A balloon contains 1.2 L of air at a particular temperature and 90 cm Hg pressure. Keeping the temperature constant if the pressure of air is reduced to 70 cm Hg, then what will be the volume of that air?

4.54L.

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1.54L

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5.54L.

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0.54L.

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Solution

The correct option is B
1.54L

As the temperature is constant, here Boyle's law is applied,

According to Boyle's law : $P_{1} V_{1} = P_{2} V_{2}$

where $P_{1} = 90 c m H g : P_{2} = 70 c m H g : V_{1} = 1.2 L$

$V_{2} = \frac{P_{1} V_{1}}{P_{1}}$

$\Rightarrow V_{2} = \frac{90 \times 1.2}{70} \Rightarrow V_{2} = 1.54 L$

Therefore, Volume of air at 70 cm Hg= 1.54L.

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A balloon contains 1.2 L of air at a particular temperature and 90 cm Hg pressure. Keeping the temperature constant if the pressure of air is reduced to 70 cm Hg, then what will be the volume of that air?

The correct option is B 1.54L As the temperature is constant, here Boyle's law is applied, According to Boyle's law : P1V1=P2V2 where P1=90cm Hg : P2=70cm Hg: V1=1.2L V2=P1V1P1 ⇒V2=90×1.270⇒V2=1.54L Therefore, Volume of air at 70 cm Hg= 1.54L.