Here's a nice alternative to converting the pressure to #"Pa"# and the volume to #"m"^3# in order to get the answer in Joules.
In your case, the expression of the work done by the system is
#W = -P_("exterior") * DeltaV#
You can first calculate the work in #"L" * "atm"# and then convert to #"Joules"# by using two expressions of the ideal gas constant, #"R"#.
Since #"R"# is both equal to
#"R" = 8.314 "J"/("mol" * "K")#, (1) and
#"R" = 0.08206 ("atm" * "L")/("mol" * "K")#, (2)
you can divide the two values and come up with a conversion factor that'll take you from atm L to Joules
#((1))/((2)) => (8.314 "J"/(cancel("mol" * "K")))/(0.082("atm" * "L")/(cancel("mol" * "K"))) = "8.314 J"/("0.082 atm" * "L")#
So, if you calculate the work done by the system this way you'll get
#"W" = -"4.1 atm" * (70.8-24.8) * 10^(-3)"L" = "-0.18856 atm" * "L"#
In Joules, this would be
#-0.1886cancel("atm" * "L") * "8.314 J"/(0.08206cancel("atm" * "L")) = "-19.1 J"#
So, if you ever need a quick way to get to Joules from atm L, use the ratio of the two values for the ideal gas constant.
