The molar volume of dinitrogen at 273.15 K and 1 atm pressure is 22.4 L . The molar volume of argon , a mono atomic gas at 273. 15 K and 1 atm present is

Options

A. 44.8 L
B. 22.4 L
C. 11.2 L
D. 5.6 L

Explanation

To solve the problem of finding the molar volume of argon at 273.15 K and 1 atm pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Molar Volume**: Molar volume is defined as the volume occupied by one mole of a substance at a given temperature and pressure. For ideal gases, this volume is often the same under the same conditions of temperature and pressure.

2. **Given Data**:
- Molar volume of dinitrogen (N₂) at 273.15 K and 1 atm = 22.4 L
- Temperature (T) = 273.15 K
- Pressure (P) = 1 atm
- Number of moles (n) = 1 (since we are considering the molar volume)

3. **Using Ideal Gas Law**: The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \(P\) = pressure
- \(V\) = volume
- \(n\) = number of moles
- \(R\) = ideal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature in Kelvin

4. **Calculate Volume for Argon**: Since we are looking for the molar volume of argon under the same conditions (T and P are the same), we can conclude that:
- If the temperature and pressure are the same for both gases, and since argon is also an ideal gas, its molar volume will also be the same as that of dinitrogen.

5. **Conclusion**: Therefore, the molar volume of argon at 273.15 K and 1 atm is also 22.4 L.

### Final Answer:
The molar volume of argon at 273.15 K and 1 atm is **22.4 L**.

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