Ncert Exercises & Solutions

The pressure of a 1 : 4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. The partial pressure of dioxygen is-

1. $0.8 \times 10^{5}$ $atm$

2. $0.008$ ${Nm}^{- 2}$

3. $8 \times 10^{4}$ ${Nm}^{- 2}$

4. $0.25$ $atm$

Hint:

Partial

pressure

of

O_{2} = Mole

fraction

of

O_{2} \times total

pressure

of

mixture

Step 1:

The formula of partial pressure of

O_{2}

is as follows:

Partial

pressure

of

O_{2} = Mole

fraction

of

O_{2} \times total

pressure

of

mixture

Step 2:

The given values are as follows:

The pressure of a 1 : 4 mixture of H₂ and O₂ enclosed in a vessel is one atmosphere. This suggests that the moles ratio of H₂ and O₂ is 1 : 4.

Thus, partial pressure of dioxygen (O₂) is given as

∴ Partial

pressure

of

O_{2} = Mole

fraction

of

O_{2} \times total

pressure

of

mixture

= \frac{4}{1 + 4} \times 1 = \frac{4}{5} \times 1

atm

= 0.8

atm = 0.8 \times 10^{5}

{Nm}^{- 2} = 8 \times 10^{4}

{Nm}^{- 2}