Question

Nitrogen at an initial state of 420 K, 116 kPa, and 3 m3 is compressed slowly in an isothermal process to a final pressure of 455 kPa. Determine the work done during this process in kJ

Answer 1

-47.5716 kJ

Step-by-step explanation:

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[
`(P_(1) )/(P_(2))`] ---------------------(i)

From the question;

Nitrogen gas =
`N_(2)`

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 [
`(116*10^(3) )/(455*10^(3) )`]

W = 116 x 10³ x 3 ln [
`(116)/(455)`]

W = 348000 ln [
`(116)/(455)`]

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.

Answer 2

-475.68kJ

Step-by-step 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