MediumChemistryMiddle Stage (6–8)Chemical Thermodynamics › Systems, Work, Heat and the First LawNumericalJEE Main
A sample of 2.00 mol of an ideal gas at 300 K expands from 10.0 L to 40.0 L, while its temperature…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A sample of of an ideal gas at expands from to , while its temperature stays at . Use the NCERT sign convention: is the work done on the gas and the heat given to it, so . Take .
- What is if the expansion is reversible?
- What is if, instead, the gas expands in one step against a constant external pressure equal to its final pressure?
- What is if the gas expands freely into a vacuum?
- What is for the reversible expansion?
- What is the entropy change of the gas, which is the same for all three ways?
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. The temperature is fixed, so for an ideal gas in every case. Only the work, and so the heat, depends on the way.
(a) Reversible.
(b) One step. The final pressure is (about ). Then
(c) Into a vacuum. Nothing pushes back: , so .
(d) Heat. , so : the gas takes in from its surroundings all the energy it gives out as work.
(e) Entropy. . Entropy is a state function, so the same start and end give the same whichever way the gas goes.
Check. Adding up along the isotherm in two hundred thousand small steps gives the same work as (a) to within . The reversible way gives the most work, the free expansion none.
Watch out. In NCERT's convention, work done by the gas makes negative. Do not drop the minus sign.
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