# Question #0c25b

##### 1 Answer

#### Explanation:

The trick here is to realize that because the pressure *and* the number of moles of gas, i.e. the amount of gas present inside the toy, are **constant**, the volume of the toy is **directly proportional** to its temperature *Charles' Law* here.

In other words, when the temperature **increases**, the volume of the toy increases as well. Similarly, when the temperature **decreases**, the volume of the toy decreases as well.

Mathematically, you can write this as

#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

So, start by converting the two temperatures to *Kelvin*

#10^@"C" = 10^@"C" + 273.15 = "283.15 K"#

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

Rearrange the equation to solve for

#V_/1T_1 = V_2/T_2 implies V_2 = T_2/T_1 * V_1#

Plug in your values to get

#V_2 = (293.15 color(red)(cancel(color(black)("K"))))/(283.15color(red)(cancel(color(black)("K")))) * "2.0 L" = color(darkgreen)(ul(color(black)("2.1 L")))#

I'll leave the answer rounded to two **sig figs**, but keep in mind that you only have one significant figure for the two temperatures in degrees Celsius.

Now, you *could* use the ideal gas law equation to write--keep in mind that **constant**

#P * V_1 = n * R * T_1 -># for the initial state of the gas

#P * V_2 = n * R * T_2 -># for the final state of the gas

Divide these two equations

#(color(red)(cancel(color(black)(P))) * V_1)/(color(red)(cancel(color(black)(P))) * V_2) = (color(red)(cancel(color(black)(n))) * color(red)(cancel(color(black)(R))) * T_1)/(color(red)(cancel(color(black)(n))) * color(red)(cancel(color(black)(R))) * T_2)#

to get

#V_1/T_1 = V_2/T_2 -># the equation forCharles' Law