Question

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"`?

Answer 1

`V_2 = "1.64 L"`.

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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 = -PDeltaV = -P(V_2 - V_1)`

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 `"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 `DeltaV > 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:

`("8.314472 J/"cancel("mol"cdot"K"))/("0.082057 L"cdot"atm/"cancel("mol"cdot"K")) = "101.326 J"/("L"cdot"atm")`

Therefore, the right-hand side now has units of `"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:

`(-230 cancel"J")/(-1.70 cancel"atm") xx ("L"cdotcancel"atm")/(101.326 cancel"J") = V_2 - underbrace("0.300 L")_(V_1)`

Add over the initial volume:

`"1.335 L" + "0.300 L" = V_2`

Thus, the final volume is `color(blue)("1.64 L")` to three sig figs.