Question

Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 3 m3, is stirred until its temperature is 600 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. (c) the amount of entropy produced, in kJ/K.

Answer 1

Answer:

a)P₂ =4 bar

b)W= - 1482.48 KJ

It means that work done on the system.

c)S₂ - S₁ = 3.42 KJ/K

Explanation:

Given that

T₁ = 300 K   ,V₁ = 3 m³  ,P₁=2 bar

T₂ = 600 K ,V₂=V₁ 3 m³

Given that tank is rigid and insulated.It means that volume of the gas will remain constant.

Lets take the final pressure = P₂

For ideal gas  P V = m R T

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

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

$P_2=\dfrac{600}{300}\times 2$

P₂ =4 bar

Internal energy

ΔU = m Cv ΔT

Cv=0.71 KJ/kg.k for air

$m=\dfrac{PV}{RT}$

$m=\dfrac{200\times 3}{0.287\times 300}\ kg$

m= 6.96 kg

ΔU= 6.96 x 0.71 x (600 - 300)

ΔU=1482.48 KJ

From first law

Q= ΔU + W

Q= 0  Insulated

W = - ΔU

W= - 1482.48 KJ

It means that work done on the system.

Change in the entropy

$S_2-S_1=mC_v\ \ln\dfrac{T_2}{T_1}$

$S_2-S_1=6.96\times 0.71\ \ln\dfrac{600}{300}$

S₂ - S₁ = 3.42 KJ/K