# The gas inside of a container exerts 5 Pa of pressure and is at a temperature of 120 ^o C. If the temperature of the gas changes to 180 ^oK with no change in the container's volume, what is the new pressure of the gas?

Feb 21, 2016

${P}_{2} \cong 2 , 29 P a$

#### Explanation:

$\text{we can use the Gay-Lussac's law}$
${P}_{1} / {T}_{1} = {P}_{'} / {T}_{2}$
$\text{change "120 ^o C " to " ^o K} 120 + 273 = {393}^{o} K$
$\frac{5}{393} = {P}_{2} / 180$
${P}_{2} = \frac{180 \cdot 5}{393}$
${P}_{2} \cong 2 , 29 P a$