Question

At a pressure P and temperature T, 7 gram of oxygen occupies a volume V. The equation of state will be (Molecular weight of Oxygen \(= 32\))

• A. \(PV = (\frac{2}{7})RT\)
• B. \(PV = (\frac{7}{2})RT\)
• C. \(PV = (\frac{7}{16})RT\)
• D. \(PV = (\frac{7}{32})RT\)

Hint:

Number of moles is mass divided by molar mass, so n = 7/32. Then PV = nRT.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The ideal gas equation for a gas of \(n\) moles is \(PV = nRT\). We need the number of moles in 7 g of oxygen.

Step 2: Key Formula or Approach:
\[ n = \frac{\text{mass}}{\text{molecular weight}} \]

Step 3: Detailed Explanation:
The molecular weight of oxygen is given as 32 g/mol.
\[ n = \frac{7}{32} \]
Substitute into the ideal gas equation:
\[ PV = \frac{7}{32}RT \]
Option (A) \(\tfrac27\) and (B) \(\tfrac72\) are not ratios of the mass and the molar mass. Option (C) \(\tfrac{7}{16}\) would result from dividing by the atomic weight 16 instead of the molecular weight 32, which is the standard slip when the gas is diatomic.

Final Answer:
The equation of state is \(PV = \dfrac{7}{32}RT\), option (D). \[ \boxed{PV=\frac{7}{32}RT \text{ (D)}} \]