LECTURE 10

 

Polytropic Process

 

W = ς cdv/ vn

 

w = (P1v1- P2v2)/(n-1)

 

du = dq – dw

 

u2­ – u1 = q - (P1v1- P2v2)/(n-1)

 

u2­ – u1 = Cv  (T2 – T1) = q – w

 

q = R(T2 – T1)/(g-1) + (P1v1- P2v2)/(n-1)

 

= R (T1 – T2){1/(n-1) – 1/(g-1)}

 

=(P1v1- P2v2)/(n-1) {(g -n)/(g-1)}

 

=w.{ (g -n)/(g-1)}

 

 

 

 

 

 

Problem: Air (ideal gas with g = 1.4) at 1 bar and 300K is compressed till the final volume is one-sixteenth of the original volume, following a polytropic process Pv1.25 = const. Calculate (a) the final pressure and temperature of the air,  (b) the work done and (c) the energy transferred as heat per mole of the air.

 

Solution: (a) P1v11.25 = P2v21.25

 

P2 = P1(v1/v2)1.25 = 1(16)1.25 = 32 bar

T2 = (T1P2v2)/(P1v1) = (300 x 32 x 1)/(1x16)

                                 = 600K

 

(b) w =   (P1v1- P2v2)/(n-1)

      =  Ru(T1 – T2)/(n-1)

     = 8.314 (300 – 600)/(1.25-1) = -9.977 kJ/mol

 

(c) q = w.{ (g -n)/(g-1)}

        = -9.977 (1.4 – 1.25)/(1.4-1)

        = -3.742 kJ/mol

 

 

 

Unresisted or Free expansion

 

 

 

 

 

 

In an irreversible process, w Ή ς Pdv

 

Vessel A:  Filled with fluid at pressure

 

Vessel B:  Evacuated/low pressure fluid

 

Valve is opened:  Fluid in A expands and fills

both vessels A and B. This is known as unresisted expansion or free expansion. 

 

No work is done on or by the fluid.

 

No heat flows (Joule’s experiment) from the boundaries as they are insulated.

 

 

 

 

U2  = U1  (U = UA + UB)

 

 

 

 

 

 

 

 

 

 

 

 

 

            

 

 

 

 

 

 

 

 

Problem: A rigid and insulated container of 2m3 capacity is divided into two equal compartments by a membrane. One compartment contains helium at 200kPa and 127oC while the second compartment contains nitrogen at 400kPa and 227oC. The membrane is punctured and the gases are allowed to mix. Determine the temperature and pressure after equilibrium has been established. Consider helium and nitrogen as perfect gases with their Cv  as 3R/2 and 5R/2 respectively.

 

Solution:  Considering the gases contained in both the compartments as the system, W= 0 and Q = 0. Therefore, DU = 0 (U2 = U1)

 

Amount of helium = NHe = PAVA/RuTA

                               = 200 x 103 x 1/(8.314 x400)

                             = 60.14 mol.

Amount of nitrogen = NN2 = PBVB/RuTB

                              = 400 x 103 x 1/(8.314x500)

                              = 96.22 mol.

 

 

Let Tf be the final temperature after equilibrium has been established. Then,

 

[NCv(Tf-400)]He + [NCv(Tf-500)]N2 = 0

 

Ru[60.14(Tf-400)3 + 96.22(Tf-500)5 ] /2 = 0

 

Or,  Tf  = 472.73 K

 

The final pressure of the mixture can be obtained by applying the equation of state:

 

PfVf = (NHe + NN2)Ru­ Tf

 

2Pf  = (60.14 + 96.22) 8.314 (472.73)

 

or, Pf  = 307.27 kPa