# 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# ispressurein#"atm"# .#V# isvolumein#"L"# .#n# is#bb"mols"# of ideal gas.#R = "0.082057 L"cdot"atm/mol"cdot"K"# is theuniversal gas constantif#P# is in#"atm"# and#V# is in#"L"# .#T# istemperaturein#"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.