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Polytropic Process & Free Expansion

The document discusses several thermodynamic processes including: 1) A polytropic process where work done is related to pressure and volume changes. 2) An example problem solving for the final pressure, temperature, work and heat of compressing air in a polytropic process. 3) Free expansion or unresisted expansion where a gas expands into an evacuated vessel, doing no work and exchanging no heat. 4) An example problem solving for the final temperature and pressure of gases mixing by free expansion after a membrane is punctured.

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285 views6 pages

Polytropic Process & Free Expansion

The document discusses several thermodynamic processes including: 1) A polytropic process where work done is related to pressure and volume changes. 2) An example problem solving for the final pressure, temperature, work and heat of compressing air in a polytropic process. 3) Free expansion or unresisted expansion where a gas expands into an evacuated vessel, doing no work and exchanging no heat. 4) An example problem solving for the final temperature and pressure of gases mixing by free expansion after a membrane is punctured.

Uploaded by

yeng botz
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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LECTURE 10

Polytropic Process

W =  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)/(-1) + (P1v1- P2v2)/(n-1)

= R (T1 – T2){1/(n-1) – 1/(-1)}

=(P1v1- P2v2)/(n-1) {( -n)/(-1)}

=w.{ ( -n)/(-1)}
Problem: Air (ideal gas with  = 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.{ ( -n)/(-1)}


= -9.977 (1.4 – 1.25)/(1.4-1)
= -3.742 kJ/mol
Unresisted or Free expansion

In an irreversible process, w   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 C v 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, U = 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

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