At 1.0 atm, 2.0 litres is occupied by air which is trapped in a glass tube. At constant temperature, its pressure is increased to 4 atm by compressing it and its new volume of the air (in L) is:

0.25 L

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4.0 L

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0.5 L

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2.0 L

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8.0 L

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Solution

The correct option is C 0.5 L
$P_{1} V_{1} = P_{2} V_{2}$
$P_{1} =$ initial pressure $=$ 1.0 atm
$V_{1} =$ initial volume $=$ 2.0 liters
$P_{2} =$ 4.0 atm
$V_{2} =$ ??
$1.0 \times 2.0 = 4.0 \times V_{2}$
$V_{2} = 0.5$ liters.

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At 1.0 atm, 2.0 litres is occupied by air which is trapped in a glass tube. At constant temperature, its pressure is increased to 4 atm by compressing it and its new volume of the air (in L) is:

The correct option is C 0.5 LP1V1=P2V2P1= initial pressure = 1.0 atmV1= initial volume = 2.0 litersP2= 4.0 atmV2= ??1.0×2.0=4.0×V2V2=0.5 liters.

The correct option is C 0.5 L
$P_{1} V_{1} = P_{2} V_{2}$
$P_{1} =$ initial pressure $=$ 1.0 atm
$V_{1} =$ initial volume $=$ 2.0 liters
$P_{2} =$ 4.0 atm
$V_{2} =$ ??
$1.0 \times 2.0 = 4.0 \times V_{2}$
$V_{2} = 0.5$ liters.