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Heat Capacity

The Total pre...

Question

The Total pressure in a mixture of $8 g$ of dioxygen and $4 g$ of dihydrogen confined in a vessel of $1 d m^{3} 4$ at $27^{o} C$ is ?

R = 83 b a r d m^{3} k^{- 1} m o l^{- 1}

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R = 93 b a r d m^{3} k^{- 1} m o l^{- 1}

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R = 56 b a r d m^{3} k^{- 1} m o l^{- 1}

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R = 63 b a r d m^{3} k^{- 1} m o l^{- 1}

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Solution

The correct option is C $R = 56 b a r d m^{3} k^{- 1} m o l^{- 1}$
Mass of oxygen= $8 g = 8 / 32 = 0.25 m o l$
Mass of hydrogen= $4 g = 4 / 2 = 2 m o l$
From $P V = n R T$
$\Rightarrow P \times 1 = (0.25 + 2) 0.083 \times 300 = 56.02$ bar

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Similar questions

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. 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}

)

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. 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}

^{- 1}

mol

^{- 1}

)

Q. Calculate the value of molar gas constant,

^{'} R^{'}

in (i)

c c a t m K^{- 1} m o l^{- 1}

(ii)

t o r r c c K^{- 1} m o l^{- 1}

(iii)

k P a l i t r e K^{- 1} m o l^{- 1}

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The Total pressure in a mixture of 8g of dioxygen and 4g of dihydrogen confined in a vessel of 1dm34 at 27oC is ?

The correct option is C R=56bar dm3k−1mol−1Mass of oxygen= 8g=8/32=0.25molMass of hydrogen= 4g=4/2=2molFrom PV=nRT⇒P×1=(0.25+2)0.083×300=56.02 bar

The Total pressure in a mixture of $8 g$ of dioxygen and $4 g$ of dihydrogen confined in a vessel of $1 d m^{3} 4$ at $27^{o} C$ is ?

The correct option is C $R = 56 b a r d m^{3} k^{- 1} m o l^{- 1}$
Mass of oxygen= $8 g = 8 / 32 = 0.25 m o l$
Mass of hydrogen= $4 g = 4 / 2 = 2 m o l$
From $P V = n R T$
$\Rightarrow P \times 1 = (0.25 + 2) 0.083 \times 300 = 56.02$ bar