`1` mole of an ideal diatomic gas undergoes a reversible polytropic process `(PV^(2)="constant")`. The gas expand from initial volume of `1` litre and temp `300 K` to final volume `3` lit. Claculate change in internal energy (approx).

Options

A. `-40 KJ`
B. `-4.2 KJ`
C. `-4.4 KJ`
D. `-4.6 KJ`

Explanation

To calculate the change in internal energy (ΔU) for the given process, we can follow these steps:

### Step 1: Identify the given data
- Number of moles (n) = 1 mole
- Initial volume (V1) = 1 L = 0.001 m³ (conversion to cubic meters for calculations)
- Final volume (V2) = 3 L = 0.003 m³
- Initial temperature (T1) = 300 K
- For a diatomic gas, the molar heat capacity at constant volume (C_v) = (5/2)R

### Step 2: Calculate the final temperature (T2)
Using the relation for a polytropic process, we have:
\[ P_1 V_1^n = P_2 V_2^n \]
Since \( PV = nRT \), we can express pressure in terms of temperature and volume:
\[ P_1 = \frac{nRT_1}{V_1} \]
\[ P_2 = \frac{nRT_2}{V_2} \]

Equating the two expressions:
\[ \frac{nRT_1}{V_1} V_1^n = \frac{nRT_2}{V_2} V_2^n \]

For a polytropic process where \( n = 2 \):
\[ T_1 V_1^{n-1} = T_2 V_2^{n-1} \]
\[ T_1 V_1 = T_2 V_2 \]

Now substituting the known values:
\[ 300 \times 1 = T_2 \times 3 \]
\[ T_2 = \frac{300}{3} = 100 \text{ K} \]

### Step 3: Calculate the change in temperature (ΔT)
\[ \Delta T = T_2 - T_1 = 100 - 300 = -200 \text{ K} \]

### Step 4: Calculate the change in internal energy (ΔU)
Using the formula for change in internal energy:
\[ \Delta U = n C_v \Delta T \]
Where \( C_v = \frac{5}{2}R \) and \( R = 8.314 \, \text{J/mol·K} \):
\[ C_v = \frac{5}{2} \times 8.314 = 20.785 \, \text{J/mol·K} \]

Now substituting the values:
\[ \Delta U = 1 \times 20.785 \times (-200) \]
\[ \Delta U = -4157 \, \text{J} \]

### Final Answer
The change in internal energy (ΔU) is approximately **-4157 J**.

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