# Question #dd7e8

Jun 27, 2016

$\text{140 atm}$

#### Explanation:

The trick here is to realize that when volume and number of moles of gas are kept constant, pressure and temperature have a direct relationship as described by Gay Lussac's Law.

In other words, when volume and number of moles are constant, increasing a gas' temperature will cause its pressure to increase, and decreasing the gas' temperature will cause its pressure to decrease. In your case, the temperature of the gas increases from $\text{428 K}$ to $\text{684 K}$, which can only mean that its pressure increased as well.

Mathematically, Gay Lussac's Law can be written as

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

${P}_{1}$, ${T}_{1}$ - the pressure and temperature of the gas at an initial state
${P}_{2}$, ${T}_{2}$ - the pressure and temperature of the gas at final state

Rearrange the equation to solve for ${P}_{2}$

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

Plug in your values to find

${P}_{2} = \left(684 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{K"))))/(428color(red)(cancel(color(black)("K")))) * "86 atm" = color(green)(|bar(ul(color(white)(a/a)color(black)("140 atm}} \textcolor{w h i t e}{\frac{a}{a}} |}}\right)$

The answer is rounded to two sig figs, the number of sig figs you have for the initial pressure of the gas.