Question

If `10/3 L` of a gas at room temperature exerts a pressure of `42 kPa` on its container, what pressure will the gas exert if the container's volume changes to `12/7 L`?

Answer 1

The new pressure is `=81.7kPa`

Explanation:

Apply Boyle's Law

`"Pressure" xx"Volume"="Constant"` at constant temperature

`P_1V_1=P_2V_2`

The initial pressure is `P_1=42kPa`

The initial volume is `V_1=10/3L`

The final volume is `V_2=12/7L`

The final pressure is

`P_2=V_1/V_2*P_1=(10/3)/(12/7)*42=81.7kPa`

Answer 2

`81.7 \ "kPa"`

Explanation:

We use Boyle's law, if the volume and the number of moles is constant, and we get,

`Pprop1/V` or `P_1V_1=P_2V_2`

Now, we plug in the given values, and we get

`42 \ "kPa"*10/3color(red)cancelcolor(black)"L"=P_2*12/7color(red)cancelcolor(black)"L"`

`P_2~~81.7 \ "kPa"`

So, the new pressure is approximately `81.7` kilopascals.