# If a gaseous system does "230 J" of work on its surroundings against an external pressure of "1.70 atm", to what final volume does the gas expand from "0.300 L"?

Sep 19, 2017

${V}_{2} = \text{1.64 L}$.

You can see that this is a work problem (what gave it away?). The two common errors are sign conventions and unit conversion.

Recall that work (in General Chemistry) is defined as:

$w = - P \Delta V = - P \left({V}_{2} - {V}_{1}\right)$

where $P$ is the external pressure (assumed constant) and $V$ is the volume of the ideal gas in the piston.

You know that the system does $\text{230 J}$ of work on its surroundings. Since the system does work, i.e. work was done by the gas, the work is negative:

$w < 0$

Thus, since pressure is always positive, we expect that $\Delta V > 0$. That makes sense because the piston expands, in the first sentence.

Therefore, the final volume is found by some algebra. The initial setup is then:

-"230 J" = -overbrace("1.70 atm")^(P) xx overbrace((V_2 - "0.300 L"))^(DeltaV)

Do note that the units MUST work out. They do not right now. You can use the conversion factor:

$\left(\text{8.314472 J/"cancel("mol"cdot"K"))/("0.082057 L"cdot"atm/"cancel("mol"cdot"K")) = "101.326 J"/("L"cdot"atm}\right)$

Therefore, the right-hand side now has units of $\text{J}$ as required:

-"230 J" = -overbrace("1.70 atm")^(P) xx overbrace((V_2 - "0.300 L"))^(DeltaV) xx "101.326 J"/("L"cdot"atm")

Divide through by $P$ and the conversion factor:

${\left(- 230 \cancel{\text{J")/(-1.70 cancel"atm") xx ("L"cdotcancel"atm")/(101.326 cancel"J") = V_2 - underbrace("0.300 L}}\right)}_{{V}_{1}}$

$\text{1.335 L" + "0.300 L} = {V}_{2}$
Thus, the final volume is $\textcolor{b l u e}{\text{1.64 L}}$ to three sig figs.