Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm $^{3}$ at 27 $^{\circ} C$ . (R = 0.083 bar dm $^{3}$ K $^{- 1}$ mol $^{- 1}$ )

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Solution

The partial pressure of oxygen is $P = \frac{n R T}{V} = \frac{8 \times 0.083 \times 300}{32 \times 1} = 0.225 b a r$ .
The partial pressure of hydrogen gas is $P^{'} = \frac{n^{'} R T}{V^{'}} = \frac{2 \times 0.083 \times 300}{1} = 49.8 b a r$ .
Total pressure of the mixture $= 6.225 + 49.8 = 56.025 b a r$ .

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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.

Q. Calculate the total pressure in mixture of

8 g

of dioxygen and

8 g

of dihydrogen confined in a vessel of

1 {dm}^{3}

27^{o} C

.
[Take

R = 0.083 {bar dm}^{3} K^{- 1} {mol}^{- 1}

]

Q. The total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1

{d m}^{3}

27^{o} C

is :

(R = 0.083 bar dm

^{3}

^{- 1}

mol

^{- 1}

)

Calculate the total pressure of a mixture of 8 g of oxygen and 4g of hydrogen confined in vessel of

1 d m^{3}

27^{0} C

(R = 0.083 b a r d m^{3} k^{- 1} {m o l}^{- 1})

Q. calculate the total pressure of a mixture of 8grams of O2 and 4grams of H2 and enclosed in a vessel of 1 dm^3 at 27°C ?
Given that R = 0.083 bar dm^3 K^-1 mol^-1

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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)

The partial pressure of oxygen is P=nRTV=8×0.083×30032×1=0.225 bar.The partial pressure of hydrogen gas is P′=n′RTV′=2×0.083×3001=49.8 bar.Total pressure of the mixture=6.225+49.8=56.025 bar.

Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm $^{3}$ at 27 $^{\circ} C$ . (R = 0.083 bar dm $^{3}$ K $^{- 1}$ mol $^{- 1}$ )