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a quantity of ideal gas undergoes an expansion that doubles its volume. is the work done by gas higher if the expansion is at constant pressure or at constant temperature? assume that the initial state is the same for both cases.?

Posted on December 14, 2022December 14, 2022 by ahmed mustafa
a quantity of ideal gas undergoes an expansion that doubles its volume. is the work done by gas higher if the expansion is at constant pressure or at constant temperature? assume that the initial state is the same for both cases.?

Welcome to our dear students on Boxnell website. Now we offer you a solution to this question. Work or work in thermodynamics is the amount of energy transferred from one system to another without being accompanied by a transfer to entropy.

The principle of work in thermodynamics is a generalization of the principle of physical work.

a quantity of ideal gas undergoes an expansion that doubles its volume. is the work done by gas higher if the expansion is at constant pressure or at constant temperature? assume that the initial state is the same for both cases.?

The answer to this was based on the graph of the two cases and comparison of the two pV graphs. Which gave a conclusion that W at constant temperature gives a greater value for the given case.

But when I tried to solve it mathematically, I got greater value of W at constant pressure which is plain wrong. I don’t understand where did I go wrong.

For constant pressure my result was:

W = p(2V−V) = nRTV(2V−V) = nRT.
For constant temperature my result was:
W = nRTln(2VV) = nRTln(2).
So, I got lesser value for constant temperature than constant pressure.

A quantity of ideal gas undergoes an expansion that doubles its volume. Does the gas do more work on its surroundings if the expansion is at constant pressure or at constant temperature?

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