# The gas inside of a container exerts 48 Pa of pressure and is at a temperature of 270 ^o K. If the pressure in the container changes to 60 Pa with no change in the container's volume, what is the new temperature of the gas?

Mar 2, 2016

$337.5 K$

#### Explanation:

Recall that Gay-Lussac's Law states that pressure is directly proportional to temperature. The formula that represents this is:

$\frac{\textcolor{b r o w n}{{P}_{1}}}{\textcolor{b l u e}{{T}_{1}}} = \frac{\textcolor{p u r p \le}{{P}_{2}}}{\textcolor{\mathmr{and} a n \ge}{{T}_{2}}}$

where:
$\textcolor{b r o w n}{{P}_{1}} =$initial pressure
$\textcolor{b l u e}{{T}_{1}} =$inital temperature (Kelvin)
$\textcolor{p u r p \le}{{P}_{2}} =$final pressure
$\textcolor{\mathmr{and} a n \ge}{{T}_{2}} =$final temperature (Kelvin)

Finding the Final Temperature
$1$. Since all other values have been given, the only variable you need to solve for is ${T}_{2}$, the final temperature. Thus, start by rearranging the formula to solve for ${T}_{2}$.

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

${T}_{2} = \frac{{P}_{2} {T}_{1}}{P} _ 1$

$2$. Substitute your known values into the rearranged formula.

$\textcolor{p u r p \le}{{P}_{2} = 60 P a} \textcolor{w h i t e}{X X X} \textcolor{b l u e}{{T}_{1} = 270 K} \textcolor{w h i t e}{X X X} \textcolor{b r o w n}{{P}_{1} = 48 P a}$

${T}_{2} = \frac{\left(60 P a\right) \left(270 K\right)}{\left(48 P a\right)}$

$3$. Solve for ${T}_{2}$.

$\textcolor{g r e e n}{{T}_{2} = 337.5 K}$

$\therefore$, the new temperature of the gas is $337.5 K$.