Question

At 300 K temperature 20 g Hz, 220 g CO2 and 140 g N, are filled in a
vessel having volume 2L. find the total pressure in bar unit and which
gas is removed from the vessel so that pressure can be reduced by 50%.​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

The total pressure = 249.44 bar

Explanation:

Given,

$T = 300 K, V = 2 L\\m_{H_{2} } = 20 g, m_{CO_{2} } = 220 g, m_{N_{2} } = 140 g$

No. of moles of $H_{2} = \frac{20}{2} = 10$ moles

No. of moles of $CO_{2} = \frac{220}{44} = 5$ moles

No. of moles of $N_{2} = \frac{140}{28} = 5$ moles

So, according to Dalton's law of partial pressure,

Pressure ∝ Mole fraction of $H_{2}$

​So, as the mole fraction is 0.5, when hydrogen is removed half of the pressure is removed.

We know that,

$PV = nRT$

Total number of moles, n = 10 + 5 + 5 = 20 moles

Gas constant, R = 0.08206 $\frac{L-atm}{mole} K$

$P$×$2 = 20$×$0.08206$×$300$

$P = 246.18 atm$ = 246.18×1.01325 = 249.44 bar