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**ADIABATIC EXPANSION**

In adiabatic expansion, no heat is allowed to enter or leave the system, hence, q = 0. When this value is substituted in first law of thermodynamics, ∆U= q + w, we get ∆U = w.

In expansion,

- Work is done by the system on the surroundings, hence, w is negative.
- Accordingly ∆U is also negative, i.e., internal energy decreases and therefore, the temperature of the system falls.

In case of compression,

- ∆U is positive, i.e., internal energy increases and therefore,
- The temperature of the system rises.

The molar specific heat capacity at constant volume of an ideal gas is given by.

d_{U}=C_{v} ·d_{T}

∆U=C_{v}∆T

So, w=∆U=C_{v}∆T

The value of ∆T depends upon the process whether it is reversible or irreversible.

**REVERSIBLE ADIABATIC EXPANSION**

Let P be the external pressure and ∆V the increase in volume. Thus, the work done by the system is

w = ∆U = – PdV

We know C_{P} – C_{V} = R

Thus, knowing Ɣ, V_{1} , V_{2} and initial temperature, T_{1} , the final temperature, T_{2}, can be readily evaluated.

Thus, knowing Ɣ, P_{1}, P_{2} and initial temperature, T_{1}, the final temperature, T_{2}, can be readily evaluated.

**IRREVERSIBLE ADIABATIC EXPANSION**

In free expansion, the external pressure is zero, i.e. , work done is zero. Accordingly, ∆U which is equal to w is also zero. If ∆U is zero, ∆T should be zero. Thus, in free expansion (adiabatically),

∆T = 0, ∆U= 0, w = 0 and ∆H = O.

In intermediate expansion, the volume changes from V_{1} to V_{2} against external pressure, P_{ext}

or**Example 1. Two moles of an ideal monoatomic gas at NTP are compressed adiabatically and reversibly to occupy a volume of 4.48 dm ^{3}. Calculate the amount of work done, ∆U, final temperature and pressure of the gas. Cv for ideal gas 12.45J K ^{-1} mol^{-1}.**

Initial volume, V

Initial pressure, P

Initial temperature, T

Final volume, V

Let the final pressure be P

Applying

or

or

P_{2} = (10)^{1.667}(P_{1} = 1 given)

log P_{2} = 1.667 log 10= 1.667

P_{2} = antilog 1.667= 46.45 atm

Final temperature

= 1268 K

Work done on the system = n.C_{v}.ΔT

= 2 × 12.45 × (1268 - 273)

= 2 × 12.45 × 995 = 24775.5 J

From the first law of thermodynamics,

ΔE = q + w = 0 + 24775.5 = 24775.5 J**Example 2. A certain volume of dry air at NTP is expanded reversibiy to four times its volume (a) isothermally (b) adiabatically. Calculate thefinal pressure and temperature in each case, assuming ideal behaviour.****(C _{P} \ C_{V} for air = 1.4)**

Let V

Since P

dH = dU + d(PV)

ΔH = ΔU + nRΔT

du = dQ + dw**Calculation of q, w , ∆H, ∆U for a reversible isothermal process involving an ideal gas :**

ΔU = q + w = 0 ⇒ -w

**Calculation of q, w, ∆H, ∆U for an Irreversible isothermal process involving an ideal gas:**

For isothermal process involving

ΔH = ΔU = 0 ∵ ΔT = 0

Also,

For isobaric process

= -nR(T_{2} - T_{1}) (∵ P_{ext} = P)

= -nRΔT

**Calculation of q, w, ∆H, ∆U for an IRREVERSIBLE ISOCHORIC process involving an ideal gas:**

w= 0 ∵ dV = 0

**Calculation of q, w, ∆H, ∆U for reversible adiabatic process**For an adiabatic process,

dq = 0 ⇒ dU = dw

for a reversible change

Now substituting V = nRT/P in equation

substituting T = PV/nR in eq...

⇒ Equation (i), (ii) and (iii) is valid only for reversible adiabatic process, for irreversible adiabatic process these equations are not applicable.

- Expression for work:

- Expression for ΔH and ΔU

**CALCULATION OF Q, W, ∆H, ∆U FOR IRREVERSIBLE ADIABATIC ****PROCESS INVOLVING AN IDEAL GAS:**

Operation wise adiabatic process and isothermal process are similar hence all the criteria that is used for judging an isothermal irreversible process are applicable to adiabatic process.

If large amount of dust particles are removed abruptly an irreversible adiabatic expansion take place.

In an irreversible adiabatic process, an ideal gas is subjected to compression or expansion in a thermally insulated vessel.

The heat absorbed in the process =0

⇒ dU = w_{irr }....(i)

If P_{ext}, = P_{2} = P_{final}

Then

eq. (ii) or (iii) can be solved for T_{2}

Expression for w

**Note:** If two states A and B are connected by a reversible path then they can never be connected by an irreversible path.

If the two states are linked by an adiabatic reversible and irreversible path then w_{rev} = ∆U_{rev}

But as U is a state function

Therefore, ∆U_{ irrev} = ∆U_{rev}

w_{irrev} = w_{rev}

as work is a path function.

If we assume that

w_{irrev} = w_{rev}

It implies that which again is a contradiction as U is a state function.

∆U_{ irrev} ≠ ∆U_{rev}

Two states A and B can never lie both on a reversible as well as irreversible adiabatic path.

There lies only one unique adiabatic path linkage between two states A and B.

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