Question

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 one quarter from where it started, what would be the new pressure of the gas

Answer 1

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.