# The gas inside of a container exerts 84 Pa of pressure and is at a temperature of 420 ^o K. If the pressure in the container changes to 64 Pa with no change in the container's volume, what is the new temperature of the gas?

Mar 25, 2017

The new temperature is $= 320 K$

#### Explanation:

We apply Gay Lussac's Law

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

Pressure, ${P}_{1} = 84 P a$

Temperature, ${T}_{1} = 420 K$

Pressure, ${P}_{2} = 64 P a$

Temperature,

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

$= \frac{64}{84} \cdot 420$

$= 320 K$