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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 a volume of $100 m L$ and contains hydrogen. After opening the gas from $A$ to the evacuated bulb $B$ , the pressure falls down to $40$ %. The volume (mL) of $B$ must be

75

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150

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125

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200

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250

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Solution

The correct option is B $150$
Moles of gases will remain same. Using mole conservation, we get:
$P_{1} V_{1} (A) = P_{2} V_{2} (B) = P_{R} (V_{1} + V_{2})$
$100 \times 100 + 0 \times V_{2} = 40 (100 + V_{2})$
$V_{2} = 250 - 100 = 150 m L$

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

The correct option is B 150Moles of gases will remain same. Using mole conservation, we get:P1V1(A)=P2V2(B)=PR(V1+V2)100×100+0×V2=40(100+V2)V2=250−100=150 mL

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

The correct option is B $150$
Moles of gases will remain same. Using mole conservation, we get:
$P_{1} V_{1} (A) = P_{2} V_{2} (B) = P_{R} (V_{1} + V_{2})$
$100 \times 100 + 0 \times V_{2} = 40 (100 + V_{2})$
$V_{2} = 250 - 100 = 150 m L$