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

##### 1 Answer

#### Explanation:

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

#color(blue)(ul(color(black)(PV = nRT)))#

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 theuniversal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is theabsolute temperatureof 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

#color(blue)(ul(color(black)("1 atm = 760 mmHg")))#

Start by converting the mass of chlorine to *moles* by using the **molar mass** of chlorine gas,

#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

#PV = nRT implies T = (PV)/(nR)#

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 = "500.02 K"#

To convert this to *degrees Celsius*, use the fact that

#color(blue)(ul(color(black)(t[""^@"C"] = T["K"] - 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.