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# Adiabatic Process - Thermodynamic Chemistry Notes | EduRev

## Physical Chemistry

Created by: Asf Institute

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

The document Adiabatic Process - Thermodynamic Chemistry Notes | EduRev is a part of the Chemistry Course Physical Chemistry.
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Adiabatic reversible work, heat and energy

We know that         PVr = constant = K

we know,

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

PV= K
P1V1r = K
P2V2r = K
P1 = KV1â€“r
P2 = KV2â€“r
P1V1 = RT1
and P2V2 = RT2

Another way to find the work done in adiabatic process.

We know that dâˆµ = dU â€“ Ï‰
dâˆµ = 0 because process is adiabatic
dU = Ï‰
â‡’ Ï‰ = dV = CVdT
and

dâˆµ = dU â€“ w
dâˆµ = 0 for adiabat ic process

From first law
dq = dU â€“ Ï‰
âˆµ dq = 0
â‡’ 0 = dUâ€“ Ï‰

â‡’

and

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 = â€“P(V2 â€“ V1)
= â€“1 atm (20 dm3 â€“ 10 dm3) = â€“10 dm3 atm
âˆµ R = 8.314 JKâ€“1 molâ€“1 = 8.02 Ã— 10â€“2 dm3 atm Kâ€“1 molâ€“1
âˆ´ 1 dm3 atm =

1 dm3 atm = 1.013 Ã— 102 J

Put this value in equation we get

W = â€“ 10 dm3 atm
= â€“ 10 Ã— 1.013 Ã— 10J = â€“1013 J
i.e. W = â€“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, Î”U & Î”H.
Sol.
(a) For an adiabat ic process q = 0

i.e.

CV for a diatomic molecule =

(assuming no contribution for vibration)

Substituting the values, we get

Solving for T2, we get

T2 = 270 K

thus

(270 K -300 K)

= â€“1247.1 J

= â€“1247.1 J + 2 mol Ã— 8.314 JKâ€“1 molâ€“1 (â€“30 K)
= â€“1745.9 J

or              CP â€“ CV = R

â‡’

and

= â€“1745.9 J

Note

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

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

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

Total energy (U) for non linear molecular is:
(at high temperature)
(at room temperature)

We know that
â‡’

CV for Na:

CV for Cl2

(for linear molecule)

= 2.7 RT
CV = 2R

Cfor SO2

for 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, Î”H & Î”U.
Sol.
Amount of

i.e., n = 0.714 mol
T1 = 300 K
V= 20 dm3
V2 = 10 dm3
T= ?

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

â‡’

Thus

and for linear molecule

Thus

T1 = 300 K Ã— 1.32 = 396 K
Hence,

= 1424.69 J
Î”H = nCP(T2 - T1)

= 0.714 mol  96 K

= 1994.56 J

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