Question

`"2.50 L"` of a gas at `"303 K"` and `"100 kPa"` are heated. The new volume is `"3.75 L"` and the new pressure is `"90.0 kPa"`. What is the new temperature?

Answer 1

`"409 K"` to three significant figures.

Explanation:

This question is an example of the combined gas law. The formula is:

`(P_1V_1)/T_1=(P_2V_2)/T_2`

Organize the data:

Known

`P_1="100 kPa"`

`V_1="2.50 L"`

`T_1="303 K"`

`P_2="90.0 kPa"`

`V_2="3.75 L"`

Unknown

`T_2`

Solution

Rearrange the formula to isolate `T_2`. Insert your data and solve.

`T_2=(P_2V_2T_1)/(P_1V_1)`

`T_2=((90.0color(red)cancel(color(black)("kPa"))xx3.75color(red)cancel(color(black)("L"))xx303"K"))/((100color(red)cancel(color(black)("kPa"))xx2.50color(red)cancel(color(black)("L"))))="409K"` rounded to three significant figures

(`"100 kPa"` is an exact number so it has an infinite number of significant figures.)