# 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"#?

##### 1 Answer

#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 **work was done by the gas**, the work is

*negative*:

#w < 0#

Thus, since pressure is *always* positive, we expect that *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

#-"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

#(-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