For a perfectly crystalline solid `C_(p.m.)=aT^(3)`, where ais 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. 1.12 J/K mol
D. zero

Explanation

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

### Step 1: Understand the relationship between heat capacity and entropy
The molar entropy \( S \) can be calculated using the formula:
\[ S = \int_0^T \frac{C_{p.m.}}{T} dT \]
This means we need to integrate the heat capacity divided by temperature from 0 to the given temperature \( T \).

### Step 2: Substitute the expression for \( C_{p.m.} \)
Given that \( C_{p.m.} = aT^3 \), we can substitute this into the entropy formula:
\[ S = \int_0^T \frac{aT^3}{T} dT = \int_0^T aT^2 dT \]

### Step 3: Perform the integration
Now, we can integrate \( aT^2 \):
\[ S = a \int_0^T T^2 dT = a \left[ \frac{T^3}{3} \right]_0^T = a \frac{T^3}{3} \]

### Step 4: Calculate the value of \( a \)
We know that at \( T = 10 \, \text{K} \), \( C_{p.m.} = 0.42 \, \text{J/K-mol} \). Thus,
\[ 0.42 = a(10^3) \]
From this, we can solve for \( a \):
\[ a = \frac{0.42}{1000} = 0.00042 \, \text{J/K-mol-K}^4 \]

### Step 5: Substitute \( a \) back into the entropy equation
Now we can substitute \( a \) back into the entropy equation:
\[ S = 0.00042 \cdot \frac{(10)^3}{3} = 0.00042 \cdot \frac{1000}{3} \]

### Step 6: Calculate the molar entropy
Calculating this gives:
\[ S = 0.00042 \cdot 333.33 \approx 0.14 \, \text{J/K-mol} \]

### Final Answer
The molar entropy at 10 K is approximately:
\[ \boxed{0.14 \, \text{J/K-mol}} \]