$14$ g of nitrogen gas and $7$ g of hydrogen gas is present in a closed vessel at $27^{0}$ C. If the volume of the container is $8.2$ L then the pressure of the gas mixture is:

6

atm

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12

atm

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8

atm

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10

atm

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Solution

The correct option is C $12$ atm
Moles of $N_{2} = \frac{14}{28} = 0.5$
Moles of $H_{2} = \frac{7}{2} = 3.5$
$∴ n = 0.5 + 3.5 = 4.0$ moles
$P V = n R T$
$\Rightarrow P = \frac{n R T}{V} = \frac{4 \times 0.0821 \times 300}{8.2}$
$= 12.01$ atm

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14g of nitrogen gas and 7 g of hydrogen gas is present in a closed vessel at 270C. If the volume of the container is 8.2L then the pressure of the gas mixture is:

The correct option is C 12 atmMoles of N2=1428=0.5Moles of H2=72=3.5∴n=0.5+3.5=4.0 molesPV=nRT⇒P=nRTV=4×0.0821×3008.2 =12.01 atm

$14$ g of nitrogen gas and $7$ g of hydrogen gas is present in a closed vessel at $27^{0}$ C. If the volume of the container is $8.2$ L then the pressure of the gas mixture is:

The correct option is C $12$ atm
Moles of $N_{2} = \frac{14}{28} = 0.5$
Moles of $H_{2} = \frac{7}{2} = 3.5$
$∴ n = 0.5 + 3.5 = 4.0$ moles
$P V = n R T$
$\Rightarrow P = \frac{n R T}{V} = \frac{4 \times 0.0821 \times 300}{8.2}$
$= 12.01$ atm