The equation of state for 15 gram of oxygen at pressure P, volume V and temperature T is given by

Options

A. PV = 15 RT
B. `PV = (15RT)/2`
C. `PV = (15)/(32)RT`
D. PV = RT

Explanation

To derive the equation of state for 15 grams of oxygen at pressure \( P \), volume \( V \), and temperature \( T \), we will use the ideal gas equation, which is given by:

\[ PV = nRT \]

where:
- \( P \) = pressure of the gas
- \( V \) = volume of the gas
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant (approximately \( 8.314 \, \text{J/(mol K)} \))
- \( T \) = temperature in Kelvin

### Step 1: Calculate the number of moles of oxygen
The molecular weight of oxygen (O₂) is approximately \( 32 \, \text{g/mol} \) (since each oxygen atom has a mass of about \( 16 \, \text{g/mol} \), and there are two atoms in an O₂ molecule).

To find the number of moles \( n \) in 15 grams of oxygen, we use the formula:

\[ n = \frac{\text{mass}}{\text{molecular weight}} = \frac{15 \, \text{g}}{32 \, \text{g/mol}} \]

Calculating this gives:

\[ n = \frac{15}{32} \, \text{mol} \]

### Step 2: Substitute \( n \) into the ideal gas equation
Now that we have the number of moles, we can substitute \( n \) back into the ideal gas equation:

\[ PV = nRT \]

Substituting \( n = \frac{15}{32} \):

\[ PV = \left(\frac{15}{32}\right)RT \]

### Step 3: Rearranging the equation
To express the equation of state for the gas, we can rearrange the equation to isolate \( P \):

\[ P = \frac{15RT}{32V} \]

This is the equation of state for 15 grams of oxygen at pressure \( P \), volume \( V \), and temperature \( T \).

### Final Answer
The equation of state for 15 grams of oxygen is:

\[ P = \frac{15RT}{32V} \]