Question

Nitrogen at an initial state of 300 K, 150 kPa, and 0.45 m3 is compressed slowly in an isothermal process to a final pressure of 800 kPa. Determine the work done during this process

Answer 1

W = -113.0 kJ.

Step-by-step explanation:

The work done during the isothermal process is the following:

`W_(a-b) = p_(a)V_(a)ln((p_(a))/(p_(b)))`

Where:

`p_(a)` = 150 kPa

`V_(a)` = 0.45 m³

`p_(b)` = 800 kPa

`W_(a-b) = 150 kPa*0.45 m^(3)ln((150 kPa)/(800 kPa)) = -113.0 \cdot 10^(3) J`

Therefore, the work done during this process is -113.0 kJ.

I hope it helps you!