A mixture of one mole each of H 2, He and O 2 each are enclosed in a cylinder of volume V at temperature T. If the partial pressure of H 2 is 2 atm, the total pressure of the gases in the cylinder is :

Question

A mixture of one mole each of H 2, He and O 2 each are enclosed in a cylinder of volume V at temperature T. If the partial pressure of H 2 is 2 atm, the total pressure of the gases in the cylinder is : (A) 14 atm (B) 22 atm (C) 6 atm (D) 38 atm

Tardigrade Admin · Accepted Answer

According to Dalton's law of partial pressure

p_{i} = x_{i} \times P_{T}

p_{i} =

partial pressure of the

i^{t h}

component

x_{i} =

mole fraction of the

i^{t h}

component

p_{T}

= total pressure of mixture

\Rightarrow 2 a t m = (\frac{n _{H_{2}}}{n _{H_{2} +} n _{H_{e}} + n _{O_{2}}}) \times p_{T}

\Rightarrow p_{T} = 2 a t m \times \frac{3}{1} = 6 a t m

Q. A mixture of one mole each of $H_{2},$ He and $O_{2}$ each are enclosed in a cylinder of volume $V$ at temperature $T$ . If the partial pressure of $H_{2}$ is 2 atm, the total pressure of the gases in the cylinder is :

$14 a t m$

$22$ atm

$6 a t m$

$38 a t m$

Q. A mixture of one mole each of H2​, He and O2​ each are enclosed in a cylinder of volume V at temperature T. If the partial pressure of H2​ is 2 atm, the total pressure of the gases in the cylinder is :

14atm

22 atm

6atm

38atm

According to Dalton's law of partial pressure pi​=xi​×PT​ pi​= partial pressure of the ith component xi​= mole fraction of the ith component pT​ = total pressure of mixture ⇒2atm=(nH2​+​nHe​​+nO2​​nH2​​​)×pT​ ⇒pT​=2atm×13​=6atm

Q. A mixture of one mole each of $H_{2},$ He and $O_{2}$ each are enclosed in a cylinder of volume $V$ at temperature $T$ . If the partial pressure of $H_{2}$ is 2 atm, the total pressure of the gases in the cylinder is :

$14 a t m$

$22$ atm

$6 a t m$

$38 a t m$