Amount of work done in an isothermal expansion

[pingo]

Homework Statement

500 Cal are added to a gas inside a cylinder with a piston (containing one mole) by an external heating device. The volume of the gas doubles without any change in it's temperature of 300K. How much work is done on the piston?

Homework Equations

The Attempt at a Solution

W = -nRT ln(2Vi/Vi)

where ln is the natural logarithm

W = -nRT ln(2)
W = -(1)(1.987)(300) ln(2) = -413.2 Cal

R = 1.987 as the problem is in Calories not Joules.

As it's an isothermal expansion I thought this would be correct but the answer given in the textbook is as follows:

W = -500 - (1)(R)(300) ln2 = -500 - (1)(1.987)(300) ln2 = -913.185 Cal

If 500 Cal is added to the gas and the gas expands isothermally why are there 2 terms in the solution? Basically I don't understand why we need to include the -500 Cal here.

 

[rude man]

Seems like by the 1st law the answer should be 500 cal:

dU = δQ - δW
W = work done by piston = -work done on piston
But dU = 0 for isothermal process for ideal gas.

So U2 - U1 = Q - W = 0
W = Q = 500 cal. = work done by piston
So work done on piston = -500 cal.

You can't double the volume AND stipulate exactly 500 cal. heat flow into the cylinder in a reversible process for an ideal gas. In fact, the heat transferred would be 413 cal., not 500. So this is an irreversible process OR it's not an ideal gas, but if it isn't an ideal gas then the problem can't be solved.

The 1st law is correct for any kind of process since it's a statement of the conservation of energy. But an ideal gas is still assumed if we stipulate dU = 0 which I have done.

Anyway, both our answers are a lot closer to each other than the given answer.

Refutations welcome!

 

[hjelmgart]

Maybe you are forgetting, that the energy added comes from an external source. The system then does work on its surroundings in order to keep the total internal energy constant. Thus you have to substract this external energy, which is the 500 cal.

 

[rude man]

Maybe you are forgetting, that the energy added comes from an external source. The system then does work on its surroundings in order to keep the total internal energy constant. Thus you have to substract this external energy, which is the 500 cal.

I do not understand your reasoning.

 

[Chestermiller]

Hi Pingo. Welcome to Physics Forums.

The way your book did the problem makes no sense whatsoever. In addition, your analysis and Rude Man's rationale make total sense, and are flawless. Congratulations for have the guts to say that "the emperor has no clothes."

Chet

 

[hjelmgart]

I do not understand your reasoning.

The total work is the change in kinetic energy of the system. Here we have the heat transferred from outside, and we have the internal change in kinetic energy done by the gas on the piston.

The internal kinetic energy is not zero, as the volume expands.

Well, that is what they calculated in the actual "answer". Now when I read it one more time, I see they are actually asking for the work done on the piston, so you are of course correct, my bad. :-)