Arturo O. answered • 11/20/17
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For an ideal monatomic gas,
U(T) = (3/2)nRT ⇒
ΔU = (3/2)nRΔT
T = absolute temperature
R = universal gas constant
n = number of moles
ΔT = TB - TA = PBVB/(nR) - PAVA/(nR)
ΔU = (3/2)nRΔT = (3/2)nR [PBVB/(nR) - PAVA/(nR)] = (3/2)(PBVB - PAVA)
Plug in the correct PB, VB, PA, and VA. You can finish from here.
Arturo O.
For those values of pressure and volume, yes.
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11/20/17
Sav J.
Can you explain to me why the middle changes in the graph does take part to effect this problem in anyway? The pressure raising and lowering doesn't matter, only the starting and ending point?
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11/20/17
Arturo O.
If the amount of gas is constant, the internal energy of an ideal gas depends only on the temperature of the gas. That is one of the properties of an ideal gas. So the change in internal energy depends only on the change in temperature, and the change in temperature is just the difference between the final and initial temperatures. In addition, recall from thermodynamics that internal energy is an example of a state function. The change in a state function depends only on the initial and final conditions. It does not depend on the path that was followed from one state to the other. An example of a function that is not a state function is work. The work done in going from state A to state B depends on the path from A to B. But the change in the internal energy of an ideal going from from state A to state B depends only on the temperatures of states A and B, as long as the amount of gas is constant.
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11/20/17
Sav J.
Oh I see. I'm remembering a lot of these rules from chemistry now. Thank you for all the help!
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11/20/17
Arturo O.
You are welcome, Sav.
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11/20/17
Sav J.
11/20/17