At NTP the volume of a gas is $40 m L$ . If pressure is increased to $800 m m$ of $H g$ at the same temperature, what will be the volume of the gas?

38 m L

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22400 m L

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240 m L

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431 m L

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Solution

The correct option is C $38 m L$
At $N T P$ , the pressure of the gas $P_{1} = 760 m m o f H g$

By applying Boyle's law,

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

$760 \times 40 = 800 \times V_{2}$

Therefore, $V_{2} = \frac{760 \times 40}{800} = 38 m L$

So, the correct option is $A$

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At NTP the volume of a gas is 40 mL. If pressure is increased to 800 mm of Hg at the same temperature, what will be the volume of the gas?

The correct option is C 38 mLAt NTP, the pressure of the gas P1=760 mm of HgBy applying Boyle's law, P1V1=P2V2 760×40=800×V2Therefore, V2=760×40800=38 mLSo, the correct option is A

At NTP the volume of a gas is $40 m L$ . If pressure is increased to $800 m m$ of $H g$ at the same temperature, what will be the volume of the gas?

The correct option is C $38 m L$
At $N T P$ , the pressure of the gas $P_{1} = 760 m m o f H g$

By applying Boyle's law,

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

$760 \times 40 = 800 \times V_{2}$

Therefore, $V_{2} = \frac{760 \times 40}{800} = 38 m L$

So, the correct option is $A$