Question

An insulated piston-cylinder device initially contains 300 L of air at 120 kPa and 17 oC. Air is now heated for 15 min by a 200-w resistance heater placed inside the cylinder. The pressure of air is maintained constant during this process. Determine the entropy change of air assuming (a) ) constant specific heats and (b) ) variable specific heats.

Answer 1

Answer:

a)  $\Delta S = 0.386 \ \frac{kJ}{K}$

b)  $\Delta S =0.395 \ \frac{kJ}{K}$

Explanation:

a)

Specific heat formula, we have:

$mc(\theta_1-\theta_2)=Wt$

Where

m is mass

c is specific heat capacity of air at specific temp

$\theta$ is the temperature change

W is the power

t is the time

The ideal gas law is:

$PV=nRT$

Where

P is the pressure

V is the volume

n is number of moles

T is temperature

R is the ideal gas constant

First, lets solve for $\theta_2$ in the 1st equation, remembering to use ideal gas law in it to have the variables that are given in the problem. Shown below:

$mc(\theta_2-\theta_1)=Wt\\mc\theta_2-mc\theta_1=Wt\\mc\theta_2=mc\theta_1 + Wt\\\theta_2=\frac{mc\theta_1 + Wt}{mc}\\\theta_2=\theta_1+\frac{Wt}{mc}\\\theta_2=\theta_1+\frac{WtR\theta_1}{PVc}\\\theta_2=\theta_1(1+\frac{WtR}{PVc})$

Now, we know

Theta_1 is 17 celsius, which in Kelvin is 17 + 270 = 290K

Power is 200

Time is 15 mins = 15 * 60 = 900 seconds

R is 287 J/kg K

P is 120

V is 300 L

Specific heat of air at 290K is about 1005

Substituting we get:

$\theta_2=\theta_1(1+\frac{WtR}{PVc})\\\theta_2=290(1+\frac{200*900*287}{120*300*1005})\\\theta_2=704 \ K$

Now, Entropy Change is:

$\Delta S =mc Ln(\frac{\theta_2}{\theta_1})$

Again using the substitution equations, we have:

$\Delta S = \frac{PV}{R\theta_1}cLn(1+\frac{WRt}{PVc})$

We know what the variables mean, we substitute the respective values, to get:

$\Delta S = 0.386 \ \frac{kJ}{K}$

b)

For variable specific heats, we need entropy value from entropy table:

s_2 = 2.58044

s_1 = 1.66802  [1.05 * 290 K]

The formula is:

$\Delta S = m(s_2-s_1)\\\Delta S = \frac{PV}{R\theta_1}(s_2-s_1)$

Substituting the values, we find the answer:

$\Delta S = \frac{PV}{R\theta_1}(s_2-s_1)\\\Delta S = \frac{120*300}{287*290}(2.58044-1.66802)\\\Delta S =0.395 \ \frac{kJ}{K}$