I started another post about this but I made things needlessly complicated by talking about non-ideal gasses. My question is equally valid for ideal gasses. Sorry for double-posting, but I can't figure out how to delete the old post.(adsbygoogle = window.adsbygoogle || []).push({});

If you do an energy balance for a control mass where there is no heat input, potential energy effects are negligible, and the only work interaction is work done by the systems on the surroundings through the expansion of the gas, you get

Q-W=dU+dKE+dPE

0-intPdV = dU + dKE + 0

0 = dU + int PdV + dKE

I see how if P is constant, the integral from state 1 to state 2 of PdV is

P*V(2)-P*V(1), and that

0 = P*V(2)-P*V(1) + U(2)-U(1) + KE(2) - KE(1)

which, since H=U+PV, reduces to

0 = H(2)-H(1) + KE(2)-KE(1)

But what happens when P is not constant? For example, when a gas escapes from a pressurized tank through a throttling valve? For this situation, if you look at some mass of air that escapes the tank, you can do an energy balance between state 1, when the air is motionless in the tank, and state 2, when the air is in motion and moving away from the valve. In this case P(1) does not equal P(2). Can you still use enthalpy the same way we did above?

Q-W=dU+dKE+dPE

0-intPdV = dU + dKE + 0

0 = dU + int PdV + dKE

can we say that dh=du+intPdV and write

0 = dH + dKE = H(2)-H(1) + KE(2) - KE(1)?

I don't understand the mathematics behind this, but it makes sense that you can look up enthalpy in tables and compare values at different pressures the same way you would under constant pressure.

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# Using enthalpy when pressure is not constant

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