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: Count the moles:
One mole of $\text{O}_2$ weighs 32 g. For 7 g, the fraction of a mole is $\frac{7}{32}$.

Step 2: Use the gas law:
$PV = nRT$ gives $PV = \frac{7}{32}RT$.

Step 3: Why not 16:
Oxygen gas is $\text{O}_2$, so the mass of one mole is 32 g. Using 16 g would give the number of moles of oxygen atoms, not molecules.

Step 4: Check the units:
With $P$ in pascal, $V$ in cubic metre and $R = 8.314$ J/(mol K), the right side $\frac{7}{32}RT$ has the units of energy, the same as $PV$. Using 16 instead of 32 would give the number of oxygen atoms, which is not what the equation of state needs.

Final Answer:
Option (D). \[ \boxed{PV=\frac{7}{32}RT \text{ (D)}} \]