P₁V₁ = P₂V₂;

                             (6.5atm)(1.0L) = (1.0atm)V₂,

                   which gives V₂ = 6.5L.

          (a) Work is done only in the constant pressure process:

                              W = P_a( V_b - V_a)   

                                  = (5atm)(1.013 x 10⁵N/m².atm)[(0.660 - 0.400) x 10m^-3] 

                                  = 1.3 x 10²J.

           (b) Because the initial and final temperature are the same we know that DU = 0.

                We use the first law of thermodynamics to find the heat flow for the entire process.

                             DU = Q - W;

                             0 = Q - ( + 1.3 x 10²J), which gives Q = + 1.3 x 10²J.