Question:
The volume V of an enclosure contains a mixture of gases like 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. The pressure of the mixture of gases is (R is universal gas constant)
The volume V of an enclosure contains a mixture of gases like 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. The pressure of the mixture of gases is (R is universal gas constant)
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Total pressure depends on total number of moles, regardless of the gas type.
Updated On: Jul 4, 2026
- 3RT/V
- 4RT/V
- 5RT/2V
- 88RT/V
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The Correct Option is C
Solution and Explanation
Step 1: Concept
According to the ideal gas law for a mixture, total pressure $P = \frac{n_{total}RT}{V}$. Total moles $n_{total} = n_{O_{2}} + n_{N_{2}} + n_{CO_{2}}$.
Step 2: Meaning
Calculate moles: $n = \text{mass} / \text{molar mass}$. Molar masses are $O_{2} = 32$, $N_{2} = 28$, $CO_{2} = 44$.
Step 3: Analysis
$n_{O_{2}} = 16/32 = 0.5$. $n_{N_{2}} = 28/28 = 1$. $n_{CO_{2}} = 44/44 = 1$. Total moles $n = 0.5 + 1 + 1 = 2.5$.
Step 4: Conclusion
$P = \frac{2.5RT}{V} = \frac{5}{2} \frac{RT}{V}$. Final Answer: (C)
According to the ideal gas law for a mixture, total pressure $P = \frac{n_{total}RT}{V}$. Total moles $n_{total} = n_{O_{2}} + n_{N_{2}} + n_{CO_{2}}$.
Step 2: Meaning
Calculate moles: $n = \text{mass} / \text{molar mass}$. Molar masses are $O_{2} = 32$, $N_{2} = 28$, $CO_{2} = 44$.
Step 3: Analysis
$n_{O_{2}} = 16/32 = 0.5$. $n_{N_{2}} = 28/28 = 1$. $n_{CO_{2}} = 44/44 = 1$. Total moles $n = 0.5 + 1 + 1 = 2.5$.
Step 4: Conclusion
$P = \frac{2.5RT}{V} = \frac{5}{2} \frac{RT}{V}$. Final Answer: (C)
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