Calculate the total pressure in a mixture of 4g of $O_{2}$ and 2g of $H_{2}$ confined in a bulb of 1 litre at $0$ .

Open in App

Solution

v=1L
T=273K
let us find the partial pressure of $O_{2}$
no. of moles = $\frac{4}{32}$ =0.125 moles
$P_{1}$ = $\frac{n R T}{v}$ = $\frac{0.125 \times 0.082 \times 273}{1}$ =2.80 atm
Now, let us find the partial pressure of $H_{2}$

$P_{2}$ = $\frac{n R T}{v}$ = $\frac{1 \times 0.082 \times 273}{1}$ =22.386 atm

$2.80 + 22.38 = 25.18 a t m$

Suggest Corrections

Similar questions

4 g

O_{2}

2 g

H_{2}

1

0^{o} C

4 g

O_{2}

2 g

H_{2}

1 l i t r e

0^{\circ} C

Pressure of a mixture of 4 g of O₂ and 2 g of H₂ confined in a bulb of 1 litre at 0°C is___.

The total pressure exerted by a mixture of 4g of $O_{2}$ and 2g of $H_{2}$ confined in a bulb of 1 L at $O^{\circ}$ is –

4 g

3 g

0^{\circ} C

Join BYJU'S Learning Program