# The gas inside of a container exerts 42 Pa of pressure and is at a temperature of 420 ^o K. If the pressure in the container changes to 72 Pa with no change in the container's volume, what is the new temperature of the gas?

Apr 25, 2018

$\text{final temperature is } {720}^{o} K$

#### Explanation:

$\text{As the volume of the container does not change, the increase of}$
$\text{the gas pressure causes the temperature to increase.}$

$\text{We need to use the Gay-Lussac law to find the temperature.}$

${P}_{i} = 42 P a \left(\text{initial pressure of the gas}\right)$
${P}_{f} = 72 P a \left(\text{final pressure of the gas}\right)$
${T}_{1} = {420}^{o} K \left(\text{initial temperature of the gas}\right)$
T_f=?("final temperature of the gas")

${P}_{i} / {T}_{i} = {P}_{f} / {T}_{f}$

$\frac{\cancel{42}}{\cancel{420}} = \frac{72}{T} _ f$

$\text{let's divide both numerator and denominator by 42 }$

$\frac{1}{10} = \frac{72}{T} _ f$

$1 \cdot {T}_{f} = 72 \cdot 10$

${T}_{f} = {720}^{o} K$