Question

Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3 , is stirred until its temperature is 500 K. Assuming the ideal gas model for the air, and ignoring kinetic and potential energy, determine (a) the final pressure, in bar, (b) the work, in kJ, and (c) the amount of entropy produced, i

Answer 1

Answer:

a)P₂ = 3.333 bar

b)W= 0 J  

c)ΔS=1.568  KJ/K

Explanation:

Given that

T₁= 300 K

P₁=2 bar

V₁=V₂= 2 m³

T₂ = 500 K

Tank is rigid it ,means that ,volume is constant

We know that ideal gas equation

P V = m R T

P₁ V₁ = m R T₁

$m=\dfrac{200\times 2}{0.287\times 300}\ kg$

m=4.64 kg

For constant volume

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

Now by putting the values

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

$\dfrac{2}{300}=\dfrac{P_2}{500}$

P₂ = 3.333 bar

We know that

Work done W= P.ΔV

In constant volume process ΔV = 0

W= 0 J

Change in entropy at constant volume given as

$\Delta S=mC_vln\dfrac{T_2}{T_1}$

$\Delta S=4.64\times 0.71ln\dfrac{500}{300}$

ΔS=1.568  KJ/K