Answer:
Explanation:
Given
Initial Volume [tex]v_1=1\ m^3[/tex]
final Volume [tex]v_2=3\ m^3[/tex]
Heat added at constant Pressure [tex]Q=200\ kJ[/tex]
Decrease in Energy of System [tex]\Delta U=-340\ kJ[/tex]
According to First law of thermodynamics
[tex]Q=\Delta U+W[/tex]
[tex]W=Q-\Delta U[/tex]
[tex]W=200-(-340)[/tex]
[tex]W=540\ kJ[/tex]
Work done in a constant Pressure Process is given by
[tex]W=P\Delta V[/tex]
where P is the constant Pressure
[tex]540=P\times (3-1)[/tex]
[tex]P=270\ kPa[/tex]