Monday: Derivation of Isothermal, Reversible Expansion


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Hai-Lin Yeh 1J
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Monday: Derivation of Isothermal, Reversible Expansion

Postby Hai-Lin Yeh 1J » Tue Feb 05, 2019 9:39 pm

In Monday's Lecture Notes, after talking about entropy, the slide moved onto deriving the equation for isothermal, reversible expansion. There were a bunch of equations after "need to know". Can someone explain the importance of those equations, especially ∆U = 3/2nR∆T = 0. Also, what does isothermal, reversible expansion have to do with entropy, considering we talked about entropy before getting the equation or is there no relevance?

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Re: Monday: Derivation of Isothermal, Reversible Expansion

Postby 2c_britneyly » Tue Feb 05, 2019 11:46 pm

I don't think it has anything to do with entropy, this is just calculating work with different conditions.

The basis of deriving a new equation is due to the slowly decreasing external pressure when the system is expanding. Because P is not constant, we derive a new P by rearranging PV=nRT to P=nRT/V. The original equation of work is w=-PΔV, so replacing P we get w=-(nRT/V)ΔV, and when we integrate, we end up with w=-nRT*ln(V2/V1).

Iona Pay
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Re: Monday: Derivation of Isothermal, Reversible Expansion

Postby Iona Pay » Wed Feb 06, 2019 12:06 am

Isothermal (temperature of system remains constant, so change in temperature is 0), reversible (can be reversed by an infinitesimally small change in the system). In an isothermal expansion of a gas, pressure falls as the gas expands because temperature remains constant (no extra energy is added to maintain pressure). However, if we want to make this isothermal expansion reversible, we must make sure external pressure of the environs and internal pressure of the gas are the same at every stage. So, we must gradually reduce pressure of the environs as the pressure of the system drops - meaning the system must do a great deal of work.

There's a good explanation of this mathematically on page 248 in the 7th edition if you're looking for the derivation itself!

∆U = 3/2nR∆T = 0 is used to find the change in molar internal energy of a monatomic gas at a constant volume. This relies on the equipartition theorem (4B in the 7th edition). It's somewhat complicated, and I think page 265-266 of the 7th edition does a better job at explaining it than I can.

I don't recall being asked to relate the two, but you certainly could if you wished.

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