# "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?

Aug 9, 2017

$\text{409 K}$ to three significant figures.

#### Explanation:

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

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

Organize the data:

Known

${P}_{1} = \text{100 kPa}$

${V}_{1} = \text{2.50 L}$

${T}_{1} = \text{303 K}$

${P}_{2} = \text{90.0 kPa}$

${V}_{2} = \text{3.75 L}$

Unknown

${T}_{2}$

Solution

Rearrange the formula to isolate ${T}_{2}$. Insert your data and solve.

${T}_{2} = \frac{{P}_{2} {V}_{2} {T}_{1}}{{P}_{1} {V}_{1}}$

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

($\text{100 kPa}$ is an exact number so it has an infinite number of significant figures.)