The density of `O_(2)` is 16 at STP. At what temperature (in `.^@C`) its density will be 14 ? Consider that the pressure remais constant.

Answer

To solve the problem, we will use the relationship between density, temperature, and pressure for a gas. The formula we will use is derived from the ideal gas law, which can be expressed in terms of density:

\[ \frac{\rho_1}{T_1} = \frac{\rho_2}{T_2} \]

Where:
- \(\rho_1\) = initial density
- \(T_1\) = initial temperature (in Kelvin)
- \(\rho_2\) = final density
- \(T_2\) = final temperature (in Kelvin)

### Step 1: Identify the known values
- Given:
- \(\rho_1 = 16\) (density at STP)
- \(T_1 = 273 \, \text{K}\) (standard temperature)
- \(\rho_2 = 14\) (density we want to find the temperature for)

### Step 2: Rearrange the formula to solve for \(T_2\)
From the relationship, we can rearrange the formula to find \(T_2\):

\[ T_2 = \frac{\rho_2 \cdot T_1}{\rho_1} \]

### Step 3: Substitute the known values into the equation
Now, we will substitute the known values into the equation:

\[ T_2 = \frac{14 \cdot 273}{16} \]

### Step 4: Calculate \(T_2\)
Now we perform the calculation:

\[ T_2 = \frac{3822}{16} = 238.875 \, \text{K} \]

### Step 5: Convert \(T_2\) from Kelvin to Celsius
To convert from Kelvin to Celsius, we subtract 273 from the Kelvin temperature:

\[ T_2 (\text{in } ^\circ C) = T_2 (\text{in K}) - 273 = 238.875 - 273 = -34.125 \, ^\circ C \]

### Final Answer
Thus, the temperature at which the density of oxygen will be 14 is approximately:

\[ T_2 \approx -34.13 \, ^\circ C \]