Entropy of an ideal gas Consider an ideal gas obeying the state equation PV = nR

Question

Entropy of an ideal gas Consider an ideal gas obeying the state equation PV = nRT, with the change in the internal energy with respect to temperature given by dU = (3/2)nRdT. With this information determine dQ, using the first law of thermodynamics for reversible processes. Integrate dQ/T to determine the change in the entropy of the system if the initial temperature is T_0 and the final temperature is T_F. Similarly V_0 and V_F are defined. Verify that or confirm that 5 is a state function just by observation of the calculated result.

Explanation / Answer

First law of thermodynamics

dQ=dU+PdV

given dU=(3/2)nRdT

So

dQ=(3/2)nRdT+PdV

since PV=nRT so P=nRT/V

hence equation changes to

dQ=(3/2)nRdT+nRT(dV/V)

Divide by T gives

dQ/T=(3/2)nR(dT/T)+nR(dV/V)

Now integration of dQ/T=S gives

S=(3/2)nR*ln(Tf/T0)+nR*ln(Vf/V0)

S=nR(1.5*ln(Tf/T0)+ln(Vf/V0))

Thus S is a function T and V not the path

thus S is a state function .

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