To solve the problem step by step, we will follow the outlined approach in the video transcript.
### Step 1: Identify Given Data
- \( C_{v,m} = \frac{5}{2} R \)
- Initial volume \( V_1 = 1 \, \text{L} \)
- Final volume \( V_2 = 32 \, \text{L} \)
- Initial temperature \( T_1 = 327^\circ C = 600 \, \text{K} \) (after converting to Kelvin)
- \( R = 2 \, \text{Cal K}^{-1} \text{mol}^{-1} \)
### Step 2: Calculate \( C_{p,m} \)
Using the relation \( C_p = C_v + R \):
\[
C_{p,m} = C_{v,m} + R = \frac{5}{2} R + R = \frac{5}{2} R + \frac{2}{2} R = \frac{7}{2} R
\]
Substituting \( R = 2 \, \text{Cal K}^{-1} \text{mol}^{-1} \):
\[
C_{p,m} = \frac{7}{2} \times 2 = 7 \, \text{Cal K}^{-1} \text{mol}^{-1}
\]
### Step 3: Determine \( \gamma \)
Using the relation \( \gamma = \frac{C_p}{C_v} \):
\[
\gamma = \frac{C_{p,m}}{C_{v,m}} = \frac{\frac{7}{2} R}{\frac{5}{2} R} = \frac{7}{5}
\]
### Step 4: Use the Adiabatic Relation
For a reversible adiabatic process, we have:
\[
T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}
\]
Rearranging gives:
\[
T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1}
\]
Substituting the values:
\[
T_2 = 600 \left( \frac{1}{32} \right)^{\frac{7}{5} - 1} = 600 \left( \frac{1}{32} \right)^{\frac{2}{5}}
\]
### Step 5: Calculate \( T_2 \)
Calculating \( \left( \frac{1}{32} \right)^{\frac{2}{5}} \):
\[
\left( \frac{1}{32} \right)^{\frac{2}{5}} = \frac{1}{32^{\frac{2}{5}}} = \frac{1}{4} = 0.25
\]
Thus,
\[
T_2 = 600 \times 0.25 = 150 \, \text{K}
\]
### Step 6: Calculate \( \Delta T \)
Now, calculate \( \Delta T \):
\[
\Delta T = T_2 - T_1 = 150 - 600 = -450 \, \text{K}
\]
### Step 7: Calculate \( \Delta H \)
Using the formula for enthalpy change:
\[
\Delta H = C_{p,m} \Delta T = 7 \, \text{Cal K}^{-1} \text{mol}^{-1} \times (-450 \, \text{K}) = -3150 \, \text{Cal}
\]
Since we need the magnitude:
\[
|\Delta H| = 3150 \, \text{Cal}
\]
### Final Answer
The magnitude of molar enthalpy change for the process is \( \boxed{3150} \, \text{Cal} \).
