# A balloon is filled with 14 L of gas at 302 K. What is its temperature in Kelvin when the volume expands to 20 L, assuming pressure remains constant?

Jul 20, 2016

$\text{430 K}$

#### Explanation:

The thing to remember about volume and temperature is that they have a direct relationship when pressure and number of moles of gas remain constant.

This relationship is known as Charles' Law and states that when those conditions are met, increasing the temperature of the gas will result in an increase in volume.

Similarly, decreasing the temperature of the gas will result in a decrease in volume. Mathematically, this is expressed as

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

${V}_{1}$, ${T}_{1}$ - the volume and temperature of the gas at an initial state
${V}_{2}$, ${T}_{2}$ - the volume and temperature of the gas at a final stat

In your case, the volume expanded from an initial value of $\text{14 L}$ to a final value of $\text{20 L}$, which can only mean that the temperature of the gas increased as well, i.e. ${T}_{2} > {T}_{1}$.

Rearrange the above equation to solve for ${T}_{2}$

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

Plug in your values to find

T_2 = (20 color(red)(cancel(color(black)("L"))))/(14color(red)(cancel(color(black)("L")))) * "302 K" = "431.43 K"

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the final volume of the gas

${T}_{2} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{430 K}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$