# The gas inside of a container exerts 48 Pa of pressure and is at a temperature of 320 ^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?

Apr 15, 2017

The new temperature is $= 426.7 K$

#### Explanation:

We apply Gay Lussac's Law

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

The initial pressure is ${P}_{1} = 48 P a$

The initial temperature is ${T}_{1} = 320 K$

The final pressure is ${P}_{2} = 64 P a$

The final temperature is

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

$= \frac{64}{48} \cdot 320$

$= 426.7 K$

Apr 15, 2017

use gay-lussac's law .

#### Explanation:

gay-lusaac's law gives you the relation between pressure and temperature of a gas inside a container with a fixed volume :
${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$
${P}_{1}$ is the pressure exerted by a gas at temperature ${T}_{1}$ and ${P}_{2}$ is the pressure exerted by the gas at temperature ${T}_{2}$ .
now, substituting the values :
$\frac{48}{320} = \frac{64}{T}$
$T = 426.67 k$