Question

Under standard temperature (T) and pressure (P) conditions, 128 g of an ideal gas molecule A occupies a volume of 1 L. The gas molecule A obeys the relationship $RT = 0.25PV$. R and V are universal gas constant and ideal gas volume, respectively. The molecule A is:

• A. CO$_2$
• B. H$_2$
• C. N$_2$
• D. O$_2$

Hint:

When gases obey modified gas equations, adjust the molar volume first, then compare mass versus effective molar volume to identify the molecule.

Answer 1

The Correct Option is D

At STP, 1 mole of any ideal gas occupies 22.4 L. Here 128 g of gas occupies only 1 L, meaning it is much heavier than typical gases.

Step 1: Use the modified gas equation.
Given: \[ RT = 0.25PV $\Rightarrow$ PV = 4RT. \] Thus the gas behaves as if 1 mole occupies: \[ V = 4 \times 22.4 = 89.6 \text{ L per mole}. \]

Step 2: Find effective molar mass.
If 128 g occupies 1 L, then:
\[ \text{Moles} = \frac{1}{89.6} $\Rightarrow$ \frac{128}{89.6} = 1.428. \] Thus molar mass ≈ 89.6 g/mol. Closest diatomic gas with molecular mass near 32 g/mol fits scaled relation → O₂.

Step 3: Conclusion.
The only viable option is O₂.