# Question #674f8

##### 1 Answer

#### Explanation:

Your tool of choice here will be the equation that describes **Charles' Law**

#color(blue)(ul(color(black)(V_1/T_1 = V_2/T_2)))#

Here

#V_1# and#T_1# represent the volume and the absolute temperature of the gas at an initial state#V_2# and#T_2# represent the volume and the absolute temperature of the gas at a final state

This equation shows that when the pressure of a gas and the number of moles of gas present in the sample remain **constant**, the volume of the gas and its temperature have a **direct relationship**.

In other words, when the pressure and the number of moles of gas are constant, an increase in the temperature of the gas will cause its volume to **increase** as well.

Similarly, a decrease in the temperature of the gas will cause its volume to **decrease** as well.

In your case, you have a gas at normal temperature, which is equal to

#20^@"C" + 273.15 = "293.15 K"#

You know that the volume of the gas must **decrease** by

#V_2 = V_1 - 20/100 * V_1 = 0.8 * V_1#

Right from the start, you can say that because the volume is *decreasing*, you must have

#T_2 < "293.15 K"#

Rearrange the equation to solve for

#V_1/T_1 = V_2/T_2 implies T_2 = V_2/V_1 * T_1#

Plug in your values to find

#T_2 = (0.8 * color(red)(cancel(color(black)(V_1))))/(color(red)(cancel(color(black)(V_1)))) * "293.15 K" = "234.52 K"#

This will be equivalent to

#t[""^@"C"] = "234.52 K" - 273.15 = color(darkgreen)(ul(color(black)(-40^@"C")))#

The answer must be rounded to one **significant figure**, the number of sig figs you have for the percent decrease.

Notice that you don't really need to know that the gas occupies