Question

Question #0c25b

Answer 1

`"2.1 L"`

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 `->` think 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_2`

`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 `P` and `n` are 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 for Charles' Law