For a perfectly crystalline solid `C_(p,m)=aT^(3)+bT`, where a and b are constant. If `C_(p,m)` is `0.40` J/K mol at 10 K and `0.92` J/K mol at 20 K, then molar entropy at 20 K is :

Options

A. `0.92` J/K mol
B. `8.66` J/K mol
C. `0.813` J/K mol
D. None of these

Explanation

To find the molar entropy at 20 K for a perfectly crystalline solid with a specific heat capacity given by \( C_{p,m} = aT^3 + bT \), where \( a \) and \( b \) are constants, we will follow these steps:

### Step 1: Set up the equations using the given data
We know:
- \( C_{p,m} = 0.40 \, \text{J/K mol} \) at \( T = 10 \, \text{K} \)
- \( C_{p,m} = 0.92 \, \text{J/K mol} \) at \( T = 20 \, \text{K} \)

Using the formula \( C_{p,m} = aT^3 + bT \):
1. For \( T = 10 \, \text{K} \):
\[ 0.40 = a(10^3) + b(10) \implies 0.40 = 1000a + 10b \quad \text{(Equation 1)} \]

2. For \( T = 20 \, \text{K} \):
\[ 0.92 = a(20^3) + b(20) \implies 0.92 = 8000a + 20b \quad \text{(Equation 2)} \]

### Step 2: Solve the equations for \( a \) and \( b \)
We have two equations:
1. \( 1000a + 10b = 0.40 \)
2. \( 8000a + 20b = 0.92 \)

To eliminate \( b \), we can manipulate the equations. Multiply Equation 1 by 2:
\[ 2(1000a + 10b) = 2(0.40) \implies 2000a + 20b = 0.80 \quad \text{(Equation 3)} \]

Now subtract Equation 3 from Equation 2:
\[ (8000a + 20b) - (2000a + 20b) = 0.92 - 0.80 \]
This simplifies to:
\[ 6000a = 0.12 \implies a = \frac{0.12}{6000} = 0.00002 = 2 \times 10^{-5} \]

Now substitute \( a \) back into Equation 1 to find \( b \):
\[ 1000(2 \times 10^{-5}) + 10b = 0.40 \]
\[ 0.02 + 10b = 0.40 \implies 10b = 0.40 - 0.02 = 0.38 \implies b = 0.038 \]

### Step 3: Calculate the molar entropy \( S_m \) at 20 K
The molar entropy change from 0 K to 20 K is given by:
\[ \Delta S_m = \int_{0}^{20} \frac{C_{p,m}}{T} dT \]
Substituting \( C_{p,m} \):
\[ \Delta S_m = \int_{0}^{20} \frac{aT^3 + bT}{T} dT = \int_{0}^{20} (aT^2 + b) dT \]
This can be separated into two integrals:
\[ \Delta S_m = \int_{0}^{20} aT^2 dT + \int_{0}^{20} b \, dT \]

Calculating each integral:
1. For \( aT^2 \):
\[ \int_{0}^{20} aT^2 dT = a \left[ \frac{T^3}{3} \right]_{0}^{20} = a \cdot \frac{20^3}{3} = 2 \times 10^{-5} \cdot \frac{8000}{3} = \frac{160000}{3} \times 10^{-5} = \frac{1.6}{3} \times 10^{-2} \approx 0.00533 \]

2. For \( b \):
\[ \int_{0}^{20} b \, dT = b \cdot T \bigg|_{0}^{20} = b \cdot 20 = 0.038 \cdot 20 = 0.76 \]

Adding both contributions:
\[ \Delta S_m = 0.00533 + 0.76 \approx 0.76533 \, \text{J/K mol} \]

### Final Answer
The molar entropy at 20 K is approximately:
\[ \Delta S_m \approx 0.813 \, \text{J/K mol} \]