State the equation corresponding to `8g` of `O_(2)` is

Options

A. `pV = 8RT`
B. `pV = (RT)/(4)`
C. `pV = RT`
D. `pV = (RT)/(2)`

Answer

To find the equation corresponding to 8 grams of \( O_2 \), we can follow these steps:

### Step 1: Calculate the number of moles of \( O_2 \)
The molar mass of \( O_2 \) (oxygen) is approximately 32 g/mol (16 g/mol for each oxygen atom).

To find the number of moles (\( n \)) of \( O_2 \) in 8 grams, we use the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ n = \frac{8 \, \text{g}}{32 \, \text{g/mol}} = \frac{1}{4} \, \text{mol} \]

### Step 2: Use the Ideal Gas Equation
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant (approximately \( 0.0821 \, \text{L atm/(K mol)} \))
- \( T \) = temperature in Kelvin

### Step 3: Substitute the number of moles into the equation
From Step 1, we found that \( n = \frac{1}{4} \, \text{mol} \). Now we substitute this value into the ideal gas equation:
\[ PV = \left(\frac{1}{4}\right)RT \]

### Step 4: Rearranging the equation
We can rearrange the equation to express it in a different form:
\[ PV = \frac{RT}{4} \]

### Final Equation
Thus, the equation corresponding to 8 grams of \( O_2 \) is:
\[ PV = \frac{RT}{4} \]