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Two gases A and B, having the mole ratio of 3 : 5 in a container, exert a pressure of 8 atm. If A is removed, what would be the pressure due to B only, temperature remaining constant ?

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1 atm

5 atm

4 atm

2 atm

answer is D.

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Detailed Solution

$A : B = 3 : 5$

$P_{T} = 8 atm$

$∴ P_{A} = 3 atm$

$P_{B} = 5 atm$

If ‘A’ is removed

$P_{A} + P_{B} = P_{T}$

3+5=8

$P_{B} = 5 atm$

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