Question

for an isobaric process the ratio of delta q to the delta w is

Answer 1

In a reversible isobaric process, you are adding heat to the gas Q and it is expanding at constant pressure P to do reversible work on its surroundings. So:

W=PΔV

ΔU=Q−W=Q−PΔV

But, from the ideal gas law, PV=nRT, so

W=nRΔT(1)

and

ΔU=Q−nRΔT

So,

Q=ΔU+nRΔT

But for an ideal gas:

ΔU=nCvΔT(2)

So,

Q=nCvΔT+nRΔT=nR(Cv+R)ΔT=nCpΔT(3)

Therefore, from Eons. 1-3, we have:

Q:ΔU:W=nCpΔT:nCvΔT:nRΔT