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 the perfectly crystalline solid with the given heat capacity function \( C_{p,m} = aT^3 + bT \), we will follow these steps:

### Step 1: Set up the equations based on the given data
We know that:
- \( C_{p,m} = aT^3 + bT \)

At \( T = 10 \, \text{K} \):
\[ C_{p,m} = 0.40 \, \text{J/K mol} \implies 0.40 = a(10^3) + b(10) \]
This gives us our first equation:
\[ 0.40 = 1000a + 10b \quad \text{(Equation 1)} \]

At \( T = 20 \, \text{K} \):
\[ C_{p,m} = 0.92 \, \text{J/K mol} \implies 0.92 = a(20^3) + b(20) \]
This gives us our second equation:
\[ 0.92 = 8000a + 20b \quad \text{(Equation 2)} \]

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

To eliminate \( b \), we can multiply Equation 1 by 2:
\[ 0.80 = 2000a + 20b \quad \text{(Equation 3)} \]

Now, subtract Equation 2 from Equation 3:
\[ (0.80 - 0.92) = (2000a + 20b) - (8000a + 20b) \]
This simplifies to:
\[ -0.12 = -6000a \]
Solving for \( a \):
\[ a = \frac{0.12}{6000} = 2 \times 10^{-5} \, \text{J/K}^4 \text{mol} \]

### Step 3: Substitute \( a \) back to find \( b \)
Now substitute \( a \) back into Equation 1:
\[ 0.40 = 1000(2 \times 10^{-5}) + 10b \]
Calculating:
\[ 0.40 = 0.02 + 10b \]
Subtract \( 0.02 \) from both sides:
\[ 0.38 = 10b \implies b = \frac{0.38}{10} = 0.038 \, \text{J/K mol} \]

### Step 4: Calculate the molar entropy at 20 K
The molar entropy \( S \) can be calculated using the formula:
\[ S = \int_0^T \frac{C_{p,m}}{T} dT \]
Substituting \( C_{p,m} = aT^3 + bT \):
\[ S = \int_0^{20} \frac{aT^3 + bT}{T} dT = \int_0^{20} (aT^2 + b) dT \]
This can be integrated as:
\[ S = \left[ \frac{aT^3}{3} + bT \right]_0^{20} \]
Substituting \( a \) and \( b \):
\[ S = \left[ \frac{(2 \times 10^{-5}) (20^3)}{3} + (0.038)(20) \right] \]
Calculating:
\[ = \left[ \frac{(2 \times 10^{-5}) (8000)}{3} + 0.76 \right] = \left[ \frac{0.16}{3} + 0.76 \right] = \left[ 0.0533 + 0.76 \right] = 0.813 \, \text{J/K mol} \]

### Final Answer
The molar entropy at 20 K is \( S = 0.813 \, \text{J/K mol} \).

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