Here's a nice alternative to converting the pressure to "Pa"Pa and the volume to "m"^3m3 in order to get the answer in Joules.
In your case, the expression of the work done by the system is
W = -P_("exterior") * DeltaVW=−Pexterior⋅ΔV
You can first calculate the work in "L" * "atm"L⋅atm and then convert to "Joules"Joules by using two expressions of the ideal gas constant, "R"R.
Since "R"R is both equal to
"R" = 8.314 "J"/("mol" * "K")R=8.314Jmol⋅K, (1) and
"R" = 0.08206 ("atm" * "L")/("mol" * "K")R=0.08206atm⋅Lmol⋅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.
