# Question d0747

May 16, 2017

Here's what I got.

#### Explanation:

The idea here is that the volume of a gas varies directly with its temperature when the number of moles of gas and the pressure are kept constant $\to$ this is known as Charles' Law.

You can describe this as

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}}}}$

Here

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

In your case, the volume of the gas is increasing

$\text{0.43 mL " ->" 1 mL}$

which means that the temperature must have increased.

Rearrange the equation to solve for ${T}_{2}$, the temperature at which the gas occupies $\text{1 mL}$

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

Plug in your values to find

T_2 = (1 color(red)(cancel(color(black)("mL"))))/(0.43color(red)(cancel(color(black)("mL")))) * "299 K" = "695.3 K"

To convert this to degrees Celsius, use the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{t \left[\text{^@"C"] = T["K}\right] - 273.15}}}$

You will end up with

t[""^@"C"] = "695.3 K" - 273.15 = color(darkgreen)(ul(color(black)(420^@"C")))#

I'll leave the answer rounded to two sig figs, but keep in mind that only have one significant figure for the second volume of the gas.