Answer:
-475.68kJ
Explanation:
P1=116kPa
V1=3m³
T1=420K
P2=455kPa
Formula for workdone in isothermal process=P1×V1×In(p1/p2)
Workdone=116×3×in(116/455)
Workdone=-475.68kJ
The result is negative because work is done on the system
-47.5716 kJ
In an isothermal process, temperature is constant and the workdone (W) in compressing a gas from some initial pressure(P₁) and volume (V₁) to some other pressure (P₂) and volume (V₂) is given by;
W = P₁V₁ ln[[tex]\frac{P_{1} }{P_{2}}[/tex]] ---------------------(i)
From the question;
Nitrogen gas = [tex]N_{2}[/tex]
Initial and constant temperature = 420K
initial pressure P₁ = 116 kPa = 116 x 10³ Pa
initial volume V₁ = 3m³
P₂ = 455 kPa = 455 x 10³ Pa
Substitute these values into equation (i) as follows;
W = 116 x 10³ x 3 ln [[tex]\frac{116*10^{3} }{455*10^{3} }[/tex]]
W = 116 x 10³ x 3 ln [[tex]\frac{116}{455}[/tex]]
W = 348000 ln [[tex]\frac{116}{455}[/tex]]
W = 348000 x -1.367
W = -475716 J
Divide the result by 1000 to convert it to kJ
=> W = (-475716 / 1000) kJ
=> W = -47.5716 kJ
Therefore the work done in the isothermal process is -47.5716 kJ
Note: The negative value indicates that work is done on the system.
