Question

2.4 kg of nitrogen at an initial state of 285K and 150 kPa is compressed slowly in an isothermal process to a final pressure of 600 kPa. Determine the work.

Answer 1

W=-280.67 KJ

Step-by-step explanation:

Given that

Initial pressure = 150 KPa

Final pressure = 600 KPa

Temperature T= 285 K

Mass m=2.4 Kg

We know that ,work in isothermal process given as

`W=mRT\ ln(P_1)/(P_2)`

Gas constant for nitrogen gas R=0.296 KJ/kgK

Now by putting the values

`W=mRT\ ln(P_1)/(P_2)`

`W=2.4* 0.296* 285\ ln(150)/(600)`

W=-280.67 KJ

Negative sign indicates that it is compression process and work is done on the gas.