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Three m3 of air in a rigid, insulated container fitted with a paddle wheel is initially at 295 K, 200 kPa. The air receives 1546 kJ of work from the paddle wheel. Assuming the ideal gas model, determine for the air (a) the mass, in kg, (b) final temperature, in K, and (c) the amount of entropy produced, in KJ/K.

Respuesta :

Answer:

m = 7.086 kg

[tex]T_2 = 597.99 K[/tex]

[tex]S_{gen} = 3.605 kJ/K[/tex]

Explanation:

Given data:

volume of sir - 3 m^3

temperature is t = 295 K

Pressure is = 200 kPa

for air , R =  0.287 kJ/kg k

a) from ideal gas equation we have

PV = mRT

solving for m

[tex]m = \frac{PV}{RT} =\frac{200 \times 3}{0.287 \times 295}[/tex]

m = 7.086 kg

b) by energy balance principle

[tex]E_{in} -E_{out} = \Delta E[/tex]

[tex]W_{paddle} - 0 = mc_v (T_2 -T_1)[/tex]

[tex]1546 - 0 = 7.086 \times 0.72(T_2 - 295)[/tex]

[tex]T_2 = 597.99 K[/tex]

C)ENTROPY

[tex]S_{gen} = \Delta S_{system}[/tex]

             [tex]= m c_v ln \frac{T_2}{T_1}[/tex]

             [tex]= 7.086 \times 0.72 ln \frac{597.99}{295}[/tex]

[tex]S_{gen} = 3.605 kJ/K[/tex]

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