- #1
apb000
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This is probably a trivial problem, but I can't get it straight. It's well known and easy enough to show mathematically that the change in enthalpy for the reversible isothermal expansion of an ideal gas is always zero, ΔH=0. But it's also true that the enthalpy change for a process at constant P is equal to the heat that enters (or leaves) the system, ΔH=qP. Fine.
So if I take a cylinder with a piston and place it in a heat bath, and pull the piston out in a slow, reversible fashion, I can allow that the temperature remains constant and the process is isothermal; it follows that ΔH = 0. Um...but didn't heat cross the boundary of the system? In fact doesn't it HAVE to, to keep the temperature constant? The whole process was expansion against a constant Pressure, so ΔH should equal the heat that entered the cylinder, a quantity definitely not zero.
It seems to me that with ΔT=0 the internal energy is not changing, so q=-w, and here w=P dV. Again, this is a non-zero quantity.
How do I reconcile this?
(Note: Please don't derive for me that ΔH=0 from the ideal gas law. I know, and I'm convinced. But that doesn't explain why the reasoning with the cylinder is wrong.)
So if I take a cylinder with a piston and place it in a heat bath, and pull the piston out in a slow, reversible fashion, I can allow that the temperature remains constant and the process is isothermal; it follows that ΔH = 0. Um...but didn't heat cross the boundary of the system? In fact doesn't it HAVE to, to keep the temperature constant? The whole process was expansion against a constant Pressure, so ΔH should equal the heat that entered the cylinder, a quantity definitely not zero.
It seems to me that with ΔT=0 the internal energy is not changing, so q=-w, and here w=P dV. Again, this is a non-zero quantity.
How do I reconcile this?
(Note: Please don't derive for me that ΔH=0 from the ideal gas law. I know, and I'm convinced. But that doesn't explain why the reasoning with the cylinder is wrong.)