For a perfectly crystalline solid `C_(p.m) = aT^(3)`, where a is constant. If `C_(p.m)` is 0.42 J/K `-` mol at 10K, molar entropy at 10K is

Options

A. 0.42 J/K-mol
B. 0.14 J/K-mol
C. 4.2 J/K-mol
D. zero

Answer

To find the molar entropy \( S_m \) at 10 K for a perfectly crystalline solid with the given heat capacity \( C_{p,m} = aT^3 \), where \( a \) is a constant, we can follow these steps:

### Step 1: Identify the given values
- \( C_{p,m} = 0.42 \, \text{J/K} \cdot \text{mol} \) at \( T = 10 \, \text{K} \)

### Step 2: Express \( C_{p,m} \) in terms of \( a \)
From the equation \( C_{p,m} = aT^3 \), we can substitute \( T = 10 \, \text{K} \):

\[ C_{p,m} = a \cdot (10)^3 \]

### Step 3: Solve for \( a \)
Substituting the value of \( C_{p,m} \):

\[ 0.42 = a \cdot 1000 \]

\[ a = \frac{0.42}{1000} = 0.00042 \, \text{J/K} \cdot \text{mol} \]

### Step 4: Set up the integral for molar entropy
The molar entropy \( S_m \) can be calculated using the formula:

\[ S_m = \int_0^{T} \frac{C_{p,m}}{T} \, dT \]

Substituting \( C_{p,m} = aT^3 \):

\[ S_m = \int_0^{10} \frac{aT^3}{T} \, dT = \int_0^{10} aT^2 \, dT \]

### Step 5: Calculate the integral
Now we can evaluate the integral:

\[ S_m = a \int_0^{10} T^2 \, dT \]

The integral of \( T^2 \) is:

\[ \int T^2 \, dT = \frac{T^3}{3} \]

So,

\[ S_m = a \left[ \frac{T^3}{3} \right]_0^{10} = a \cdot \frac{10^3}{3} = a \cdot \frac{1000}{3} \]

### Step 6: Substitute the value of \( a \)
Now substituting \( a = 0.00042 \):

\[ S_m = 0.00042 \cdot \frac{1000}{3} = \frac{0.42}{3} = 0.14 \, \text{J/K} \cdot \text{mol} \]

### Final Answer
The molar entropy at 10 K is:

\[ S_m = 0.14 \, \text{J/K} \cdot \text{mol} \]

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