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 container and expanded reversible and adiabatically from 1 litre of 4 litre starting from initial temperature of 320 K. `DeltaE ` or `DeltaU` for the process is :

Options

A. `240` R
B. -240 R
C. 480 R
D. `-960` R

Explanation

To solve the problem, we need to calculate the change in internal energy (ΔU) for the given adiabatic expansion of two ideal gases. Here’s the step-by-step solution:

### Step 1: Identify the Given Information
- For gas A:
- Number of moles (nA) = 1 mole
- Molar heat capacity at constant volume (Cv,m) = 3R
- For gas B:
- Number of moles (nB) = 2 moles
- Molar heat capacity at constant volume (Cv,m) = (3/2)R
- Initial volume (V1) = 1 L
- Final volume (V2) = 4 L
- Initial temperature (T1) = 320 K

### Step 2: Calculate the Degrees of Freedom
Using the formula for degrees of freedom (F):
- For gas A:
\[ F_A = \frac{2C_{v,A}}{R} = \frac{2 \times 3R}{R} = 6 \]
- For gas B:
\[ F_B = \frac{2C_{v,B}}{R} = \frac{2 \times \frac{3}{2}R}{R} = 3 \]

### Step 3: Calculate the Average Degree of Freedom
Using the formula for average degrees of freedom:
\[ F_{avg} = \frac{n_A F_A + n_B F_B}{n_A + n_B} \]
Substituting the values:
\[ F_{avg} = \frac{1 \times 6 + 2 \times 3}{1 + 2} = \frac{6 + 6}{3} = 4 \]

### Step 4: Calculate the Value of γ (Gamma)
Using the formula:
\[ \gamma = 1 + \frac{2}{F_{avg}} = 1 + \frac{2}{4} = 1.5 = \frac{3}{2} \]

### Step 5: Calculate the Final Temperature (T2)
Using the adiabatic relation:
\[ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \]
Substituting the known values:
\[ \frac{T_2}{320} = \left(\frac{1}{4}\right)^{\frac{1}{2}} = \frac{1}{2} \]
Thus,
\[ T_2 = 320 \times \frac{1}{2} = 160 \text{ K} \]

### Step 6: Calculate the Change in Internal Energy (ΔU)
Using the formula:
\[ \Delta U = n_A C_{v,A} \Delta T + n_B C_{v,B} \Delta T \]
Where ΔT = T2 - T1:
\[ \Delta T = 160 - 320 = -160 \text{ K} \]
Now substituting the values:
\[ \Delta U = 1 \times 3R \times (-160) + 2 \times \frac{3}{2}R \times (-160) \]
Calculating each term:
\[ = -480R + (-480R) = -960R \]

### Final Answer
Thus, the change in internal energy (ΔU) for the process is:
\[ \Delta U = -960R \]

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