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Ideal Gas Equation

12 g He and 4...

Question

$12 g$ $He$ and $4 g H_{2}$ filled in a container of volume $20 L$ maintained at temperature $300 K$ . The pressure of the mixture is nearly

3.225 atm

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5.225 atm

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6.225 atm

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1.225 atm

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Solution

The correct option is C $6.225 atm$
We know that, ideal gas equation is,
$P V = n R T = \frac{m}{M} R T$
For $He$ ,
$P_{He} = \frac{12}{4 \times 20} \times 0.083 \times 300 = 3.735 atm$
Similarly,
For $H_{2}$ ,
$P_{H_{2}} = \frac{4}{2 \times 20} \times 0.083 \times 300 = 2.490 atm$
So, Total pressure of non-reactive gas,
$P = P_{He} + P_{H_{2}} = 3.735 + 2.490 = 6.225 atm$

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Ideal Gas Equation

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12 g He and 4 g H2 filled in a container of volume 20 L maintained at temperature 300 K. The pressure of the mixture is nearly

The correct option is C 6.225 atmWe know that, ideal gas equation is, PV=nRT=mMRT For He, PHe=124×20×0.083×300=3.735 atm Similarly, For H2, PH2=42×20×0.083×300=2.490 atm So, Total pressure of non-reactive gas, P=PHe+PH2=3.735+2.490=6.225 atm

$12 g$ $He$ and $4 g H_{2}$ filled in a container of volume $20 L$ maintained at temperature $300 K$ . The pressure of the mixture is nearly