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Question

Three moles of an ideal gas (CV,m=12.5J K1 mol1) are at 300 K and 5 dm3. If the gas is heated to 320 K and the volume changed to 10 dm3, calculate the entropy change.


Solution

The correct option is D


We know that
ΔS =qT
& we know, from first law,
ΔU=q+ω
So, now, if we take a look at the question, temperature and volume is changing.
So, do you remember entropy change as a function of T and V?
If you don't lets derive it.
To calculate ΔsysS for a reversible process,
As per definition: ΔS=dqrevT
T dS = dqrev
or T dS = dU + (-dw) [ From the first law of thermodynamics]
or T dS = nCVdT + P dV [dU = nCVdT and dw = -P dV]
or T ds = nCVdT+nRTVdV [P=nRTV]
dS=nCVdTT+nRVdV
ΔsysS=n lnT2T1(CV+R)+nR lnP1P2
ΔsysS=nCp lnT2T1+nR lnP1P2
Here,
You can see entropy change in terms of T, V and T, P. Here, we need it as a function of T and V as CV is constant.
Therefore,
ΔS =nCV,mlnT2T1+nR lnV2V1
ΔS =(3mol)×(12.5 J K1mol1)×2.303
log 320300+(3mol)×(8.314 J K1mol1)×2.303 log105
=2.42 J K1+17.29 J K1=19.71 J K1

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