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Dalton's Law of Partial Pressures

A vessel cont...

Question

A vessel contains 0.5 mole each of ${S O}_{2} {, H}_{2}$ and $C H_{4}$ . Its outlet was made open and closed after sometime. Thus, order of partial pressure will be:

{P S O}_{2} > {P C H}_{4} > {P H}_{2}

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{P H}_{2} > {P C H}_{4} > {P S O}_{2}

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{P H}_{2} > {P S O}_{2} > {P C H}_{4}

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{P H}_{2} = {P S O}_{2} = {P C H}_{4}

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Solution

The correct option is A ${P S O}_{2} > {P C H}_{4} > {P H}_{2}$
Given gases are ${S O}_{2}, {C H}_{4}$ and Hydrogen.
Molecular weight of given gases are as follows.
${S O}_{2} > {C H}_{4} > H_{2}$
As the molecular weight is high, Partial pressure is also high.
So, ${P S O}_{2} > {P C H}_{4} > {P H}_{2}$

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Similar questions

0.5

S O_{2}, H_{2}

C H_{4}

(p H_{1}),

(p H_{2}),

p H_{3})

p H_{4})

0.5 m o l

H_{2}, S O_{2} a n d C H_{4}

H, S O_{2}

C H_{4}

A

x : y

p H = (p H)

B

y : x

p H = (p H)_{2}

(p H)_{2} - (p H)_{1} = 1

(p H)_{1} + (p H)_{2} = 9.5 u n i t s

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