A vessel o f volume 24.6 L contains 1.5 moles of H 2, and 2.5 m oles of N 2 at 300 K.Calculate the partial pressure o f N 2 in the vessel

Question

A vessel o f volume 24.6 L contains 1.5 moles of H 2, and 2.5 m oles of N 2 at 300 K.Calculate the partial pressure o f N 2 in the vessel (A) 1.5 atm (B) 2.0 atm (C) 2.5 atm (D) 3.0 atm

Tardigrade Admin · Accepted Answer

Partial pressure In a mixture of gases, each constituent gas has a partial pressure, which is the notional pressure of that constituent gas, if it alone occupied the entire volume of the original mixture at the same temperature.
Dalton's law Total pressure of an ideal gas mixture is the sum of the partial pressure of the gases in the mixture.
Partial pressure due to

n_{1} = (p_{1}^{\circ} = \frac{n _{1} RT}{V})

Similarly, due to

n_{2} = (p_{2}^{\circ} = \frac{n _{2} RT}{V})

Total pressure

p = p_{1}^{\circ} + p_{2}^{\circ} = \frac{( n _{1} + n _{2} ) RT}{V}

Given, total volume

= 24.6 L

Moles of

H_{2} = 1.5 m o l

Moles of

N_{2} = 2.5 m o l

Temperature

= 300 K

Partial pressure of

N_{2} = ?

Partial pressure of

N_{2} = \frac{Moles of N _{2} \times RT}{V}

= \frac{2.5 moles \times 8.314 \times 1 0 ^{3} L P a K ^{- 1} m o l ^{- 1} \times 300 K}{24.6 L}

= 253475.61 P a

(convert into atm)

(1 a t m = 101325 P a)

= 2.5 a t m (2.50160)

Q. A vessel o f volume $24.6 L$ contains $1.5$ moles of $H_{2}$ , and $2.5 m$ oles of $N_{2}$ at $300 K$ .Calculate the partial pressure o $f N_{2}$ in the vessel

$1.5 a t m$

$2.0 a t m$

$2.5 a t m$

$3.0 a t m$

Q. A vessel o f volume 24.6L contains 1.5 moles of H2​, and 2.5m oles of N2​ at 300K.Calculate the partial pressure o fN2​ in the vessel

1.5atm

2.0atm

2.5atm

3.0atm

Q. A vessel o f volume $24.6 L$ contains $1.5$ moles of $H_{2}$ , and $2.5 m$ oles of $N_{2}$ at $300 K$ .Calculate the partial pressure o $f N_{2}$ in the vessel

$1.5 a t m$

$2.0 a t m$

$2.5 a t m$

$3.0 a t m$