# The gas inside of a container exerts 24 Pa of pressure and is at a temperature of 90 ^o K. If the pressure in the container changes to 52 Pa with no change in the container's volume, what is the new temperature of the gas?

Apr 22, 2016

${T}_{2} = 195 {\text{ }}^{o} K$

#### Explanation:

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

${P}_{1} = 24 \text{ Pa, "T_1=90^o K ,} {P}_{2} = 52 P a$

$\frac{24}{90} = \frac{52}{T} _ 2$

${T}_{2} = \frac{90 \cdot 52}{24}$

${T}_{2} = 195 {\text{ }}^{o} K$