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Standard XII

Physics

Ideal Gas Law

Two gas bulbs...

Question

Two gas bulbs A and B are connected by a tube having a stop cock. Bulb A has a volume of 100 ml and contains $H_{2}$ . After stopcock is opened, $H_{2}$ from A travels to evacuated bulb B and the pressure falls to $40 %$ . The volume of B in ml is –

75 ml

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150 ml

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200 ml

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250 ml

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Solution

The correct option is B
150 ml

Let $P_{A}$ = 1 atm $P_{B}$ = 0 atm

$V_{A}$ = 100 ml $V_{B}$ = ?

Applying Dalton’s Law of partial pressure ;

$P_{A} V_{A} + P_{B} V_{B} = P_{t o t a l} (V_{A} + V_{B}) [(1) (100)] + [0 \times V_{B}] = 0.4 (100 + V_{B}) V_{B} = 150 m l$

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Two gas bulbs A and B are connected by a tube having a stop cock. Bulb A has a volume of 100 ml and contains H2 . After stopcock is opened, H2 from A travels to evacuated bulb B and the pressure falls to 40%. The volume of B in ml is –

The correct option is B 150 ml Let PA = 1 atm PB = 0 atm VA = 100 ml VB = ? Applying Dalton’s Law of partial pressure ; PAVA+PBVB=Ptotal(VA+VB)[(1)(100)]+[0×VB]=0.4(100+VB)VB=150 ml

Two gas bulbs A and B are connected by a tube having a stop cock. Bulb A has a volume of 100 ml and contains $H_{2}$ . After stopcock is opened, $H_{2}$ from A travels to evacuated bulb B and the pressure falls to $40 %$ . The volume of B in ml is –

The correct option is B
150 ml

Let $P_{A}$ = 1 atm $P_{B}$ = 0 atm

$V_{A}$ = 100 ml $V_{B}$ = ?

Applying Dalton’s Law of partial pressure ;

$P_{A} V_{A} + P_{B} V_{B} = P_{t o t a l} (V_{A} + V_{B}) [(1) (100)] + [0 \times V_{B}] = 0.4 (100 + V_{B}) V_{B} = 150 m l$