# A closed food jar has a fixed volume at STP. What would the new pressure be at 45°C?

Mar 15, 2016

$318 k P a$

#### Explanation:

Recall that Gay-Lussac's Law states that pressure is directly proportional to temperature , as long as the volume and number of moles of the gas remain constant.

Gay-Lussac's Law can be expressed mathematically with the formula,

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{1} / {T}_{1} = {P}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where:
${P}_{1} =$initial pressure
${T}_{1} =$inital temperature (Kelvin)
${P}_{2} =$final pressure
${T}_{2} =$final temperature (Kelvin)

Determining the Final Pressure
$1$. Start by determining the values for each variable in the formula.

STP conditions (initial):
initial pressure $\textcolor{\mathmr{and} a n \ge}{\left({P}_{1}\right)}$: $\textcolor{\mathmr{and} a n \ge}{101.325 k P a}$
initial temperature $\textcolor{p u r p \le}{\left({T}_{1}\right)}$: $\textcolor{p u r p \le}{273.15 K}$

New conditions (final)
final pressure $\textcolor{t e a l}{\left({P}_{2}\right)}$: $\textcolor{t e a l}{{P}_{2}}$
final temperature $\textcolor{m a \ge n t a}{\left({T}_{2}\right)}$: ${45}^{\circ} C + 273.15 = \textcolor{m a \ge n t a}{318.15 K}$

$2$. Rearrange Gay-Lussac's formula in terms of $\textcolor{t e a l}{{P}_{2}}$.

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

$\textcolor{t e a l}{{P}_{2}} = \left(\textcolor{m a \ge n t a}{{T}_{2}}\right) \left(\frac{\textcolor{\mathmr{and} a n \ge}{{P}_{1}}}{\textcolor{p u r p \le}{{T}_{1}}}\right)$

$3$. Using these values, substitute them into the rearranged formula.

$\textcolor{t e a l}{{P}_{2}} = \left(\textcolor{m a \ge n t a}{318.15 K}\right) \left(\frac{\textcolor{\mathmr{and} a n \ge}{101.325 k P a}}{\textcolor{p u r p \le}{273.15 K}}\right)$

$4$. Solve for $\textcolor{t e a l}{{P}_{2}}$.

$\textcolor{t e a l}{{P}_{2}} = \left(\textcolor{m a \ge n t a}{318.15 \textcolor{red}{\cancel{\textcolor{m a \ge n t a}{K}}}}\right) \left(\frac{\textcolor{\mathmr{and} a n \ge}{101.325 k P a}}{\textcolor{p u r p \le}{273.15 \textcolor{red}{\cancel{\textcolor{p u r p \le}{K}}}}}\right)$

$\textcolor{t e a l}{{P}_{2}} = 118.0177512 k P a$

$\textcolor{t e a l}{{P}_{2}} \approx \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 318 k P a \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\therefore$, the final pressure is $318 k P a$.