# Question #53733

##### 1 Answer

#### Answer:

#### 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 **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

#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.