# The gas inside of a container exerts 27 Pa of pressure and is at a temperature of 150 ^o K. If the pressure in the container changes to 40 Pa with no change in the container's volume, what is the new temperature of the gas?

Nov 20, 2017

The new temperature is $= 222.2 K$

#### Explanation:

Apply Gay Lussac's law

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$ at constant volume

The initial pressure is $= {P}_{1} = 27 P a$

The initial temperature is $= {T}_{1} = 150 K$

The final pressure is ${P}_{2} = 40 P a$

The final temperature is

${T}_{2} = {P}_{2} / {P}_{1} \cdot {T}_{1} = \frac{40}{27} \cdot 150 = 222.2 K$

${220}^{\circ} K$

#### Explanation:

Using the combined Gas Law, $\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$, you can simplify it to ${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$ since volume is constant.

Since to want to find ${T}_{2}$(Temperature 2), you rearrange the formula to become ${T}_{2} = \frac{{T}_{1} \cdot {P}_{2}}{P} _ 1$

Then you can plug and chug
${T}_{2} = \frac{{150}^{\circ} K \cdot 40 P a}{27 P a}$
$= 222.222222222 {\ldots}^{\circ} K$
$= {220}^{\circ} K$ (with Significant Figures)