Q.20 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 = 0.25 PV. R and V are universal gas constant and ideal gas volume, respectively.
The molecule A is
(A) CO2
(B) H2
(C) N2
(D) O2
RT = 0.25 PV. This deviates from the standard ideal gas law PV = nRT, indicating a modified behavior tied to molar mass.
Calculate Molar Mass
Start with the given condition: 128 g in V = 1 L at STP. The relation RT = 0.25 PV rearranges to PV = 4 RT. Compare to ideal gas law PV = (m/M) RT, where m = 128 g.
Equating gives (128 / M) RT = 4 RT, so 128 / M = 4, hence M = 128 / 4 = 32 g/mol.
✅ Correct Answer: O2 (Option D)
Molar mass of 32 g/mol matches O2 (oxygen, 16×2=32). Precise derivation: From PV = (m RT)/M for ideal gas, but given PV = 4 RT, so m/M = 4, M = 128/4 = 32 g/mol exactly.
Option Analysis
- CO2 (44 g/mol): Too heavy; 128g would be ~2.9 mol, occupying ~65 L at STP, not 1 L.
- H2 (2 g/mol): Too light; 128g =64 mol, volume ~1434 L, far exceeds 1 L.
- N2 (28 g/mol): 128g ≈4.57 mol, volume ~102 L, doesn’t fit.
- O2 (32 g/mol): 128g =4 mol exactly. Matches
PV/RT = 4 = nfrom the given relation.
Why This Matters
This tests gas law modifications in exams like GATE/NEET. The key phrase “ideal gas 128g 1L RT=0.25PV” solves to O2 by deriving M=32 g/mol directly from equating the given relation with the ideal gas law.
The modified equation RT = 0.25 PV effectively means the gas behaves as if it has 4 times the number of moles than expected under standard conditions for that mass and volume.
