Answer:
Work: 4.0 kJ, heat: 4.25 kJ
Explanation:
For a gas transformation at constant pressure, the work done by the gas is given by
[tex]W=p(V_f -V_i)[/tex]
where in this case we have:
[tex]p = 2 bar = 2\cdot 10^5 Pa[/tex] is the pressure
[tex]V_i = 0.1 m^3[/tex] is the initial volume
[tex]V_f = 0.12 m^3[/tex] is the final volume
Substituting,
[tex]W=(2\cdot 10^5)(0.12-0.10)=4000 J = 4.0 kJ[/tex]
The 1st law of thermodynamics also states that
[tex]\Delta U = Q-W[/tex]
where
[tex]\Delta U[/tex] is the change in internal energy of the gas
Q is the heat absorbed by the gas
Here we know that
[tex]\Delta U = +0.25 kJ[/tex]
Therefore we can re-arrange the equation to find the heat absorbed by the gas:
[tex]Q=\Delta U + W = 0.25 kJ + 4.0 kJ = 4.25 kJ[/tex]