Ncert Exercises & Solutions

5.15 Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K–1 mol–1.

Given,

Mass of dioxide

(O_{2})

= 8g

Thus, number of moles of

O_{2} = \frac{8}{32} = 0.25 m o l e

Mass of dihydrogen

(H_{2})

= 4 g

Thus, the number of moles of

H_{2} = \frac{4}{2} = 2 m o l e

Therefore, total number of moles in the mixture = 0.25 + 2 = 2.25 mole

Given, V =

1 d m^{3} n = 2.25 m o l

R = 0.083 bar

d m^{3} K^{- 1} m o l^{- 1}

T =

27 ° C

= 300 K

Total pressure (p) can be calculated as: pV

= nRT

\Rightarrow p = \frac{n R T}{V}

= \frac{225 \times 0.083 \times 300}{1}

= 56.025 b a r

Hence, the total pressure of the mixture is 56.025 bar.

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