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

0.920 J/K mol

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8.66 J/K mol

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0.813 J/K mol

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None of these

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Solution

The correct option is C $0.813 J/K mol$
As data given:
$C_{p} a t 10 K = 0.40 J / K m o l$
$0.40 = a T^{3} + b T$
$0.40 = a (10)^{3} + b (10) (1)$
$C_{p} a t 20 K = 0.92 J / K m o l$
$0.92 = a (20)^{3} + b (20) (2)$
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
$Δ S = S_{T (K)} - S_{0 (K)} = \int_{0}^{20} \frac{n C_{p} \times d T}{T}$
$. . . S_{0 (K)} E n t r o p y a t a b s o l u t e t e m p = 0$
$Δ S = \int_{0}^{20} \frac{(a T^{3} + b T)}{T} d T$
$Δ S = \frac{a T^{3}}{3} + b T |_{0}^{20} = 0.813 J / K m o l$

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For a perfectly crystalline solid Cp.m=aT3+bT, where a and b are constants. If Cp.m is 0.40 J/Kmol at 10 K and 0.92 J/K mol at 20K, then the molar entropy at 20 K is:

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