0.5 mole each of two ideal gases`A (C_(v,m)=(5)/(2)R)` and `B (C_(v,m)=3R)` are taken in a container and expanded reversibly and adiabatically, during this process temperature of gaseous mixture decreased from 350 K to 250 K. Find `DeltaH` (in cal/mol) for the process :

Options

A. `-100 R`
B. `-137.5 R`
C. `-375 R`
D. None of these

Explanation

To find the change in enthalpy (ΔH) for the process involving the two ideal gases A and B, we can follow these steps:

### Step 1: Identify Given Data
- Moles of gas A (n_A) = 0.5 moles
- Moles of gas B (n_B) = 0.5 moles
- Molar heat capacity at constant volume for gas A (C_v,mA) = \( \frac{5}{2} R \)
- Molar heat capacity at constant volume for gas B (C_v,mB) = \( 3R \)
- Initial temperature (T1) = 350 K
- Final temperature (T2) = 250 K

### Step 2: Calculate Change in Temperature (ΔT)
\[ \Delta T = T_2 - T_1 = 250 \, \text{K} - 350 \, \text{K} = -100 \, \text{K} \]

### Step 3: Calculate Molar Heat Capacity at Constant Pressure (C_p)
Using the relation \( C_p = C_v + R \):
- For gas A:
\[ C_{pA} = C_{vA} + R = \frac{5}{2}R + R = \frac{7}{2}R \]
- For gas B:
\[ C_{pB} = C_{vB} + R = 3R + R = 4R \]

### Step 4: Calculate Change in Enthalpy (ΔH) for Each Gas
Using the formula \( \Delta H = n C_p \Delta T \):
- For gas A:
\[ \Delta H_A = n_A \cdot C_{pA} \cdot \Delta T = 0.5 \cdot \frac{7}{2}R \cdot (-100) = -175R \]
- For gas B:
\[ \Delta H_B = n_B \cdot C_{pB} \cdot \Delta T = 0.5 \cdot 4R \cdot (-100) = -200R \]

### Step 5: Total Change in Enthalpy (ΔH)
\[ \Delta H = \Delta H_A + \Delta H_B = -175R - 200R = -375R \]

### Step 6: Convert ΔH to Calories
Since \( R \) in calories is approximately \( 2 \, \text{cal/mol K} \):
\[ \Delta H = -375R \approx -375 \cdot 2 \, \text{cal/mol} = -750 \, \text{cal/mol} \]

### Final Answer
The change in enthalpy (ΔH) for the process is:
\[ \Delta H = -750 \, \text{cal/mol} \]

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