Question

Question #53733

Answer 1

`"60. L"`

Explanation:

The idea here is that pressure and volume have an inverse relationship when temperature and number of moles of gas are being kept constant, as described by Boyle's Law.

When two parameters have an inverse relationship, increasing the value of one will cause the value of the other to decrease. Similarly, decreasing the value of one will cause the value of the other to increase.

You thus have

`color(blue)(uarr) color(white)(.)"one parameter " stackrel(color(white)(color(red)("inverse relationship")aaa))(->) " the other"color(white)(.)color(blue)(darr)`

and

`color(blue)(darr) color(white)(.)"one parameter " stackrel(color(white)(color(red)("inverse relationship")aaa))(->) " the other"color(white)(.)color(blue)(uarr)`

In your case, the pressure of the gas is decreasing from `"12.7 kPa"` to `"8.4 kPa"`, which means that you expect the volume of the gas to increase.

Mathematically, Boyle's Law can be written as

`color(blue)(ul(color(black)(P_1V_1 = P_2V_2)))`

Here

• `P_1`, `V_1` are the pressure and the volume of the gas at an initial state
• `P_2`, `V_2` are the pressure and the volume of the gas at a final state

You want to find the volume of the gas after the pressure is decreased, so rearrange to solve for `V_2`

`P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * V_1`

Plug in your values to find

`V_2 = (12.7 color(red)(cancel(color(black)("kPa"))))/(8.4color(red)(cancel(color(black)("kPa")))) * "40.0 L" = color(darkgreen)(ul(color(black)("60. L")))`

The answer must be rounded to two sig figs because you only have two sig figs given for the second pressure of the gas.

As predicted, the volume of the gas increased as a result of a decrease in pressure.