Chapter 6 - Sections 6.1-6.10 - Exercises - Problems by Topic - Page 288: 51

$-2.8\times10^{2}\,J$

$w=-P\Delta V= -1.1\,atm\times(2.5-0.0)L=-2.75\,atm\,L$ $=-2.75\,atm\,L\times\frac{10^{-3}m^3}{1 L}\times\frac{101325\,Pa}{1\,atm}\times\frac{1\,N/m^2}{1\,Pa}=-2.8\times10^{2}\,Nm=-2.8\times10^{2}\,J$