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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 $"0.300 L"$?

Md Ariful Islam

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Md Ariful Islam

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$V_2 = "1.64 L"$.

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:

$w = -PDeltaV = -P(V_2 - V_1)$

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.

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