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**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 V_{1} to V_{2}

PV^{r }= K

P_{1}V_{1}^{r} = K

P_{2}V_{2}^{r} = K

P_{1} = KV_{1}^{â€“r}

P_{2} = KV_{2}^{â€“r}

P_{1}V_{1} = RT_{1}

and P_{2}V_{2} = RT_{2}

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

**(B) Adiabatic reversible heat:**

dâˆµ = dU â€“ w

dâˆµ = 0 for adiabat ic process

**(C) Adiabatic reversible energy:**

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(V_{2} â€“ V_{1})

= â€“1 atm (20 dm^{3} â€“ 10 dm^{3}) = â€“10 dm^{3} atm

âˆµ R = 8.314 JK^{â€“1} mol^{â€“1} = 8.02 Ã— 10^{â€“2 }dm^{3} atm K^{â€“1 }mol^{â€“1}

âˆ´ 1 dm^{3} atm =

1 dm^{3} atm = 1.013 Ã— 10^{2} J

Put this value in equation we get

W = â€“ 10 dm^{3} atm

= â€“ 10 Ã— 1.013 Ã— 10^{2 }J = â€“10^{13 }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.

C_{V} for a diatomic molecule =

(assuming no contribution for vibration)

Substituting the values, we get

Solving for T_{2}, we get

T_{2} = 270 K

thus

(270 K -300 K)

= â€“1247.1 J

= â€“1247.1 J + 2 mol Ã— 8.314 JK^{â€“1} mol^{â€“1} (â€“30 K)

= â€“1745.9 J

or C_{P} â€“ C_{V} = 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 C _{V} for Na, Cl_{2} & SO_{2}?** Total internal energy (U) for linear mo lecule is:

Sol.

(at high temperature)

(at room temperature)

Total energy (U) for non linear molecular is:

(at high temperature)

(at room temperature)

We know that

â‡’

C_{V} for Na:

**C _{V} for Cl_{2}: **

(for linear molecule)

= 2.7 RT

C_{V} = 2R

**C _{V }for SO_{2}: **

for non linear molecule)

= 3.3 RT

C_{V} = 3.3 R

**Problem. 20 gm of N _{2} at 300 K is compressed reversible and adiabatically from 20 dm^{3} to 10 dm^{3}. Calculate the final temperature, q, w, Î”H & Î”U.**Amount of

Sol.

i.e., n = 0.714 mol

T_{1} = 300 K

V_{1 }= 20 dm^{3}

V_{2} = 10 dm^{3}

T_{2 }= ?

For adiabatic reversible process:

T_{1}(V_{1})^{r-1} = T_{2} (V_{2})^{r -1}

â‡’

Thus

and for linear molecule

Thus

T_{1} = 300 K Ã— 1.32 = 396 K

Hence,

= 1424.69 J

Î”H = nCP(T_{2} - T_{1})

= 0.714 mol 96 K

= 1994.56 J

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