# What is the Celsius temperature of 100.0 g of chlorine gas in a 55.0-L container at 800 mm Hg?

Feb 28, 2017

${230}^{\circ} \text{C}$

#### Explanation:

Your tool of choice here will be the ideal gas law equation, which looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Now, it's important to realize that the unit you have for pressure must match the unit used by the universal gas constant.

In your case, you must convert the pressure from mmHg to atm by using the conversion factor

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 atm = 760 mmHg}}}}$

Start by converting the mass of chlorine to moles by using the molar mass of chlorine gas, ${\text{Cl}}_{2}$

100.0 color(red)(cancel(color(black)("g"))) * "1 mol Cl"_2/(70.906color(red)(cancel(color(black)("g")))) = "1.4103 moles Cl"_2

Rearrange the ideal gas law equation to solve for $T$

$P V = n R T \implies T = \frac{P V}{n R}$

and plug in your values to find the absolute temperature of the gas

T = ( 800/760 color(red)(cancel(color(black)("atm"))) * 55.0color(red)(cancel(color(black)("L"))))/(1.4103color(red)(cancel(color(black)("moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * "K"))

$T = \text{500.02 K}$

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 thus have

t = "500.02 K" - 273.15 = color(darkgreen)(ul(color(black)(230^@"C")))

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