|Feb23-12, 03:16 AM||#1|
Thermodynamics - Internal Energy in Isochoric and Isobaric Processes
Assuming that the gas in these processes is an ideal one, is the change in internal energy in an isochoric process (volume remaining constant) the same as the change in internal energy of isobaric process (pressure remaining constant)? Mathematically I can derive that they're equivalent, but what's the physical reason behind that? Does it have something to do with the temperature of the gas? My reasoning is this: if you start from the a point on a PV diagram (I'll call it A) and move up in an isochoric process to a point B, and then starting from A if you move right to a point C in an isobaric process, you end up at an isotherm that both A and C lie on if you add equal amounts of heat Q in both AC and AB, and that's why their changes in internal energy are the same. Is my reasoning correct?
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|Feb23-12, 05:34 AM||#2|
In an ideal gas, internal energy, U, depends only on temperature. For n moles of ideal gas, U = ncvT
in which cv is the ‘molar heat capacity at constant volume’.
So ΔU will be the same in an isochoric or isobaric change, provided that ΔT is the same.
What you've said is good up to "you end up at an isotherm that both A and C lie on". After that you go adrift. In the isobaric process you have to put in more heat for the same temperature change than in the isochoric change. The reason for this is that in the isobaric change the gas does external work (pushing out a piston or whatever) and this, as well as the rise in internal energy, has to be paid for as heat. For the isochoric change there's no external work done, so only ΔU has to be paid for by heat.
|Feb23-12, 10:20 PM||#3|
Oh all right that makes sense. Thanks!
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