Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

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Chemistry : Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

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ADIABATIC PROCESS

Adiabatic reversible work, heat and energy 

We know that         PVr = constant = K

 Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

we know,

 Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

Because process is reversible than integration will possible from V1 to V2

 Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

PV= K
P1V1r = K
P2V2r = K
P1 = KV1–r
P2 = KV2–r
P1V1 = RT1
and P2V2 = RT2

Another way to find the work done in adiabatic process.

 We know that d∵ = dU – ω
d∵ = 0 because process is adiabatic
dU = ω
⇒ ω = dV = CVdT
and

 Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

(B)  Adiabatic reversible heat:

d∵ = dU – w
d∵ = 0 for adiabat ic process

(C) Adiabatic reversible energy:
From first law
dq = dU – ω
∵ dq = 0
⇒ 0 = dU– ω

⇒ Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

and Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

 

Problem. A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 dm3 to a volume of 20 dm3. In this process it absorbs 800 J of thermal energy from its surrounding calculate DU for the process in joules (because process is irreversible) 

Sol.
W = –P(V2 – V1)
= –1 atm (20 dm3 – 10 dm3) = –10 dm3 atm
∵ R = 8.314 JK–1 mol–1 = 8.02 × 10–2 dm3 atm K–1 mol–1
∴ 1 dm3 atm =  Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

1 dm3 atm = 1.013 × 102 J

Put this value in equation we get

W = – 10 dm3 atm
= – 10 × 1.013 × 10J = –1013 J
i.e. W = –1013 J
 

Problem.  (a)  2.0 mole of an ideal diatomic gas at 300 K and 0.507 MPa are expanded adiabatically to a final pressure of 0.203 MPa against a constant pressure of 0.101 MPa. Calculate the final temperature, q, w, ΔU & ΔH.
 Sol.
(a) For an adiabat ic process q = 0

Adiabatic Process - Thermodynamic Chemistry Notes | EduRevAdiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

i.e.Adiabatic Process - Thermodynamic Chemistry Notes | EduRev Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

CV for a diatomic molecule =   Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

(assuming no contribution for vibration)

 Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

Substituting the values, we get

 Adiabatic Process - Thermodynamic Chemistry Notes | EduRevAdiabatic Process - Thermodynamic Chemistry Notes | EduRev

Solving for T2, we get

 T2 = 270 K

thus 

Adiabatic Process - Thermodynamic Chemistry Notes | EduRev(270 K -300 K)

= –1247.1 J

Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

= –1247.1 J + 2 mol × 8.314 JK–1 mol–1 (–30 K)
= –1745.9 J

or              CP – CV = R

Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
andAdiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

= –1745.9 J

 

Note 

Degree of freedom: There are three types of degree of freedom

(1) Translat ional (2) Rotational (3) Vibrational

⇒ If total no. of atoms are N then no. of degree of freedo m is 3 N.
⇒ For any atom or molecule the degree of freedom for translat ional is three (3).
⇒ For any atom or linear mo lecule the degree o f freedom for is 2 (two) and for non linear it is three (3).
⇒ For any atom or linear mo lecule the degree of freedom for vibrat ional is (3N – 5) and for non linear it is (3N – 6).
⇒ The energy contribution for each degree of freedom is 1/2 KT/molecule or 1/2 RT/mole.
⇒ The vibrational degree of freedo m is affected by temperature and it is 0.2 RT at room temperature and RT at high temperature.

 

Problem.  Calculate CV for Na, Cl2 & SO2?
 Sol.
Total internal energy (U) for linear mo lecule is:
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev (at high temperature)
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev (at room temperature)

Total energy (U) for non linear molecular is:
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev (at high temperature)
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev (at room temperature)

We know that
⇒     Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

 CV for Na:

Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

CV for Cl2 

Adiabatic Process - Thermodynamic Chemistry Notes | EduRev(for linear molecule)

= 2.7 RT
CV = 2R

 

Cfor SO2

Adiabatic Process - Thermodynamic Chemistry Notes | EduRevfor non linear molecule)
= 3.3 RT
CV = 3.3 R

 

Problem.  20 gm of N2 at 300 K is compressed reversible and adiabatically from 20 dm3 to 10 dm3. Calculate the final temperature, q, w, ΔH & ΔU.
 Sol. 
Amount of

 Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

i.e., n = 0.714 mol
T1 = 300 K
V= 20 dm3
V2 = 10 dm3
T= ?
For adiabatic reversible process:

T1(V1)r-1 = T2 (V2)r -1

 Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

Thus   Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

and for linear molecule  Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

Thus
Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

T1 = 300 K × 1.32 = 396 K
Hence, 

 Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

= 1424.69 J
ΔH = nCP(T2 - T1)

= 0.714 mol Adiabatic Process - Thermodynamic Chemistry Notes | EduRev 96 K

= 1994.56 J

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