# 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"?

Aug 8, 2017

${P}_{2} = \text{1955.37 torr}$

What is this pressure in $\text{atm}$? And why do you suppose the balloon is thick-walled?

You can always start from the ideal gas law for ideal gases:

$P V = n R T$

• $P$ is pressure in $\text{atm}$.
• $V$ is volume in $\text{L}$.
• $n$ is $\boldsymbol{\text{mols}}$ of ideal gas.
• $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$ is the universal gas constant if $P$ is in $\text{atm}$ and $V$ is in $\text{L}$.
• $T$ is temperature in $\text{K}$.

If you read the question, you should find that $\Delta P \ne 0$, $\Delta V \ne 0$, and $\Delta T \ne 0$. Thus, we can write two states...

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = n R = \frac{{P}_{2} {V}_{2}}{{T}_{2}}$

giving the so-called "Combined Gas Law".

And so, the pressure must be:

$\textcolor{b l u e}{{P}_{2}} = \frac{{P}_{1} {V}_{1}}{{T}_{1}} \cdot {T}_{2} / {V}_{2}$

= (("2.95 atm")(7.456 cancel"L"))/(379 cancel"K") cdot (212 cancel"K")/(4.782 cancel"L")

$=$ $\textcolor{b l u e}{\underline{\text{2.57 atm}}}$

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.