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

Answer:

W=-280.67 KJ

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

$W=mRT\ ln\dfrac{P_1}{P_2}$

Gas constant for nitrogen gas R=0.296 KJ/kgK

Now by putting the values

$W=mRT\ ln\dfrac{P_1}{P_2}$

$W=2.4\times 0.296\times 285\ ln\dfrac{150}{600}$

W=-280.67 KJ

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