The approximate temperature at which 1 mol `L^(-1)` of a sample of pure ideal gas exhibits a pressure of 101.325 k Pa is

Options

A. 12.2 K
B. 122 K
C. 244 K
D. 300 K

Explanation

To find the approximate temperature at which 1 mol `L^(-1)` of a sample of pure ideal gas exhibits a pressure of 101.325 kPa, we can use the ideal gas equation:

**Ideal Gas Equation:**
\[ PV = nRT \]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature (in Kelvin)

### Step 1: Convert Pressure to atm
The pressure given is 101.325 kPa. We need to convert this to atm:
\[ 1 \text{ atm} = 101.325 \text{ kPa} \]
Thus,
\[ P = \frac{101.325 \text{ kPa}}{101.325 \text{ kPa/atm}} = 1 \text{ atm} \]

### Step 2: Identify the Volume and Number of Moles
The volume \( V \) is given as 1 L, and the number of moles \( n \) is given as 1 mol.

### Step 3: Use the Ideal Gas Constant
The value of the universal gas constant \( R \) is:
\[ R = 0.0821 \text{ atm L mol}^{-1} \text{ K}^{-1} \]

### Step 4: Rearrange the Ideal Gas Equation to Solve for Temperature
We can rearrange the ideal gas equation to solve for temperature \( T \):
\[ T = \frac{PV}{nR} \]

### Step 5: Substitute the Values into the Equation
Now, substituting the values we have:
\[ T = \frac{(1 \text{ atm})(1 \text{ L})}{(1 \text{ mol})(0.0821 \text{ atm L mol}^{-1} \text{ K}^{-1})} \]

### Step 6: Calculate the Temperature
Calculating the above expression:
\[ T = \frac{1}{0.0821} \]
\[ T \approx 12.2 \text{ K} \]

### Final Answer
The approximate temperature at which 1 mol `L^(-1)` of a sample of pure ideal gas exhibits a pressure of 101.325 kPa is approximately **12.2 K**.

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