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October 21, 2014

Homework Help: physics

Posted by tmaria on Friday, June 22, 2007 at 3:04pm.

When 1180 J of heat are added to one mole of an ideal monatomic gas, its temperature increases from 272 to 292 K. Find the work done by the gas during this process.

I know the specific heat is Q/delta T, so C = (1180 J)/(20K) = 59. This seems like any easy problem, but now I'm stuck on what to do next.

Hmmm. It seems to me you are fishing for answers. Your bait of Q/deltaT is not good enough. Recheck your text.

For a given mass,
Heat in = Work out + delta U
delta U is the change in internal energy. You can calculate it as
Delta U = Cv delta T
where Cv is the specific heat at constant volume. This formula is valid even though your problem may involve a changing volume.
In your monatomic case, Cv = 3/2 R
R = 8.317 J/ mole K is the universal gas constant.
Solve for Work out

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