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Standard XII

Physics

Real Gas Equation

Total pressur...

Question

Total pressure of a gaseous mixture containing $28 g N_{2}$ and $16 g O_{2}$ in a cylinder of volume $8.21 L$ at $300 K$ is:
[Given: $R = 0.0821 {L atm mol}^{- 1} K^{- 1}$ ]

3.5 atm

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2.5 atm

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4.5 atm

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5.5 atm

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Solution

The correct option is C $4.5 atm$
$P = \frac{n R T}{V} = \frac{(\frac{28}{28} + \frac{16}{32}) \times 0.0821 \times 300}{8.21} = 4.5 atm$

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Total pressure of a gaseous mixture containing 28 g N2 and 16 g O2 in a cylinder of volume 8.21 L at 300 K is: [Given: R=0.0821 L atm mol−1K−1]

The correct option is C 4.5 atmP=nRTV=(2828+1632)×0.0821×3008.21=4.5 atm

Total pressure of a gaseous mixture containing $28 g N_{2}$ and $16 g O_{2}$ in a cylinder of volume $8.21 L$ at $300 K$ is:
[Given: $R = 0.0821 {L atm mol}^{- 1} K^{- 1}$ ]

The correct option is C $4.5 atm$
$P = \frac{n R T}{V} = \frac{(\frac{28}{28} + \frac{16}{32}) \times 0.0821 \times 300}{8.21} = 4.5 atm$