Verified Solution • 2025–26 Session moderate

QJEEM21Ph9e76: JEE Main 2021 Physics - Class 11 JEEM21Ph9e76 Solution: The volume V of an enclosure contains a mixture of three gases, …

The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is :

Multiple Choice Options

  • A 3RTV
  • B 4RTV
  • C 88RTV
  • D 52RTV

Step-by-Step Verified Solution

No. of moles of O2 :
n1 = 1632 = 0.5 mole

No. of moles of N2 :
n2 = 2828 = 1 mole

No. of moles of CO2 :
n3 = 4444 = 1 mole

Total no. of moles in container : n = n1 + n2 + n3

∴ n = 0.5 + 1 + 1 = 52 moles

Now; PV = nRT

⇒ P = nRTV

⇒ P = 52RTV

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