The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be

Options

A. `PV = (5)/(2) RT`
B. `PV = (5)/(16) RT`
C. `PV = (5)/(32) RT`
D. `PV = 5 RT`

Explanation

To derive the equation of state for 5 g of oxygen at a pressure \( P \), temperature \( T \), and volume \( V \), we will use the ideal gas law, which is given by the equation:

\[ PV = nRT \]

where:
- \( P \) = pressure,
- \( V \) = volume,
- \( n \) = number of moles,
- \( R \) = universal gas constant,
- \( T \) = temperature.

### Step 1: Calculate the number of moles of oxygen
To find the number of moles \( n \), we use the formula:

\[ n = \frac{\text{mass}}{\text{molar mass}} \]

Given that the mass of oxygen is 5 g, we need to find the molar mass of oxygen. Oxygen exists as a diatomic molecule (\( O_2 \)), and its molar mass is:

\[ \text{Molar mass of } O_2 = 16 \, \text{g/mol} \times 2 = 32 \, \text{g/mol} \]

Now we can calculate the number of moles:

\[ n = \frac{5 \, \text{g}}{32 \, \text{g/mol}} = \frac{5}{32} \, \text{mol} \]

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

\[ PV = nRT \]

Substituting \( n = \frac{5}{32} \):

\[ PV = \left(\frac{5}{32}\right)RT \]

### Step 3: Rearranging the equation
We can rearrange the equation to express it in the standard form:

\[ PV = \frac{5}{32}RT \]

This is the equation of state for the given conditions of 5 g of oxygen at pressure \( P \), temperature \( T \), and volume \( V \).

### Conclusion
Thus, the final equation of state is:

\[ PV = \frac{5}{32}RT \]