Question

If `3/7 L` of a gas at room temperature exerts a pressure of `9 kPa` on its container, what pressure will the gas exert if the container's volume changes to `5/9 L`?

Answer 1

`P_2 = 243/35 kPa`

Explanation:

Let `P_1 = 9color(white).kPa`
Let `V_1 = 3/7color(white).L`
Let `V_2 = 9/5color(white).L`

`P_1V_1 = P_2V_2`

`P_2 = P_1V_1/V_2`

`P_2 = 243/35 kPa`

Answer 2

`243/35`kPa

Explanation:

Boyle's Law for gases states that pressure and volume vary inversely when temperature and mass are held constant. This can be expressed with the formula:

`P_1V_1=P_2V_2`

`P_1=9`kPa

`V_1=3/7 L`

`P_2=`?

`V_2=5/9 L`

`9*3/7=P_2*5/9`

`9/5*9*3/7=P_2*cancel(5)/cancel(9)*cancel(9)/cancel(5)color(white)(aaa)`Multiply both sides by 9/5

`243/35=P_2`

`P_2=243/35`kPa