Question

If `7/5 L` of a gas at room temperature exerts a pressure of `3 kPa` on its container, what pressure will the gas exert if the container's volume changes to `2/3 L`?

Answer 1

The final pressure will be `63/10"kPa"` `=` `"6.3 kPa"`

Explanation:

This is a question involving Boyle's law, which states that the volume of a gas varies inversely with its pressure, as long as the amount and temperature are held constant.

`P_1="3 kPa"`
`V_1="7/5 L"`
`V_2="2/3 L"`
`P_2="???"`

Rearrange the equation to isolate `P_2`. Substitute the known values into the equation and solve.

`P_2=(P_1V_1)/V_2`

`P_2=(3"kPa"·7/5cancel"L")/(2/3 cancel"L")`

Simplify.

`P_2=(21/5 "kPa")/(2/3)`

Multiply by the reciprocal of `2/3`.

`P_2=(21/5"kPa")·(3/2)`

Simplify.

`P_2=63/10 "kPa"` `=` `"6.3 kPa"`