JEE Main 2026 — Apr 2, Shift 2 — Physics solution

JEE Main2026Apr 2, Shift 2Physics

A mixture of carbon dioxide and oxygen has volume $8310$ cm $^{3}$ , temperature $300$ K, pressure $100$ kPa and mass $13.2$ g. The number of moles of carbon dioxide and oxygen gases in the mixture respectively are_____. (Assume both carbondioxide and oxygen gases behave like ideal gases) [ $R = 8.31$ J/mol.K]

A
$0.15$ and $0.18$
B
$0.25$ and $0.08$
C
$0.21$ and $0.12$
D
$0.13$ and $0.20$

Solution

From $P V = n R T$ : $n_{total} = \frac{( 1 0 ^{5} ) ( 8.31 \times 1 0 ^{- 3} )}{( 8.31 ) ( 300 )} = \frac{1}{3} \approx 0.333$ mol.

Mass balance with molar masses $44$ for CO $_{2}$ and $32$ for O $_{2}$ : $44 n_{1} + 32 n_{2} = 13.2$ and $n_{1} + n_{2} = 0.333$ .

Substituting $n_{2} = 0.333 - n_{1}$ : $44 n_{1} + 32 (0.333 - n_{1}) = 13.2$ , giving $12 n_{1} = 2.533$ , so $n_{1} \approx 0.211$ mol.

Then $n_{2} = 0.333 - 0.211 \approx 0.122$ mol.

Therefore CO $_{2} \approx 0.21$ mol and O $_{2} \approx 0.12$ mol.

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