Question

A gas has a volume of 3.0 l at a pressure of 3 kpa. What happens to the volume when the pressure is increased to 110 kpa? The temperature does not change

Answer 1

Answer:

The final volume is $V_2=0.08 L$.

Explanation:

The expression in terms of volume and pressure is as follows;

$P_1V_1=P_2V_2$

Here, $P_1,P_2$ are the initial pressure and the final pressure and $V_1,V_2$ are the initial volume and the final volume.

This expression is known as Boyle's law.

It is given in the problem that a gas has a volume of 3.0 l at a pressure of 3 kpa. The pressure is increased to 110 kpa.

Calculate the volume, $V_2$.

$P_1V_1=P_2V_2$

Put $V_1=3 L$, $P_1=3kpa$ and $P_2=110 kpa$.

$(3)(3)=(110)V_2$

$V_2=0.08 L$

Therefore, the final volume is $V_2=0.08 L$.