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

- physics -
**frank fred**, Sunday, March 18, 2012 at 2:18pm
to solve for work done, cv=3/2R=3/2*8.317=12.4755, deltaU=cv*deltaT =12.4755*20=249.515, Q=W+deltaU, W=Q-deltaU, W=118J-249.515J=930.49J, the workdone is 930.49J, any question call me +2348065434920 for clearification

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