The pressure of 2 mole of ideal gas at 546 K having volume 44.8 L is

Options

A. 2 atm
B. 3 atm
C. 4 atm
D. 1 atm

Explanation

To find the pressure of 2 moles of an ideal gas at 546 K with a volume of 44.8 L, we can use the Ideal Gas Law, which is expressed as:

\[ PV = nRT \]

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

### Step-by-Step Solution:

1. **Identify the values given in the problem:**
- Number of moles (\( n \)) = 2 moles
- Temperature (\( T \)) = 546 K
- Volume (\( V \)) = 44.8 L

2. **Select the appropriate value for the gas constant (\( R \)):**
- Since we want the pressure in atm, we use \( R = 0.0821 \, \text{atm} \cdot \text{L} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \).

3. **Rearrange the Ideal Gas Law to solve for pressure (\( P \)):**
\[ P = \frac{nRT}{V} \]

4. **Substitute the known values into the equation:**
\[ P = \frac{(2 \, \text{moles}) \times (0.0821 \, \text{atm} \cdot \text{L} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}) \times (546 \, \text{K})}{44.8 \, \text{L}} \]

5. **Calculate the numerator:**
\[ 2 \times 0.0821 \times 546 = 89.82612 \, \text{atm} \cdot \text{L} \]

6. **Now divide by the volume (44.8 L):**
\[ P = \frac{89.82612 \, \text{atm} \cdot \text{L}}{44.8 \, \text{L}} \approx 2.00 \, \text{atm} \]

7. **Final Answer:**
The pressure of the gas is approximately **2 atm**.