The document discusses thermodynamic processes involving gases. It provides equations relating the temperature (T), pressure (p), volume (V), and internal energy (U) of a gas during different processes. Specifically, it presents the equations for: 1) An isothermal expansion/compression process where pVγ = constant, with γ being the heat capacity ratio. 2) An adiabatic expansion/compression process where TVγ-1 = constant. 3) The work (W) done during an adiabatic expansion/compression process as the gas moves between two states, relating W to the change in temperature.

2. Not allowing
heat to enter
or leave the
gas
∆Q = 0 ∆Q = ∆U + W Thermodynamic
first law
0 = ∆U + W
∆U = –W

3. ∆U = –W
Work done by the gas
W = +ve
∆U = –W

4. ∆U = –W
Work done by the gas
Work done on the gas
W = +ve
W = –ve
∆U = –W
∆U = –(–W) = +W

7. back
γ–1 γ–1
back
T1V1 = T2V2
1–γ γ 1–γ γ
back
p1 T1 = p2 T2

8. ∆U ∆U = CV,m∆T
∆T
∆Q = 0
W = p∆V
∆Q = ∆U + W Thermodynamic
first law
0 = CV,m∆T + p∆V
back p∆V = –CV,m∆T ---------(1)

9. For one mole of an ideal gas
Initially: substitute
pV = RT
After compression:
(p + ∆p)(V + ∆V) = R(T + ∆T)
pV + p∆V + V∆p + ∆p∆V = RT + R∆T)
Assumed zero
RT + p∆V + V∆p + ∆p∆V = RT + R∆T)
p∆V + V∆p = R∆T
Substitute (1) into here V∆p = R∆T – p∆V
∆pV = R∆T + CV,m∆T
Cp,m – CV,m = R ∆pV = (R + CV,m )∆T
= Cp,m ∆T ---------(2)

10. ∆pV = Cp,m ∆T ---------(2)
(2) :
(1)
Cp,m ∆V
∆p
= – C V
p V,m
Cp,m
∆p = –γ
∆V γ =
p V CV,m
---------(3)

11. As ∆p  0 and
–γ
dp dV
= ∆V  0
p V
–γ
dV
∫ dp
p
= ∫ V
∫
1 dp = – γ 1 dV
p V ∫ To adiabatic
ln p = – γ ln V + m equation
Use log rule
ln p = ln V–γ + m
p = e ln V–γ + m Use log rule
p = e ln V–γ e m p1V1γ = p2V2γ
p = e ln V–γ (constant)
–γ
Use log rule
p = V (constant)
pVγ = constant ----------(4)

12. pVγ = constant
RT γ
V V = constant p = RT
V
γ–1 constant
RTV =
constant
TV γ–1
= R
Because R is also a constant
TVγ – 1 = constant ----------(5)
T1V1γ – 1 = T2V2γ – 1
To adiabatic
equation

13. pVγ = constant
RT γ V = RT
p p = constant p
1–γ γ γ
p R T = constant
1–γ γ constant
p T = γ
R Because R γ is also a
1–γ γ constant
p T = constant
γ γ
1– γ 1–γ
p1 T1 = p2 T2
To adiabatic
equation

14. Vf
γ
From pV = constant W=∫ p dV
= k Vi
Vf
p = kV–γ =∫ kV–γ dV
Vi
Vf
P
Adiabatic = k∫ V–γ dV
Vi
pi expansion Vf
= k V–γ+1
–γ+1 Vi
T1 k Vf
=
pf T2 –γ+1 V–γ+1
Vi
k
0 =
Vi Vf V
–γ+1 Vf–γ+1 – V –γ+1
i

15. ----(1)
1
= –γ+1 – kV –γ+1 pVγ = constant = k
1–γ kV f
piViγ = pfVfγ = k
i
1
= pfVfγ ) V –γ+1 – p V γ
1–γ ( f ( i i ) Vi–γ+1
1
= pfVf – piVi
1–γ
1
= piVi – pfVf ----(2)
γ –1
1
= nRT1 – nRT2
γ –1 pV = nRT
nR
= T1 – ----------(3)
γ –1 T2

16. nR
W = T1 – ∆T = T2 – T1
γ –1 T2
nR
=
( –∆T) γ = f+2
f+2 f
–1
f
nR Work done on
(–∆T)
= f+2 – f the gas , W = –ve
f f  W = +∆U
W = –∆U
nR
= 2 (–∆T) Means a rise in
f Work done its internal
by the gas energy
= – nfR ∆T
2  W = –∆U T
Means a
= –∆U reduction in its T
internal energy

17. Is a process which can be made to retrace its path from
one equillibrium state to another equillibrium state
through small changes at every step.
Isothermal expansion Isothermal compression
p Small p Small
step at step at
every every
instant instant
0 V 0 V

18. The wall of the container must be as thin as possible
to allow heat transfer.
The piston must be light and frictionless.
carried out very slowly through small steps
For a reversible
process to occur
in practicce
Isothermal expansion Isothermal compression
p Small p Small
step at step at
every every
instant instant
0 V 0 V

19. Thick insulator so that
heat transfer cannot occur
Piston must be light
and frictionless
Must be carried out very
quickly through small steps
For a reversible
process to occur
in practicce
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20. Example 1 :

25. Example 2 :

33. Example 3 :