2 mole of an ideal monoatomic gas undergoes a reversible process for which `PV^(2)=C`. The gas is expanded from initial volume of 1 L to final volume of 3 L starting from initial temperature of 300 K. Find `DeltaH` for the process :

Options

A. `-600` R
B. `-1000` R
C. `-3000` R
D. None of these

Explanation

To find the change in enthalpy (ΔH) for the given process, we will follow these steps:

### Step 1: Understand the process
We have 2 moles of an ideal monoatomic gas undergoing a reversible process defined by the equation \( PV^2 = C \). The gas expands from an initial volume \( V_1 = 1 \, \text{L} \) to a final volume \( V_2 = 3 \, \text{L} \) starting from an initial temperature \( T_1 = 300 \, \text{K} \).

### Step 2: Determine the relationship between pressures and volumes
From the equation \( PV^2 = C \), we can say that:
\[ P_1 V_1^2 = P_2 V_2^2 \]
This implies:
\[ \frac{P_1}{P_2} = \frac{V_2^2}{V_1^2} \]

### Step 3: Use the ideal gas law
For an ideal gas, we know:
\[ PV = nRT \]
This can be rearranged to relate pressures and temperatures:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
From this, we can express the ratio of temperatures:
\[ \frac{T_2}{T_1} = \frac{P_2 V_2}{P_1 V_1} \]

### Step 4: Substitute the pressure ratio
Substituting the expression for \( \frac{P_1}{P_2} \) into the temperature ratio:
\[ \frac{T_2}{T_1} = \frac{V_2}{V_1} \cdot \frac{V_2^2}{V_1^2} = \frac{V_2^3}{V_1^3} \]
Substituting the values \( V_1 = 1 \, \text{L} \) and \( V_2 = 3 \, \text{L} \):
\[ \frac{T_2}{T_1} = \frac{3^3}{1^3} = 27 \]
Thus:
\[ T_2 = 27 \cdot T_1 = 27 \cdot 300 \, \text{K} = 8100 \, \text{K} \]

### Step 5: Calculate the change in temperature
Now, we can calculate the change in temperature:
\[ \Delta T = T_2 - T_1 = 8100 \, \text{K} - 300 \, \text{K} = 7800 \, \text{K} \]

### Step 6: Calculate \( C_P \) for a monoatomic gas
For a monoatomic ideal gas:
\[ C_V = \frac{3R}{2}, \quad C_P = C_V + R = \frac{3R}{2} + R = \frac{5R}{2} \]

### Step 7: Calculate \( \Delta H \)
Using the formula for change in enthalpy:
\[ \Delta H = n C_P \Delta T \]
Substituting the values:
\[ \Delta H = 2 \cdot \frac{5R}{2} \cdot 7800 = 5R \cdot 7800 = 39000R \]

### Final Answer
Thus, the change in enthalpy \( \Delta H \) for the process is:
\[ \Delta H = 39000R \]

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