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

The final volume is
`V_2=0.08 L`.

Step-by-step 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`.