Question

If `9 L` of a gas at room temperature exerts a pressure of `15 kPa` on its container, what pressure will the gas exert if the container's volume changes to `4 L`?

Answer 1

`33.75kPa`

Explanation:

From the information given for this question, we can see that this kind of situation is involving Boyle's Law.

Boyle's Law states that the pressure of a fixed mass of gas at a constant temperature is inversely proportional to its volume .

In which from the definition, the equation is derived as;

`Pquadpropquad 1/V` or `P=k/V` or `PV=k`

`P=` Pressure of gas
`V=` Volume of gas
`k=` Constant

When there are two situations , given initial and final value of both pressure and volume, the equation is derived as;

`P_iV_i=P_fV_f`

From the information given in this question;

`P_i=` Initial pressure of gas `=15kPa`
`P_f=` Final pressure of gas `=?kPa`
`V_i=` Initial volume of gas `=9L`
`V_f=` Final volume of gas `=4L`

Calculating `P_f`;

`(15kPa)(9L)=P_(f)(4L)`

`P_f=(135kPacancel(L))/(4cancel(L))`

`P_f=33.75kPa`