The density of `O_(2)` gas at `25^(@)C` is 1.458 mg/lt at one atm pressure. At what pressure will `O_(2)` have the density twice the value?

Options

A. 0.5 atm
B. `2" atm/ "25^(@)C`
C. `4" atm/ "25^(@)C`
D. none of the above

Answer

To solve the problem, we need to determine the pressure at which the density of \( O_2 \) gas will be twice the given density at 1 atm pressure and 25°C.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Density of \( O_2 \) at 1 atm and 25°C = 1.458 mg/L
- We want to find the pressure at which the density is \( 2 \times 1.458 \) mg/L.

2. **Calculate the Target Density**:
\[ \text{Target Density} = 2 \times 1.458 \, \text{mg/L} = 2.916 \, \text{mg/L} \]

3. **Use the Ideal Gas Law**:
The ideal gas law is given by:
\[ PV = nRT \]
where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

4. **Relate Density to Pressure**:
The density \( d \) of a gas can be expressed in terms of its molar mass \( M \) and the ideal gas law:
\[ d = \frac{PM}{RT} \]
Rearranging gives:
\[ P = \frac{dRT}{M} \]

5. **Determine the Relationship Between Density and Pressure**:
Since \( d \) is directly proportional to \( P \) (at constant temperature and molar mass), we can write:
\[ \frac{P_2}{P_1} = \frac{d_2}{d_1} \]
where:
- \( P_1 = 1 \, \text{atm} \)
- \( d_1 = 1.458 \, \text{mg/L} \)
- \( d_2 = 2.916 \, \text{mg/L} \)

6. **Substituting the Values**:
\[ \frac{P_2}{1 \, \text{atm}} = \frac{2.916 \, \text{mg/L}}{1.458 \, \text{mg/L}} = 2 \]
Therefore,
\[ P_2 = 2 \times 1 \, \text{atm} = 2 \, \text{atm} \]

7. **Conclusion**:
The pressure at which the density of \( O_2 \) will be twice the original value is:
\[ \boxed{2 \, \text{atm}} \]