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 two ideal gases A and B, we will follow these steps:

### Step 1: Identify the 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,m,A) = (5/2)R
- Molar heat capacity at constant volume for gas B (C_v,m,B) = 3R
- Initial temperature (T_initial) = 350 K
- Final temperature (T_final) = 250 K

### Step 2: Calculate the heat capacity at constant pressure (C_p) for both gases
Using the relation \( C_p = C_v + R \):
- For gas A:
\[ C_{p,m,A} = C_{v,m,A} + R = \frac{5}{2}R + R = \frac{7}{2}R \]
- For gas B:
\[ C_{p,m,B} = C_{v,m,B} + R = 3R + R = 4R \]

### Step 3: Calculate the change in temperature (ΔT)
\[ \Delta T = T_{final} - T_{initial} = 250 K - 350 K = -100 K \]

### Step 4: Calculate the change in enthalpy (ΔH) for each gas
Using the formula for change in enthalpy:
\[ \Delta H = n C_p \Delta T \]
- For gas A:
\[ \Delta H_A = n_A C_{p,m,A} \Delta T = 0.5 \times \frac{7}{2}R \times (-100) = -\frac{350}{2}R = -175R \]
- For gas B:
\[ \Delta H_B = n_B C_{p,m,B} \Delta T = 0.5 \times 4R \times (-100) = -200R \]

### Step 5: Calculate the total change in enthalpy (ΔH_total)
\[ \Delta H_{total} = \Delta H_A + \Delta H_B = -175R - 200R = -375R \]

### Step 6: Convert ΔH to calories
Since \( R = 2 \) cal/(mol·K) (using the value of R in cal):
\[ \Delta H_{total} = -375 \times 2 = -750 \text{ cal} \]

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