Posted by tmaria on .
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,
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=QdeltaU, W=118J249.515J=930.49J, the workdone is 930.49J, any question call me +2348065434920 for clearification