Adiabatic Process & Degrees of Freedom | Physical Chemistry PDF Download

ADIABATIC PROCESS

Adiabatic reversible work, heat and energy 

We know that         PV^r = constant = K

we know,

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

PV^r = K
P₁V₁^r = K
P₂V₂^r = 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

(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

⇒ 

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 = –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 =  

1 dm³ atm = 1.013 × 10² J

Put this value in equation we get

W = – 10 dm³ atm
= – 10 × 1.013 × 10² J = –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

i.e. 

C_V 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)

= –1247.1 J

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

or              C_P – C_V = R

⇒

and

= –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 C_V 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
⇒     

 C_V for Na:

C_V for Cl₂:  

(for linear molecule)

= 2.7 RT
C_V = 2R

C_V for SO₂: 

for non linear molecule)
= 3.3 RT
C_V = 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

i.e., n = 0.714 mol
T₁ = 300 K
V₁ = 20 dm³
V₂ = 10 dm³
T₂ = ?
For adiabatic reversible process:

T₁(V₁)^r-1 = T₂ (V₂)^{r -1}

 

⇒

Thus   

and for linear molecule  

Thus

T₁ = 300 K × 1.32 = 396 K
Hence, 

= 1424.69 J
ΔH = nCP(T₂ - T₁)

= 0.714 mol  96 K

= 1994.56 J