If a gas in a balloon starts at #"2.95 atm"#, #"7.456 L"#, and #"379 K"#, what is the final pressure in #"torr"# for the gas when it compresses to #"4.782 L"# and #"212 K"#?
1 Answer
Aug 8, 2017
#P_2 = "1955.37 torr"#
What is this pressure in
You can always start from the ideal gas law for ideal gases:
#PV = nRT#
#P# is pressure in#"atm"# .#V# is volume in#"L"# .#n# is#bb"mols"# of ideal gas.#R = "0.082057 L"cdot"atm/mol"cdot"K"# is the universal gas constant if#P# is in#"atm"# and#V# is in#"L"# .#T# is temperature in#"K"# .
If you read the question, you should find that
#(P_1V_1)/T_1 = nR = (P_2V_2)/(T_2)# giving the so-called "Combined Gas Law".
And so, the pressure must be:
#color(blue)(P_2) = (P_1V_1)/(T_1) cdot T_2/V_2#
#= (("2.95 atm")(7.456 cancel"L"))/(379 cancel"K") cdot (212 cancel"K")/(4.782 cancel"L")#
#=# #color(blue)ul"2.57 atm"#
It is likely that the balloon is thick-walled to enforce conservation of mass and energy, i.e. the system is mechanically-closed and thermally-insulating.