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Arada [10]
3 years ago
7

Concerning Boyle's Law, if you had a gas at a pressure of 101 kPa and decreased the volume of the container holding the gas to o

ne quarter from where it started, what would be the new pressure of the gas
Chemistry
1 answer:
Viefleur [7K]3 years ago
7 0

Answer:

P_2=404 kPa

Explanation:

Hello,

In this case, the Boyle's is mathematically defined via:

P_1V_1=P_2V_2

Which stands for an inversely proportional relationship between volume and pressure, it means the higher the volume the lower the pressure and vice versa. In such a way, since the volume is decreased to one quarter, we can write:

V_2=\frac{1}{4} V_1

We can compute the new pressure:

P_2=\frac{P_1V_1}{V_2} =\frac{P_1V_1}{\frac{1}{4} V_1} =\frac{101kPa*V_1}{\frac{1}{4} V_1} \\\\P_2=4*101kPa\\\\\\P_2=404 kPa

Which means the pressure is increased by a factor of four.

Regards.

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