Question

a 0.8 ,^3 insulated rigid tank contains 1.54 kg of carbon dioxide at 100 kPa. Now paddle wheel work is done on the system until pressure in the tank rises to 135 kpa assuming the ideal gas model and negligible kinetic and potential energy effects determine the paddle wheel work dine during the process and the energychange during this process

Answer 1

Answer:

The answer is "$W= 100.44 \ KJ\ \ \ \W_{win}=7.23 \ KJ$"

Explanation:

Please find the complete question in the attached file.

From of the ideal gas relation that initial and the last temperatures were determined:  

$T_1 = \frac{P_1 V}{m R}$

    $= \frac{100 \times 0.8}{1.54 \times 0.1889} \\\\ = 275 \ K$

$T_2 = \frac{P_2 V}{m R}$

    $= \frac{135 \times 0.8}{1.54 \times 0.1889} \\\\ = 317 \ K$

In the initial and final states, the internal energies for given temperatures are described from A-20 by means of intelmpolation and divided by the carlxon molar mass.  

$u_1 = 141.56 \frac{KJ}{kg}\\\\u_2 = 206.78 \frac{KJ}{kg}$

The real job is just the difference between internal energies:  

$W = m(u_2 - u_1)$

    $= 1.54(206.78 -141.56) \ kJ \\\\ =100.44 \ kJ$

In the initial and final states, the zero entries are as determined as internal energies:

$S_1^{\circ} =4.788 \frac{KJ}{kg K}\\\\S_2^{\circ} =5.0478 \frac{KJ}{kg K}$

From its energy increase, the minimum work required is determined:

$W_{min} = m(u_2-u_1 - T_0(s_2 -S_1))\\\\=W-mT_0(S_2^{\circ}- S_1^{\circ} -R \In \frac{P_2}{P_1})\\\\= 100.44kJ -1.54 \times 298( 5.0478-4.788-0.1889 \In \frac{135}{100})\\\\=7.23\ KJ$