Question

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"`?

Answer 1

`P_2 = "1955.37 torr"`

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

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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.