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Ideal Gas Equation

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 some time. Thus, the order of partial 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 $P_{S O_{2}} > P_{C H_{4}} > P_{H_{2}}$

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Similar questions

{S O}_{2} {, H}_{2}

C H_{4}

(p H_{1}),

(p H_{2}),

p H_{3})

p 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

Column I Column II

(A) Maximum potential for

H-electrode at $p H_{2} = 1 a t m$ (P)-0.414 V

(B) Minimum potential for

H-electrode at $p H_{2} = 1 a t m$ (Q) -0.018 V

up in 1 M KCl and $p H_{2} = 1 a t m$ (R) -0.827 V

(D)Potential of H-electrode set up

in 1.0 M HCl and $p H_{2} = 1 a t m$ (S) 0.0 V

0.5 m o l

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

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