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
Truong-Son N.
Aug 8, 2017

#P_2 = "1955.37 torr"#

What is this pressure in #"atm"#? And why do you suppose the balloon is thick-walled?

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 #DeltaP ne 0#, #DeltaV ne 0#, and #DeltaT ne 0#. Thus, we can write two states...

#(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.

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