The transformation is isobaric (constant pressure), so the work done by the gas is the product between its pressure and the variation of volume:
[tex]W=p \Delta V=(2500 Pa)(3.0 m^3-1.0 m^2 ) = +5000 J =+5.0 kJ [/tex]
The heat transferred to the gas is
[tex]Q=+11.2 kJ[/tex]
So we can use the first law of thermodynamics to compute the variation of internal energy of the gas:
[tex]\Delta U = Q-W=11.2 kJ-5.0 kJ=+6.2 kJ = +6200 J[/tex]
where the positive sign means the internal energy of the gas has increased.
