Let 70.0 J of heat be added to a particular ideal gas. As a result, its volume changes from 65.0 cm3 to 169.0 cm3 while the pressure remains constant at 1.7 atm. By how much did the internal energy of the gas change?

Answer 1

P*deltaV= external work -work internal

1.7atm*104cm^3=-work done internally+70J

101.3E3*1.7*104E-6 J-70J=-work done internally
work internal=-.1013*1.7*104*6+70J
= -17.9+70=52.1 J

check that.