.

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  • $\begingroup$ I see... so in the formula Q=W+U2-U1, the work W is net work? $\endgroup$
    – M. Wother
    Mar 13 '17 at 22:35
  • $\begingroup$ Yes, that is correct. You include all the forms of work in specifying W. $\endgroup$ Mar 13 '17 at 22:43

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