JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Heat & Thermodynamics

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✓ Correct
(D)0.13 and 0.20

Answer: (C)

Solution

The total number of moles follows from

P V = n R T

n = \frac{P V}{R T} = \frac{1 0 ^{5} \times 8310 \times 1 0 ^{- 6}}{8.31 \times 300} = \frac{1}{3} mol

. Writing

n_{1}

and

n_{2}

for the moles of

C O_{2}

and

O_{2}

n_{1} + n_{2} = \frac{1}{3}

while the mass gives

44 n_{1} + 32 n_{2} = 13.2

. Substituting

n_{2} = \frac{1}{3} - n_{1}

yields

44 n_{1} + 32 (\frac{1}{3} - n_{1}) = 13.2

, so

12 n_{1} = 13.2 - \frac{32}{3} = 2.53

and

n_{1} = 0.21 mol

, whence

n_{2} = 0.12 mol

. The mixture contains

0.21

mol of carbon dioxide and

0.12

mol of oxygen.

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Heat & Thermodynamics

Solution

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