Question

# Three moles of an ideal gas (CV,m=12.5J K−1 mol−1) 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 K−1mol−1)×2.303 log 320300+(3mol)×(8.314 J K−1mol−1)×2.303 log105 =2.42 J K−1+17.29 J K−1=19.71 J K−1

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