There is a standard value of temperature and pressure at which the molar volume of a gas is 22.4 L . The correct values are (A) 273 K ,1atm (B)300 K , 760 mm (C)25 C ∘ , 760 mm (D)373 K , 1 atm

Options

A. 273 K ,1atm
B. 300 K , 760 mm
C. `25 C^(@)` , 760 mm
D. 373 K , 1 atm

Answer

To solve the question regarding the standard values of temperature and pressure at which the molar volume of a gas is 22.4 L, we will analyze each option given.

### Step-by-step Solution:

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

2. **Identify Molar Volume**:
The molar volume of a gas at standard temperature and pressure (STP) is 22.4 L for 1 mole of an ideal gas.

3. **Evaluate Each Option**:
We will calculate the molar volume for each option provided.

**Option A: 273 K, 1 atm**
- Here, \( n = 1 \) mole, \( R = 0.0821 \) L·atm/(K·mol), \( T = 273 \) K, and \( P = 1 \) atm.
\[ V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 273}{1} = 22.4 \, \text{L} \]
This is correct.

**Option B: 300 K, 760 mmHg**
- Convert 760 mmHg to atm: \( 760 \, \text{mmHg} = 1 \, \text{atm} \).
- Here, \( T = 300 \) K, \( P = 1 \) atm.
\[ V = \frac{1 \times 0.0821 \times 300}{1} = 24.63 \, \text{L} \]
This is incorrect.

**Option C: 25°C, 760 mmHg**
- Convert 25°C to Kelvin: \( 25 + 273 = 298 \) K.
- Convert 760 mmHg to atm: \( 760 \, \text{mmHg} = 1 \, \text{atm} \).
\[ V = \frac{1 \times 0.0821 \times 298}{1} = 24.45 \, \text{L} \]
This is incorrect.

**Option D: 373 K, 1 atm**
- Here, \( T = 373 \) K, \( P = 1 \) atm.
\[ V = \frac{1 \times 0.0821 \times 373}{1} = 30.62 \, \text{L} \]
This is incorrect.

4. **Conclusion**:
The only option that gives the molar volume of 22.4 L is **Option A: 273 K, 1 atm**.

### Final Answer:
The correct values are **(A) 273 K, 1 atm**.

---