# The gas inside of a container exerts 12 Pa of pressure and is at a temperature of 250 ^o C. If the temperature of the gas changes to 220 ^oK with no change in the container's volume, what is the new pressure of the gas?

##### 1 Answer
Feb 21, 2016

${P}_{2} \cong 5 , 048 P a$

#### Explanation:

$\text{use Gay-Lussac's Law}$
${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$
${250}^{o} C = 250 + 273 = {523}^{o} K$
$\frac{12}{523} = {P}_{2} / 220$
${P}_{2} = \frac{12 \cdot 220}{523}$
${P}_{2} \cong 5 , 048 P a$