# The pressure of 250 kPa acting on 15 cubic meters of gas at a temperature of 100 K is increased to 500 kPa and the volume is increased to 30 cubic meters. What is the temperature of the gas at this new pressure and volume?

Mar 11, 2016

The final temperature will be 400 K.

#### Explanation:

This is an example of the combined gas law. The equation to use is $\frac{{P}_{1} {V}_{1}}{{T}_{1}} = \frac{{P}_{2} {V}_{2}}{{T}_{2}}$.

Given
Initial pressure, ${P}_{1} = \text{250 kPa}$
Initial volume, ${V}_{1} = \text{15 m"^3}$
Initial temperature, ${T}_{1} = \text{100 K}$
Final pressure, ${P}_{2} = \text{500 kPa}$
Final volume, ${V}_{2} = \text{30 m"^3}$

Unknown
Final temperature, ${T}_{2}$

Solution
Rearrange the equation to isolate ${T}_{2}$. Substitute the given values into the equation and solve.

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

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

T_2=(100"K"·500cancel"kPa"·30cancel"m"^3)/(250cancel"kPa"·15cancel"m"^3)="400K"