Question

Nitrogen (N2) undergoes an internally reversible process from 6 bar, 247°C during which pν1.2 = constant. The initial volume is 0.1 m3 and the work for the process is 50 kJ. Assuming ideal gas behavior, and neglecting kinetic and potential energy effects, determine heat transfer, in kJ, and the entropy change, in kJ/K.

Answer 1

Answer:

$Q=25\ kJ$

$\Delta s= 0.2885 J.K^{-1}$

Explanation:

Given:

• Initial pressure of nitrogen, $P_1=6\times 10^5\ Pa$

• initial temperature, $T_1=247+273=520\ K$

• polytropic index, $n=1.2$

• initial volume, $V_1=0.1\ m^3$

• work done in the process, $W=50000\ J$

For heat interaction during the polytropic process we have:

$Q=W[\frac{\gamma -n}{\gamma-1} ]$

$Q=50\times[\frac{1.4-1.2}{1.4-1} ]$

$Q=25\ kJ$

For ideal gas we have the Gas Law:

$P_1.V_1=m.R.T_1$

$6\times 10^5\times 0.1=m.R\times 520$

$m.R=115.385\ J.K^{-1}$

For work we have the relation:

$W=m.R.\frac{(T_1-T_2)}{(n-1)}$

putting respective values

$50000=115.385\times \frac{(520-T_2)}{(1.2-1)}$

$T_2=433.33\ K$

We know entropy change:

$\Delta s=\frac{dQ}{dT}$

$\Delta s=\frac{25}{520-433.33}$

$\Delta s= 0.2885 J.K^{-1}$