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A bulb of unknown volume ‘V’ contains an ideal gas at 2 atm pressure. It was connected to another evacuated bulb of volume 0.5 litre through a stopcock. When the stopcock was opened the pressure in each bulb became 0.5 atm. Then V is aprroximately

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0.166 L

0.3 L

1.76 L

3.0 L

answer is A.

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Detailed Solution

P₂V₂=P₁V₁

0.5(0.5+V)=2×V

0.25+0.5V=2V

1.5V=0.25

$V = \frac{0.25 }{1.5} = 0.17 L$

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