A sample of argon gas at `1atm` pressure and `27^(@)C` expands reversibly and adiabatically from `1.25 dm^(3)` to `2.50 dm^(3)`. Calculate the enthalpy change in this process. `C_(vm)` for orgon is `12.48J K^(-1) mol^(-1)`.

Explanation

To solve the problem of calculating the enthalpy change for a sample of argon gas expanding reversibly and adiabatically, we will follow these steps:

### Step 1: Calculate the number of moles of argon gas (n)
Using the ideal gas equation:
\[ PV = nRT \]
Where:
- \( P = 1 \, \text{atm} = 101.325 \, \text{kPa} \) (for consistency, we will convert to kPa)
- \( V = 1.25 \, \text{dm}^3 = 1.25 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 27^\circ C = 300 \, \text{K} \)

Rearranging the equation to find \( n \):
\[ n = \frac{PV}{RT} \]

Substituting the values:
\[ n = \frac{(1 \, \text{atm}) \times (1.25 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (300 \, \text{K})} \]
Calculating:
\[ n = \frac{1.25}{24.63} \approx 0.05075 \, \text{mol} \]

### Step 2: Determine the final temperature (T2) after adiabatic expansion
Using the relation for adiabatic processes:
\[ \frac{T_1}{T_2} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}} \]
Where:
- \( T_1 = 300 \, \text{K} \)
- \( P_1 = \frac{nRT_1}{V_1} \) and \( P_2 = \frac{nRT_1}{V_2} \)

Calculating \( P_1 \) and \( P_2 \):
\[ P_1 = \frac{(0.05075 \, \text{mol}) \times (0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (300 \, \text{K})}{1.25 \, \text{L}} \approx 1 \, \text{atm} \]
\[ P_2 = \frac{(0.05075 \, \text{mol}) \times (0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (300 \, \text{K})}{2.50 \, \text{L}} \approx 0.5 \, \text{atm} \]

Now substituting into the adiabatic relation:
\[ \frac{300}{T_2} = \left(\frac{0.5}{1}\right)^{\frac{1.66 - 1}{1.66}} \approx 0.66 \]
Thus:
\[ T_2 = \frac{300}{0.66} \approx 188.55 \, \text{K} \]

### Step 3: Calculate \( C_p \)
Using the relation:
\[ C_p = C_v + R \]
Where:
- \( C_v = 12.48 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \)

Calculating:
\[ C_p = 12.48 + 8.314 = 20.794 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 4: Calculate the enthalpy change (\( \Delta H \))
Using the formula:
\[ \Delta H = n C_p \Delta T \]
Where:
- \( \Delta T = T_1 - T_2 = 300 - 188.55 = 111.45 \, \text{K} \)

Substituting the values:
\[ \Delta H = 0.05075 \times 20.794 \times 111.45 \]
Calculating:
\[ \Delta H \approx 0.05075 \times 20.794 \times 111.45 \approx 117.6 \, \text{J} \]

### Final Answer:
The enthalpy change in this process is approximately \( 117.6 \, \text{J} \).

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