What is the temperature of the gas in Kelvins and Celsius?

Here is the more typical way to do it, using only numbers you are allowed during an exam...

I got #T = "756 K"#. Consequently, I would get:

#"756 K" - "273.15 K" = 483^@ "C"#

Assuming ideality, it is obvious that given a mass, volume, and pressure, while also being asked for a temperature, you need to use the ideal gas law (which contains all of the above, in some form):

#PV = nRT#

where:

• #P# is the pressure in #"atm"#.
• #V# is the volume in #"L"#.
• #n# is the #bb"mols"# of ideal gas.
• #R = "0.082057 L"cdot"atm/mol"cdot"K"# is the universal gas constant.
• #T# is the temperature in #"K"#.

In 99% of all chemistry exams, you would NOT be given a form of #R# involving pressure units of #"mm Hg"# (that would be too easy). Thus, the temperature of the gas is:

#color(blue)(T) = (PV)/(nR)#

#= ((690. cancel"mm Hg" xx cancel"1 atm"/(760 cancel"mm Hg"))(50.0 cancel"L"))/((20.5 cancel("g N"_2) xx cancel"1 mol"/(28.014 cancel("g N"_2)))(0.082057 cancel"L"cdotcancel"atm""/"cancel"mol"cdot"K"))#

#=# #color(blue)("756 K") = color(blue)(483^@ "C")#

Here's a question for you... write down all of the conversion factors that I used. You should know them by heart or be able to look at a periodic table and derive them.