Solve Isothermal Expansion: Find v2 Given P1, P2 & v1

AI Thread Summary
The discussion revolves around solving for the specific volume v2 in an isothermal expansion process given initial and final pressures (P1 and P2) and an initial specific volume (v1). Participants emphasize using the ideal gas law rather than the first law of thermodynamics, as the latter is deemed inappropriate for this scenario. There is confusion regarding obtaining a negative temperature when calculating v2, which is attributed to incorrect assumptions about the process. Ultimately, the correct answer for v2 is identified as 1.0 m^3, indicating a compression rather than an expansion. The conversation highlights the importance of correctly applying thermodynamic principles in solving such problems.
rmiller70015
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Homework Statement


Let P2 = 10^6 N/m^2, P1 = 4X10^5 N/m^2 and v = 2.5m^3/kmole Find the specific volume v2.

Homework Equations


Isothermal process, ideal gas. There is also a picture but it's just a generic P = constant/V plot.

The Attempt at a Solution


The volume given is the specific volume, so when using the ideal gas law I have to convert from kmoles to moles and make sure I have my n value in the correct place.
I've tried to use the first law of thermodynamics:
d'U = d'Q + d'W
d'U - d'W = d'Q
PdV = Q = nRTdV/V
I get stuck here because once I take the integral I get the v I am looking for, but I don't know what goes on the other side of the equal sign. I don't have a specific heat or an identity for this ideal gas and I'm not sure what to do with the Q.

I've also tried using the ideal gas law P1V1/n = P2V2/n
But I keep getting values of v2 that are smaller than v1
 
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You can't use the first law to solve this problem. The ideal gas law is the way to go. Why do you feel that getting values of v2 that are smaller than v1 is incorrect?
 
Chestermiller said:
You can't use the first law to solve this problem. The ideal gas law is the way to go. Why do you feel that getting values of v2 that are smaller than v1 is incorrect?
Mostly that when I use then in the next part of the problem I get a negative temperature in kelvin.
 
rmiller70015 said:
Mostly that when I use then in the next part of the problem I get a negative temperature in kelvin.
The system is isothermal, which you already accounted for with the ideal gas law. So how could the temperature come out negative?
 
Chestermiller said:
The system is isothermal, which you already accounted for with the ideal gas law. So how could the temperature come out negative?
Well I didn't include thus because it wasn't part of this part of the problem. The isotherm has a box drawn around it to represent a cyclic process. An isobaric expansion, an isobaric compression, an isochoric increase in pressure and an isochoric decrease in pressure. I'm looking at the isobaric expansion portion
 
rmiller70015 said:
Well I didn't include thus because it wasn't part of this part of the problem. The isotherm has a box drawn around it to represent a cyclic process. An isobaric expansion, an isobaric compression, an isochoric increase in pressure and an isochoric decrease in pressure. I'm looking at the isobaric expansion portion
Well then, all I can say is that the answer to the problem that you posed is 1.0 m^3. This is obviously a compression rather than an expansion.
 
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