Study

1. 3
Expression for irreversible-work.
2

2. 4
Expression for reversible-work.
2

3. 5
The internal energy of a system can be changed in two ways:
(i) By allowing heat to flow into the system or out of the system.
(ii) By doing work on the system or by the system.
2

4. 6
FIRST LAW OF THERMODYNAMICS
First law of thermodynamics is also known as law of conservation of
energy.
Energy can neither be created nor destroyed although it may be changed
from one form to another.
2

5. 7
(i) lt is not possible to construct a perpetual machine which can do work
without the expenditure of energy.
(ii) Energy is conserved in chemical reactions also.
Justification for the First Law of Thermodynamics
2

6. 8
Mathematical Expression of First Law
Now, final internal energy U2 is given by
U2 = U1 + q + w
or U2 – U1 = q + w or U = q + w
2

7. 8
Mathematical Expression of First Law
Now, final internal energy U2 is given by
U2 = U1 + q + w
or U2 – U1 = q + w or U = q + w
2

8. Some Important Results From the First Law Equation, U = q + w
1. For isothermal reversible as well irreversible expansion of ideal gas
U = 0
 0 = q + w or q = - w
(a) For irreversible expansion, w = –PextV
q = – w = PextV
(b) For reversible expansion,
w = – 2.303 nRT log V2/V1
q = – w = 2.303 nRT log V2/V1

9. 0
Some Important Results From the First Law Equation, U = q + w
2. For expansion of ideal gas into vacuum
3

10. 0
Some Important Results From the First Law Equation, U = q + w
2. For expansion of ideal gas into vacuum
3

11. 1
Some Important Results From the First Law Equation, U = q + w
3. For adiabatic process,U = Wadia
3

12. 1
Some Important Results From the First Law Equation, U = q + w
3. For adiabatic process, q = 0
 U = Wadia
3

13. 2
4. For isochoric process, V = 0
 U = q
Some Important Results From the First Law Equation, U = q + w
3

14. 2
4. For isochoric process, V = 0
 U = q
Some Important Results From the First Law Equation, U = q + w
3

15. 3
Can you Crack it !!
1. A system gives out 30 J of heat and also does 40 joules of work. What
is the internal energy change?
3

16. 3
Can you Crack it !!
1. A system gives out 30 J of heat and also does 40 joules of work. What
is the internal energy change?
3

17. 4
Can you Crack it !!
2. 100 J of work is done on the system and at the same time 140 J of
heat is given out. What is the change in internal energy?
3

18. 4
Can you Crack it !!
2. 100 J of work is done on the system and at the same time 140 J of
heat is given out. What is the change in internal energy?
3

19. 5
Can you Crack it !!
3. A gas absorbs 121.4 J of heat and expands against the external
pressure of 1.2 atm. from a volume of 0.5 L to 1.0 L What is the
change in internal energy ?
3

20. 5
Can you Crack it !!
3. A gas absorbs 121.4 J of heat and expands against the external
pressure of 1.2 atm. from a volume of 0.5 L to 1.0 L What is the
change in internal energy ?
3

21. 6
4. 3.00 moles of ideal gas is heated at constant pressure from 27°C to 127°C
(i) Calculate the work of expansion.
(ii) If the gas were expanded isothermally in a reversible manner at 27°C
from 1 atm to 0.70 atm. Calculate the work done.
Can you Crack it !!
3

22. 6
4. 3.00 moles of ideal gas is heated at constant pressure from 27°C to 127°C
(i) Calculate the work of expansion.
(ii) If the gas were expanded isothermally in a reversible manner at 27°C
from 1 atm to 0.70 atm. Calculate the work done.
Can you Crack it !!
3

23. 7
expands
5. Calculate w, q, and U when 1 mol of ideal gas
isothermally and reversibly at 300 K from 20 L to 30 L.
Can you Crack it !!
3

24. 8
3

25. 8
3

26. 9
3

27. 9
3

28. 0
4

29. 0
4

30. 1
1
11th
Enthalpy
LECTURE

31. 9
System
1

32. 0
Sign Conventions
2

33. 1
(a) Zeroth law
(c) Second law
(b) First law
(d) None of the above
Can you Crack it !!
1. Which law of thermodynamics is used to understand the concept of
energy conservation.
2

34. 1
(a) Zeroth law
(c) Second law
(b) First law
(d) None of the above
Can you Crack it !!
1. Which law of thermodynamics is used to understand the concept of
energy conservation.
2

35. 2
Can you Crack it !!
2. During a process, a system absorbs 710 J of heat and does work. The
change in internal energy for the process is 460 J. What is the work
done by the system ?
2

36. 2
Can you Crack it !!
2. During a process, a system absorbs 710 J of heat and does work. The
change in internal energy for the process is 460 J. What is the work
done by the system ?
2

37. 3
Can you Crack it !!
3. A gas expands isothermally against constant external pressure of 1
atm from a volume of 10 dm3 to a volume of 20 dm3. In the process,
it absorbs 800 J of thermal energy from surroundings. Calculate the
value of internal energy change.
2

38. 3
Can you Crack it !!
3. A gas expands isothermally against constant external pressure of 1
atm from a volume of 10 dm3 to a volume of 20 dm3. In the process,
it absorbs 800 J of thermal energy from surroundings. Calculate the
value of internal energy change.
2

39. 4
INTERNAL ENERGY AS A STATE FUNCTION
U is a State Function
2

40. 5
(a) W = -E
(c) E = 0
(b) W = E
(d) W = 0
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4. In an adiabatic expansion of ideal gas
2

41. 5
(a) W = -E
(c) E = 0
(b) W = E
(d) W = 0
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4. In an adiabatic expansion of ideal gas
2

42. 6
(a) - 35420J
(c) -26570J
(b) -22798 J
(d) - 48900 J
Can you Crack it !!
5. What is the work done when 1 mole of a gas expands isothermally from 25
L to 250 L against a const external pressure of 1 atm and a temperature of
300 K?
2

43. 7
6. For an isothermal reversible expansion process the value of q can be
calculated by the expression
(a) q = 2.303nRT log V2/V1
(b) q = - 2.303nRT log V2/V1
(c) q = - 2.303nRT log V1/V2
(d) q = - PexpnRT log V1/V2
Can you Crack it !!
2

44. 7
6. For an isothermal reversible expansion process the value of q can be
calculated by the expression
(a) q = 2.303nRT log V2/V1
(b) q = - 2.303nRT log V2/V1
(c) q = - 2.303nRT log V1/V2
(d) q = - PexpnRT log V1/V2
Can you Crack it !!
2

45. 8
(a) Increase by 30 kJ
(c) Increase by 70 kJ
(b) Decrease by 30 kJ
(d) Decrease by 70 kJ
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7. A system absorbs 50 kJ heat and does 20 kJ of work. What is the net
change in the internal energy of the system? .
2

46. 8
(a) Increase by 30 kJ
(c) Increase by 70 kJ
(b) Decrease by 30 kJ
(d) Decrease by 70 kJ
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7. A system absorbs 50 kJ heat and does 20 kJ of work. What is the net
change in the internal energy of the system? .
2

47. 9
(a) +10 kJ
(c) +5 kJ
(b) -10 kJ
(d) -5 kJ
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8. What will be the change in internal energy when 12 kJ of work is
done on the system and 2 kJ of heat is given by the system?
2

48. 0
Can you Crack it !!
9. Which one of the following equations does not correctly represent the
first law of thermodynamics for the given processes involving an ideal
gas? (Assume non-expansion work is zero)
[Main April 8,2019 (I)]
(a) Cyclic process : q = -w
(b) Adiabatic process : U = -w
(c) Isochoric process: U = q
(d) Isothermal process: q = - w
3

49. 1
Can you Crack it !!
10. A gas undergoes change from state A to state B. In this process, the heat
absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas
is brought back to A by another process during which 3 J of heat is
evolved. In this reverse process of B to A:
[Main Online April 9,2017]
(a) 10 J of the work will be done by the gas.
(b) 6 J of the work will be done by the gas.
(c) 10 J of the work will be done by the surrounding on gas.
(d) 6 J of the work will be done by the surrounding on gas.
3

50. 2
ENTHALPY (H)-ANOTHER USEFUL STATE FUNCTION
Energy change occurring during the reaction at constant temperature and
constant volume is given by internal energy change, However, most of the
reactions in the laboratory are carried out in open beakers or test tubes,
etc. In such cases, the reacting system is open to atmosphere. Since
atmospheric pressure is almost constant, therefore, such reactions may
involve the changes in volume. The energy change occurring during such
reactions may not be equal to the internal energy change.
3

51. 3
(a) If the reaction proceeds with the increases in volume the system
has to expand against the atmospheric pressure and energy is
required for this purpose. The heat evolved in this case would be
little less than the heat evolved at constant volume because a part
of the energy has to be utilized for expansion.
3

52. 4
ENTHALPY (H)
3

53. 5
(b) If the reaction proceeds with decrease in volume at constant
pressure, the work is done on the system and heat evolved will be
greater than the heat evolved at constant volume.
Enthalpy is the total energy associated with any system which
includes its internal energy and also energy due to environmental
factors such as pressure-volume conditions.
3

54. 6
ENTHALPY (H)
3

55. 7
Enthalpy
The sum of internal energy and PV energy of any system, under given set
of conditions, is called enthalpy. It is denoted by H and is also called heat
content of the system.
H = U + PV
Some important features of enthalpy are:
• It is a state function and is an extensive property,
• It is also called heat content of the system,
• Its value depends upon amount of the substance, chemical nature
of the substance and conditions of temperature and pressure.
3

56. 8
Difference Between U and H
3

57. 8
Difference Between U and H
3

58. 9
Internal energy (U) is the energy required to create the system and does
not include the energy needed to displace the system's surroundings in
which it exists.
Enthalpy (H) is a measure of total energy of thermodynamic system which
includes its internal energy and also the energy required to make room for
it by displacing its environment and establishing its volume at the
atmospheric pressure. It may be regarded as thermodynamic potential of
the system.
3
Difference Between U and H

59. 0
can be
change in enthalpy (H) taking place during the process
experimentally determined.
The change in enthalpy may be expressed as H = Hp - HR
ENTHALPY CHANGE (H)
It is not possible to determine the absolute value of enthalpy of a system
because absolute value of internal energy (U) is not known. However,
4

60. 1
U = q – pV
or q = U +pV = (U2 – U1) + p(V2 – V1)
= (U2 + pV2) – (U1 + pV1)
= H2 – H1 = H
or H = qp
In short:
U: Heat Changes at Constant T and V.
H: Heat Changes at Constant T and P.
SIGNIFICANCE OF H
4

61. 1
U = q – pV
or q = U +pV = (U2 – U1) + p(V2 – V1)
= (U2 + pV2) – (U1 + pV1)
= H2 – H1 = H
or H = qp
In short:
U: Heat Changes at Constant T and V.
H: Heat Changes at Constant T and P.
SIGNIFICANCE OF H
4

62. 2
nA RT
and
p VA =
p VB = nB RT
p VB – p VA =
or p(VB - VA) =
(nB - nA)RT
(nB - nA)RT or pV = (n)RT
= H = U + p V
= H = U + (n) RT
RELATIONSHIP BETWEEN H AND U
4

63. 2
nA RT
and
p VA =
p VB = nB RT
p VB – p VA =
or p(VB - VA) =
(nB - nA)RT
(nB - nA)RT or pV = (n)RT
= H = U + p V
= H = U + (n) RT
RELATIONSHIP BETWEEN H AND U
4

64. 3
An, here, refers to the change in the number of moles of gaseous species
during the process. The difference between H and U will depend upon
the magnitude as well as sign of An. For example,
if n is 0;
if n is -ve ;
if n is +ve ;
H = U
H < U and
H > U.
q(P) = q(v) + (n)RT
In short,
H = U + pV or
q(P) = q(V) + pV or
H = U + ngRT
q(P) = q(V) + ngRT
Lets Learn
4

65. 3
An, here, refers to the change in the number of moles of gaseous species
during the process. The difference between H and U will depend upon
the magnitude as well as sign of An. For example,
if n is 0;
if n is -ve ;
if n is +ve ;
H = U
H < U and
H > U.
q(P) = q(v) + (n)RT
In short,
H = U + pV or
q(P) = q(V) + pV or
H = U + ngRT
q(P) = q(V) + ngRT
Lets Learn
4

66. 4
Remember the Formulae
1. H = U + pV
2. U = U + nRT
n is change in number of gaseous moles
3. R = 0.002 kCal K-1 mol-1 or 8.314 JK-1 mol-1.
4

67. 5
1 For a gaseous reaction,
2A2(g) + 5B2(g) -------------> 2A2B5(g) at 27°C the heat change at
constant pressure is found to be - 50160 J. Calculate the value of
internal energy change (U). Given that R = 8.314 JK-1 mol-1.
Can you Crack it !!
4

68. 6
2. The enthalpy change (H) for the
N2(g) + 3H2(g) --------> 2NH3(g)
is -92.38 kJ at 298 K What is U at 298 K ?
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4

69. 7
(a) -312.5 kJ
(c) -310 kJ
(b) -125.03 kJ
(d) -156 kJ
Can you Crack it !!
3. What will be the standard internal energy change for the reaction at 298 K?
OF2(g) + H2O(g) ----------> O2(g) + 2HF(g); H° = - 310 kJ
4

70. 8
4. Which of the following relationships is not correct for
the relation
between H and U?
(a) When ng = 0 then H = U
(b) When ng > 0 then AH > AU
(c) When ng < 0 then H < U
(d) When ngRT=0 then H > U
Can you Crack it !!
4