At a constant temperature a gas in initially at $2$ atm pressure. To compress it to $1 / 8$ th of its initial volume, pressure to be applied is:

4

atm

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8

atm

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12

atm

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16

atm

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Solution

The correct option is D $16$ atm
$As P_{1} V_{1} = P_{2} V_{2} [Boyle's Law] ; \frac{V_{2}}{V_{1}} = \frac{1}{8} (2) (V_{1}) = P_{2} (V_{2}) \Rightarrow (2) = P_{2} (\frac{1}{8}) \Rightarrow P_{2} = 16 a t m$

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At a constant temperature a gas in initially at 2 atm pressure. To compress it to 1/8th of its initial volume, pressure to be applied is:

The correct option is D 16 atmAs P1V1=P2V2 [Boyle's Law] ; V2V1=18 (2)(V1)=P2(V2)⇒(2)=P2(18)⇒P2=16atm

At a constant temperature a gas in initially at $2$ atm pressure. To compress it to $1 / 8$ th of its initial volume, pressure to be applied is:

The correct option is D $16$ atm
$As P_{1} V_{1} = P_{2} V_{2} [Boyle's Law] ; \frac{V_{2}}{V_{1}} = \frac{1}{8} (2) (V_{1}) = P_{2} (V_{2}) \Rightarrow (2) = P_{2} (\frac{1}{8}) \Rightarrow P_{2} = 16 a t m$