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

Answer

To find the molar entropy at 20 K for a perfectly crystalline solid with the heat capacity given by \( C_{p,m} = a T^3 + b T \), we can follow these steps:

### Step 1: Set up the equations based on the given values of \( C_{p,m} \)

We know:
- \( C_{p,m} = a T^3 + b T \)

Given:
- \( C_{p,m} = 0.40 \, \text{J/K mol} \) at \( T = 10 \, \text{K} \)
- \( C_{p,m} = 0.92 \, \text{J/K mol} \) at \( T = 20 \, \text{K} \)

From these, we can form two equations:

1. At \( T = 10 \, \text{K} \):
\[ 0.40 = a (10)^3 + b (10) \implies 0.40 = 1000a + 10b \quad \text{(Equation 1)} \]

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

### Step 2: Solve the equations simultaneously

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

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

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

### Step 3: Substitute \( a \) back to find \( b \)

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

### Step 4: Calculate the molar entropy at 20 K

The molar entropy \( S \) can be calculated using the integral:
\[ S = \int_0^T \frac{C_{p,m}}{T} dT = \int_0^{20} \frac{a T^3 + b T}{T} dT \]
This simplifies to:
\[ S = \int_0^{20} (a T^2 + b) dT \]
Calculating the integral:
\[ S = \left[ \frac{a T^3}{3} + b T \right]_0^{20} \]
Substituting \( a = 2 \times 10^{-5} \) and \( b = 0.038 \):
\[ S = \left[ \frac{2 \times 10^{-5} (20)^3}{3} + 0.038 (20) \right] \]
Calculating each term:
\[ \frac{2 \times 10^{-5} \times 8000}{3} + 0.76 = \frac{0.16}{3} + 0.76 = 0.0533 + 0.76 = 0.8133 \, \text{J/K mol} \]

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