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Standard X

Chemistry

Volume of Gases and Number of Moles

Two gas bulbs...

Question

Two gas bulbs $A$ and $B$ are connected by a tube having a stopcock. Bulb $A$ has volume of 100 mL and contains hydrogen. After opening the gas from $A$ to the evacuated bulb $B$ , the pressure falls down by 40%. The volume (mL) of $B$ must be:

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150

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125

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200

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Solution

The correct option is B 150
From Dalton's partial pressure law,
$P_{A} V_{A} + P_{B} V_{B} = P_{T} V_{(A + B)}$
$\Rightarrow P_{A} = 100 a t m$
$\Rightarrow P_{B} = 0$
$\Rightarrow P_{T} = 40 a t m$
$∴ 100 \times 100 + 0 = 40 (400 + V_{2})$
$\Rightarrow V_{2} = 250 - 100 = 150 m l$

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Two gas bulbs A and B are connected by a tube having a stopcock. Bulb A has volume of 100 mL and contains hydrogen. After opening the gas from A to the evacuated bulb B, the pressure falls down by 40%. The volume (mL) of B must be:

The correct option is B 150From Dalton's partial pressure law,PAVA+PBVB=PTV(A+B)⇒PA=100atm⇒PB=0⇒PT=40atm∴100×100+0=40(400+V2)⇒V2=250−100=150ml

Two gas bulbs $A$ and $B$ are connected by a tube having a stopcock. Bulb $A$ has volume of 100 mL and contains hydrogen. After opening the gas from $A$ to the evacuated bulb $B$ , the pressure falls down by 40%. The volume (mL) of $B$ must be:

The correct option is B 150
From Dalton's partial pressure law,
$P_{A} V_{A} + P_{B} V_{B} = P_{T} V_{(A + B)}$
$\Rightarrow P_{A} = 100 a t m$
$\Rightarrow P_{B} = 0$
$\Rightarrow P_{T} = 40 a t m$
$∴ 100 \times 100 + 0 = 40 (400 + V_{2})$
$\Rightarrow V_{2} = 250 - 100 = 150 m l$