# The gas inside of a container exerts 9 Pa of pressure and is at a temperature of 320 ^o K. If the pressure in the container changes to 32 Pa with no change in the container's volume, what is the new temperature of the gas?

Mar 2, 2017

The new temperature is $= 1137.8 P a$

#### Explanation:

The ideal gas law is

$P V = n R T$

$\frac{P}{T} = n \frac{R}{V} = k$

Where $k =$ a constant

This is Gay Lussac's Law

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

${P}_{1} = 9 P a$

${T}_{1} = 320 K$

${P}_{2} = 32 P a$

${T}_{2} = {P}_{2} / {P}_{1} \cdot {T}_{1}$

$= \frac{32}{9} \cdot 320 = 1137.8 P a$