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

Answer:

W = -113.0 kJ.

Explanation:

The work done during the isothermal process is the following:        

$W_{a-b} = p_{a}V_{a}ln(\frac{p_{a}}{p_{b}})$

Where:

$p_{a}$ = 150 kPa

$V_{a}$ = 0.45 m³

$p_{b}$ = 800 kPa

$W_{a-b} = 150 kPa*0.45 m^{3}ln(\frac{150 kPa}{800 kPa}) = -113.0 \cdot 10^{3} J$

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

I hope it helps you!