1 mole of an ideal gas A (`C_(v,m)=3R`) and 2 mole of an ideal gas B are `(C_(v.m)= (3)/(2)R)` taken in a constainer and expanded reversible and adiabatically from 1 litre of 4 litre starting from initial temperature of 320K. `DeltaE or DeltaU` for the process is (in Cal)

Answer

To solve the problem, we will calculate the change in internal energy (ΔU) for the given process of the two ideal gases A and B during an adiabatic expansion.

### Step-by-Step Solution:

1. **Identify Given Data:**
- For gas A:
- Moles (n_A) = 1 mole
- \( C_{v,m} = 3R \)
- For gas B:
- Moles (n_B) = 2 moles
- \( C_{v,m} = \frac{3}{2}R \)
- Initial Volume (V1) = 1 L
- Final Volume (V2) = 4 L
- Initial Temperature (T1) = 320 K

2. **Calculate the Degree of Freedom (f):**
- For gas A (monatomic): \( f_A = 6 \) (since \( C_{v,m} = \frac{f}{2}R \))
- For gas B (diatomic): \( f_B = 3 \) (since \( C_{v,m} = \frac{f}{2}R \))
- Total degree of freedom (F) can be calculated as:
\[ F = \frac{f_A \cdot n_A + f_B \cdot n_B}{n_A + n_B} = \frac{6 \cdot 1 + 3 \cdot 2}{1 + 2} = \frac{6 + 6}{3} = 4 \]

3. **Calculate the Heat Capacity Ratio (γ):**
- \( \gamma = \frac{C_p}{C_v} = \frac{f + 2}{f} \)
- For total degrees of freedom (F = 4):
\[ \gamma = \frac{2 + 4}{4} = \frac{3}{2} \]

4. **Calculate Final Temperature (T2) using Adiabatic Condition:**
- Using the relation for adiabatic processes:
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]
- Substitute the values:
\[ T_2 = 320 \left( \frac{1}{4} \right)^{\frac{3}{2} - 1} = 320 \left( \frac{1}{4} \right)^{\frac{1}{2}} = 320 \cdot \frac{1}{2} = 160 \, K \]

5. **Calculate Change in Internal Energy (ΔU):**
- For gas A:
\[ \Delta U_A = n_A C_{v,m} (T_2 - T_1) = 1 \cdot 3R \cdot (160 - 320) = 1 \cdot 3R \cdot (-160) = -480R \]
- For gas B:
\[ \Delta U_B = n_B C_{v,m} (T_2 - T_1) = 2 \cdot \frac{3}{2}R \cdot (160 - 320) = 2 \cdot \frac{3}{2}R \cdot (-160) = -480R \]

6. **Total Change in Internal Energy (ΔU):**
\[ \Delta U = \Delta U_A + \Delta U_B = -480R - 480R = -960R \]

7. **Convert ΔU to Calories:**
- Given that \( R = 2 \, \text{cal/K·mol} \):
\[ \Delta U = -960R = -960 \cdot 2 = -1920 \, \text{cal} \]

### Final Answer:
The change in internal energy (ΔU) for the process is **-1920 cal**.