Question

For a perfectly crystalline solid $C_{p.m}=a{T}^3+bT$, where a and b are constants. If $C_{p.m}$ is $0.40\text{J/Kmol}$ at 10 K and $0.92\text{J/K mol}$ at 20K, then the molar entropy at 20 K is:

• A. $0.920\text{J/K mol}$
• B. $8.66\text{J/K mol}$
• C. $0.813\text{J/K mol}$
• D. None of these

Answer 1

The correct option is C $0.813\text{J/K mol}$

As data given:
$C_{p}at10K=0.40J/Kmol$
$0.40=a{T}^3+bT$
$0.40=a(10{)}^3+b(10\left)\right(1)$
$C_{p}at20K=0.92J/Kmol$
$0.92=a(20{)}^3+b(20\left)\right(2)$
On solving 1 and 2 we get
$a=2\times {10}^{-5}$
$b=0.038$
The entropy of perfect crystals can be calculated at any given temperature by
$\Delta S=S_{T\left(K\right)}-S_{0\left(K\right)}={\int }_0^{20}\frac{n{C}_p\times dT}{T}$
$\stackrel{.}{..}{S}_{0\left(K\right)}Entropyatabsolutetemp=0$
$\Delta S={\int }_0^{20}\frac{(a{T}^3+bT)}{T}dT$
$\Delta S=\frac{a{T}^3}{3}+bT{|}_0^{20}=0.813J/Kmol$