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anonymous
 3 years ago
On a pv diagram, there are constant temperature lines. If the process line is parallel to that of specific volume(that is, constant pressure inside the saturation dome), is there any change in internal energy?
anonymous
 3 years ago
On a pv diagram, there are constant temperature lines. If the process line is parallel to that of specific volume(that is, constant pressure inside the saturation dome), is there any change in internal energy?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1349069900920:dw

Fellowroot
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1349076248710:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks Fellowroot. Exactly what I was looking for... I didn't need help with the problem itself. I just didn't remember one of the concepts.

Fellowroot
 3 years ago
Best ResponseYou've already chosen the best response.1Could you explain it to me how your problem works since I'm really not getting the 1, 2, and 3 on your graph. Glad I could help. I actually took thermodynamics, but i don't know everything.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok. There is a saturated water vapor inside a pistoncylinder that is designed to stop once the volume is .3m^3. The initial pressure is 500kPa, and initial volume is .5m^3. Therefore, P1=P2 (constant pressure). On that pressure line, there is a constant temperature line? Therefore, no change in internal energy.

Fellowroot
 3 years ago
Best ResponseYou've already chosen the best response.1And you are meaning that the constant pressure line is from 1 to 2 right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That is correct. Isobaric process from 1 to 2.
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