An isolated rigid container consists of two equal-volume compartments separated

Question

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

Explanation / Answer

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, pf = (n*R*Tf)/(2v) = (0.188v*8314.3*360)/(2v) = 281356 N/m2

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

Final mass = (mf*pf*2v)/(R* Tf) = (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, pf = (n*R*Tf)/(2v) = (0.229v*8314.3*360)/(2v) = 342716 N/m2

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

Final mass = (mf*pf*2v)/(R* Tf) = (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

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