An isolated rigid container consists of two equalvolume comp

An isolated rigid container consists of two equal-volume compartments separated by a diaphragm. Initially, the left compartment contains He at T_1and p_1 and the right compartment contains Air at T_2 and p_2. Subsequently, the diaphragm is burst and the mixture is allowed to reach equilibrium. Calculate the specific entropy produced, sigma/m in kJ/kg/K. where m is the total final mass of the mixture at equilibrium for the following processes and complete the table: * m_f = molecular mass of the final equilibrium mixture

Solution

Case 1:

Given,

T₁ = 400 K

T₁ = 320 K

p₁ = 5 atm = 5*101325 N/m² = 506625 N/m²

p₂ = 1 atm = 1*101325 N/m² = 101325 N/m²

Molar mass = 20.083

Let, the volume of each compartment = v m³

Hence, volume of total container = 2v m³

For ideal gas, p*v = n*R*T ; where n = number of moles and R = universal gas constant = 8314.3 Nm/kg mol K.

So, the number of moles in compartment 1, n₁ = (p₁*v)/( 8314.3* T₁) = (506625*v)/( 8314.3* 400)

                                                                            => n₁ = 0.150 v

The number of moles in compartment 2, n₂ = (p₂*v)/( 8314.3* T₂) = (101325*v)/( 8314.3* 320)

                                                                            => n₂ = 0.038v

The number of moles after mixture, n = n₁ + n₂ = 0.150 v + 0.038v = 0.188 v

Now, mole fraction in compartment1, x₁ = n₁/( n₁ + n₂ ) = 0.150v/0.188 v = 0.798

And mole fraction in compartment 2, x₂ = 0.038v/0.188 v = 0.202

Now, final pressure of the mixture, p_f = (n*R*T_f)/(2v) = (0.188v*8314.3*360)/(2v) = 281356 N/m²

final temperature, T_f = (400+ 320)/2 = 360 K

Final mass = (m_f*p_f*2v)/(R* T_f) = (20.083*281356*2v)/(8314.3* 360) = 3.771v

Hence, the entropy produced during the process = - n*R(x₁ln x₁ + x₂ln x₂)

= - 0.188v*8314.3*(0.798 ln 0.798 + 0.202 ln 0.202) = 786.49 v

So, the entropy produced per unit mass = (786.49 v )/(3.771v) = 208.563 kJ/kg K

Case 2:

the number of moles in compartment 1, n₁ = (p₁*v)/( 8314.3* T₁) = (506625*v)/( 8314.3* 300)

                                                                            => n₁ = 0.200v

n₂ = (p₂*v)/( 8314.3* T₂) = (101325*v)/( 8314.3* 320)

                                                                            => n₂ = 0.029v

The number of moles after mixture, n = n₁ + n₂ = 0.229 v

Now, mole fraction in compartment1, x₁ = n₁/( n₁ + n₂ ) = 0.200v/0.229 v = 0.873

And mole fraction in compartment 2, x₂ = 0.029v/0.229v = 0.127

Now, final pressure of the mixture, p_f = (n*R*T_f)/(2v) = (0.229v*8314.3*360)/(2v) = 342716 N/m²

final temperature, T_f = (420+ 300)/2 = 360 K

Final mass = (m_f*p_f*2v)/(R* T_f) = (16.694*281356*2v)/(8314.3* 360) = 3.135v

Hence, the entropy produced during the process = - n*R(x₁ln x₁ + x₂ln x₂)

= - 0.229v*8314.3*(0.873 ln 0.873 + 0.127 ln 0.127) = 724.74 v

So, the entropy produced per unit mass = (724.74 v )/(3.135v) = 231.177 kJ/kg K