Problems

1. First law of thermodynamics is given by [CPMT 1977, 91]

a. dQ = dU + PdV

b. dQ = dU × PdV

c. dQ = (dU + dV) P

d. dQ = PdU + dV

Solution: dQ = dU + PdV

ΔQ = ΔU + ΔW and ΔW = PΔV

2. If a system undergoes contraction of volume then the work done by the system will be [BHU 1999]

a. Zero

b. Negligible

c. Negative

d. Positive

Solution: Negative.

ΔW = PΔV,

here DV is negative so DW will be negative.

3. The internal energy of an ideal gas depends up on [RPMT 1997; MP PMT 1999; CPMT 2003]

a. Specific volume

b. Pressure

c. Temperature

d. Density

Solution: Temperature.

4. In changing the state of thermodynamics from A to B state, the heat required is Q and the work done by the system is W. The change in its internal energy is [MP PMT 1986; AMU (Med.) 2001]

a. Q + W

b. Q − W

c. Q

Solution: Q + W

ΔQ = ΔU + ΔW

ΔU = ΔQ − ΔW

= Q + W (using proper sign)

5. Heat given to a system is 35 joules and work done by the system is 15 joules. The change in the internal energy of the system will be [MP PET/PMT 1988]

a. 50 J

b. 20 J

c. 30 J

d. 50 J

Solution: 20 J

ΔU = ΔQ − W

= 35 −15

= 20 J

6. The temperature of an ideal gas is kept constant as it expands. The gas does external work. During this process, the internal energy of the gas [MP PMT 1990]

a. Decreases

b. Increases

c. Remains constant

d. Depends on the molecular motion

Solution: Remains constant

Internal energy depends only on the temperature of the gas.

7. The first law of thermodynamics is concerned with the conservation of [MP PMT 1987; CBSE PMT 1990, 92; AFMC 1997; CPMT 1999; BHU 1999; DCE 2000; BCECE 2003]

a. Momentum

b. Energy

c. Mass

d. Temperature

Solution: Energy

8. A thermodynamic system goes from states (i) P₁, V to 2P₁, V (ii) P, V to P, 2V. Then work done in the two cases is [MP PMT 1990]

a. Zero, Zero

b. Zero, PV₁

c. PV₁, Zero

d. PV₁, P₁V₁

Solution: Zero, PV₁

(i) Case → Volume = constant

⇒ = 0

(ii) Case → P = constant

⇒ = P = PV₁

9. If the amount of heat given to a system be 35 joules and the amount of work done by the system be −15joules, then the change in the internal energy of the system is [MP PMT 1989]

a. −50 joules

b. 20 joules

c. 30 joules

d. 50 joules

Solution: 50 joules

ΔQ = ΔW + ΔU

35 = −15 + ΔU

ΔU = 50J

10. Work done on or by a gas, in general depends upon the

a. Initial state only

b. Final state only

c. Both initial and final states only

d. Initial state, final state and the path

Solution: Initial state, final state and the path

Work done = , which is state dependent as well as path dependent.

11. If R = universal gas constant, the amount of heat needed to raise the temperature of 2 mole of an ideal monoatomic gas from 273K to 373K when no work is done [MP PET 1990]

a. 100 R

b. 150 R

c. 300 R

d. 500 R

Solution: 300 R

ΔQ = ΔU + ΔW (∵ ΔW = 0)

ΔQ = ΔU

= μRΔT

= 32 × 2 R(373 − 273)

= 300R.

12. Find the change in internal energy of the system when a system absorbs 2 kilocalorie of heat and at the same time does 500 joule of work [EAMCET 1984]

a. 7900 J

b. 8200 J

c. 5600 J

d. 6400 J

Solution: 7900 J

ΔQ = 2 kcal

= 2 × 10³ × 4.2 J

= 8400 J

ΔW = 500J

Hence from

ΔQ = ΔU + ΔW

ΔW = ΔQ − ΔU

= 8400 − 500

= 7900 J

13. A system performs work ΔW when an amount of heat is ΔQ added to the system, the corresponding change in the internal energy is ΔU. A unique function of the initial and final states (irrespective of the mode of change) is [CPMT 1981; J & KCET 2004]

a. ΔQ

b. ΔW

c. ΔU and ΔQ

d. ΔU

Solution: ΔU

Change in internal energy ΔU depends upon initial an find state of the function while ΔQ and ΔW are path dependent also.

14. A container of volume 1m3is divided into two equal compartments by a partition. One of these compartments contains an ideal gas at 300 K. The other compartment is vacuum. The whole system is thermally isolated from its surroundings. The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature now would be [Manipal MEE 1995]

a. 300 K

b. 239 K

c. 200 K

d. 100 K

Solution: 300 K

This is the case of free expansion and in this case ΔW=0, ΔU=0. So temperature remains same i.e. 300 K.

15. 110 J of heat is added to a gaseous system, whose internal energy change is 40 J, then the amount of external work done is [CBSE PMT 1993; DPMT 1996, 03; AFMC 1999; JIPMER 2000; MH CET 2000; Pb. PMT 2003]

a. 150 J

b. 70 J

c. 110 J

d. 40 J

Solution: 70 J

ΔQ = ΔU + ΔW

ΔW = ΔQ – ΔU

= 100 − 40

= 70 J

16. Which of the following is not thermodynamical function [CBSE PMT 1993; CPMT 2001; DCE 1996; 2001]

a. Enthalpy

b. Work done

c. Gibb's energy

d. Internal energy

Solution: Work done

Work done is not a thermodynamical function.

17. When the amount of work done is 333 calorie and change in internal energy is 167 calorie, then the heat supplied is [AFMC 1998]

a. 166 calorie

b. 333 calorie

c. 500 calorie

d. 400 calorie

Solution: 500 calorie

ΔQ = ΔU + ΔW

= 167 + 333

= 500 calorie

18. First law thermodynamics states that [KCET 1999]

a. System can do work

b. System has temperature

c. System has pressure

d. Heat is a form of energy

Solution: Heat is a form of energy

Heat always refers to energy in transit from one body to another because of temperature difference.

19. A thermo-dynamical system is changed from state (P₁, V₁) to (P₂, V₂) by two different process. The quantity which will remain same will be [RPET 1999]

a. ΔQ

b. ΔW

c. ΔQ+ΔW

d. ΔQ−ΔW

Solution: ΔQ−ΔW

Change in internal energy does not depend upon path so ΔU=ΔQ−ΔW remain constant.

20. In thermodynamic process, 200 Joules of heat is given to a gas and 100 Joules of work is also done on it. The change in internal energy of the gas is[AMU (Engg.) 1999]

a. 100 J

b. 300 J

c. 419 J

d. 24 J

Solution: 300 J

ΔQ = ΔU + ΔW;

ΔQ = 200 J

ΔW = −100 J

ΔU = ΔQ − ΔW

= 200 − (−100)

= 300 J

21. A perfect gas contained in a cylinder is kept in vacuum. If the cylinder suddenly bursts, then the temperature of the gas [MH CET 1999]

a. Remains constant

b. Becomes zero

c. Increases

d. Decreases

Solution: Remains constant

During free expansion of a perfect gas no, work is done and also no heat is supplied from outside. Therefore, no change in internal energy. Hence, temperature remain constant.

22. If 150 J of heat is added to a system and the work done by the system is 110 J, then change in internal energy will be [AMU (Engg.) 1999; BHU 2000]

a. 260 J

b. 150 J

c. 110 J

d. 40 J

Solution: 40 J

ΔQ = ΔU + ΔW

ΔU = ΔQ − ΔW

= 150 − 110

= 40 J

23. If ΔQ and ΔW represent the heat supplied to the system and the work done on the system respectively, then the first law of thermodynamics can be written as [Roorkee 2000]

a. ΔQ = ΔU + ΔW

b. ΔQ = ΔU − ΔW

c. ΔQ = ΔW − ΔU

d. ΔQ = − ΔW − ΔU

where ΔU is the internal energy

Solution: ΔQ = ΔU − ΔW

From FLOT ΔQ = ΔU + ΔW. Q Heat supplied to the system so ΔQ is Positive and work is done on the system so ΔW is Negative. Hence ΔQ = ΔU – ΔW

24. For free expansion of the gas which of the following is true [AMU (Med.) 2000]

a. Q = W = 0 and ΔE_int= 0

b. Q = 0, W > 0 and ΔE_int= −W

c. W = 0 =, Q > 0, and ΔE_int= Q

d. W > 0, Q < 0 and ΔE_int= 0

Solution: Q = W = 0 and ΔE_int=0

25. Which of the following cannot determine the state of a thermodynamic system [AFMC 2001]

a. Pressure and volume

b. Volume and temperature

c. Temperature and pressure

d. Any one of pressure, volume or temperature

Solution: Any one of pressure, volume or temperature. State of a thermodynamic state cannot determine by a single variable (P or V or T)

26. Which of the following is not a thermodynamics co-ordinate [AIIMS 2001]

a. P

b. T

c. V

d. R

Solution: R

R is the universal gas constant.

27. Which of the following is incorrect regarding the first law of thermodynamics [AIEEE 2005]

a. It introduces the concept of the internal energy

b. It introduces the concept of the entropy

c. It is not applicable to any cyclic process

d. None of the above

Solution: It introduces the concept of the entropy. Entropy is related to second law of thermodynamics.

28. If a system undergoes contraction of volume then the work done by the system will be [BHU 1999]

a. Zero

b. Negligible

c. Negative

d. Positive

Solution: Negative

ΔW=PΔV; here ΔV is negative so ΔW will be negative.

29. A perfect gas goes from state A to another state B by absorbing 8×105J of heat and doing 6.5×105J of external work. It is now transferred between the same two states in another process in which it absorbs 105J of heat. Then in the second process [BHU 1997]

a. Work done on the gas is 0.5×10⁵J

b. Work done by gas is 0.5×10⁵J

c. Work done on gas is 10⁵J

d. Work done by gas is 10⁵J

Solution: Work done on the gas is 0.5×10⁵J

In first process using ΔQ=ΔU+ΔW

8 × 10⁵=ΔU+6.5×10⁵

ΔU=1.5×10

Since final and initial states are same in both process.

So ΔU will be same in both process. For second process using

ΔQ=ΔU+ΔW

10⁵=1.5×10⁵+ΔW

ΔW=−0.5×105J

30. The state of a thermodynamic system is represented by [MH CET 2004]

a. Pressure only

b. Volume only

c. Pressure, volume and temperature

d. Number of moles

Solution: Pressure, volume and temperature

31. In a thermodynamics process, pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8J of work is done on the gas. If the initial internal energy of the gas was 30J. The final internal energy will be [DPMT 2002]

a. 18J

b. 9J

c. 4.5J

d. 36J

Solution: 18J

Given

ΔQ = −20J,

ΔW = −8J

U1 = 30J

ΔQ = ΔU+ΔW

ΔU = (ΔQ−ΔW)

(U_f− U_i) = (U_f−30)

= −20− (−8)

U_f = 18J

32. If heat given to a system is 6 kcal and work done is 6 kJ. Then change in internal energy is [BHU Med. 2000]

a. 19.1 kJ

b. 12.5 kJ

c. 25 kJ

d. Zero

Solution: 19.1 kJ

ΔQ=ΔU+ΔW

ΔU=ΔQ−ΔW

=6×4.18−6

=19.08kJ

≈19.1kJ

33. Temperature is a measurement of coldness or hotness of an object. This definition is based on [RPET 2003]

a. Zeroth law of thermodynamics

b. First law of thermodynamics

c. Second law of thermodynamics

d. Newton's law of cooling

Solution: Zeroth law of thermodynamics

34. If CV=4.96cal/mole K, then increase in internal energy when temperature of 2 moles of this gas is increased from 340 K to 342 K [RPET 1997]

a. 27.80 calorie

b. 19.84 calorie

c. 13.90 calorie

d. 9.92 calorie

Solution: 19.84 calorie

ΔU=μC_VΔT=2×4.96× (342−340)=19.84 calorie.

35. First law of thermodynamics is a special case of [CPMT 1985; RPET 2000; DCE 2000; CBSE PMT 2000; AIEEE 2002; AFMC 2002]

a. Newton's law

b. Law of conservation of energy

c. Charle's law

d. Law of heat exchange

Solution: Law of conservation of energy

Heat supplied to a gas raise its internal energy and does some work against expansion, so it is a special case of law of conservation of energy.

36. Out of the following which quantity does not depend on path [RPET 2002]

a. Temperature

b. Energy

c. Work

d. None of these

Solution: Temperature

37. Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which one of the following graphs represents the variation of temperature with time?

Solution:

At first temperature will increase then there will be state change from liquid to gas.

38. The door of an operating refrigerator is opened. Now the temperature of the room will

a. Increase

b. Decrease

c. Remain unchanged

d. Decrease in summer and increase in winter

Solution: Increase

Use first law of thermodynamics.

39. An ideal gas undergoes an isothermal change in volume with pressure. Then

d. PV = constant

Solution: PV = constant

For the special case of a gas to which Boyle's law applies, the product PV is a constant if the gas is kept at isothermal conditions. The value of the constant is nRT.

40. In changing the state of thermodynamics from A to B state, the heat required is Q and the work done by the system is W. the change is its internal energy is

a. Q + W

b. Q – W

d. Q

Solution: Q – W

Use dQ = dU + dW.

41. An adiabatic process occurs at constant

a. Temperature

b. Pressure

c. Heat

d. Temperature and pressure

Solution: Heat

42. Pressure-temperature relationship for an ideal gas undergoing adiabatic change is

c. = constant

Solution:

For adiabatic process
PV^γ = constant -- (1)
and the combined gas law: PV/T = constant -- (2)
putting V from (2) to (1)
P(T⁄P)^γ = constant
P^(γ-1)T^γ = constant

43. The gas law = constant is true for

a. Isothermal changes only

b. Adiabatic changes only

c. both isothermal and adiabatic changes

d. Neither isothermal nor adiabatic changes

Solution: Both isothermal and adiabatic changes

The Gas law is true for both isothermal and adiabatic changes.

44. A gas behaves more closely as an ideal gas at:

a. Low pressure and low temperature

b. Low pressure and high temperature

c. High pressure and low temperature

d. High pressure and high temperature

Solution: Low pressure and high temperature

45. A block of a mass 100 g slides on a rough horizontal surface. If the speed of the block decreases from 10m/s to 5m/s the thermal energy developed in the process is:

a. 3.75 J

b. 37.5 J

c. 0.375 J

d. 0.75 J

Solution: 3.75 J

Thermal energy = loss in kinetic energy

=m ( –

=3.75J

46. A thermodynamic system undergoes cyclic process ABCDA as shown in Fig. The work done by the system in the cycle is[ AIPMT 2014]

b. Zero

c. P₀V₀

d. 2 P₀V₀

Solution: P₀V₀

P₀V₀Work in COB = (-P₀) V

Work in AOD = P₀V₀

Total work = Work in COB + Work in AOD =0.

47. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of for the gas is

b. 2

Solution:

P ∝ T₃

PV = nRT

PV=C (P)

C=constant

P V=C

PV =C

On comparing above equation with P (V) γ= Constant

We get γ = 3/2.

48. The change in internal energy, when a gas is cooled from 927°C to 27°C

a. a)200%

b. b) 100%

c. c)300%

d. d) 400%

Solution: 300%

Energy of gas E ∝ T

Let T1=927 + 273=1200 K

T2=27 + 273=300%

49. A sample of Oxygen gas and a sample of Hydrogen gas both have the same mass, the same volume and the same pressure. The ratio of their absolute temperature is

a. 1:4

b. 4:1

c. 1:16

d. 16:1

Solution: 16:1

PV=nRT

PV= RT

PV= RT₁ and PV = RT₂

=16:1

50. A system goes from A to B via two process I and II as shown in figure. If ΔU1 and ΔU2 are the changes in internal energies in the process I and II respectively, then [AIEEE 2005]

a) Relation between ΔU₁ and ΔU₂ cannot be determined

b) ΔU₁ = ΔU₂

c) ΔU₂ < ΔU₁

d) ΔU₂ > ΔU₁

Solution: ΔU₁ = ΔU₂

Change in internal energy do not depend upon the path followed by the process. It only depends on initial and final states i.e. ΔU₁ = ΔU₂

51. "Heat cannot by itself flow from a body at lower temperature to a body at higher temperature" is a statement or consequence of

a. First law of thermodynamics

b. Second law of thermodynamics

c. Conservation of momentum

d. Conservation of mass.

Bottom of Form

Solution: Second law of thermodynamics.

Heat cannot flow itself from a lower temperature to a body of higher temperature. This corresponds to second law of thermodynamics.

52. The first law of thermodynamics is a restatement of the

a. Law of conservation of momentum

b. Newton's law of cooling

c. Law of conservation of mass

d. Law of conservation of energy

Solution: Law of conservation of energy

The First Law of Thermodynamics is simply a restatement of the Law of Conservation of Energy. There are a few ways of stating it, but they all mean the same thing:

"Energy is neither created nor destroyed. It can only change form."

"The change in the system's thermal energy is equal to the heat added to the system minus the work done by the system."

53. In an adiabatic process, there is no

a. change in temperature

b. exchange of heat

c. change in internal energy

d. work done

Solution: Exchange of heat

In thermodynamics, an adiabatic process is one that occurs without transfer of heat or matter between a thermodynamic system and its surrounding. In an adiabatic process, energy is transferred only as work.

54. On a cold morning, a metal surface on touching is felt colder than a wooden surface, because the metal has:

a. low thermal conductivity

b. high thermal conductivity

c. high specific heat

d. low specific heat

Solution: High thermal conductivity

Metal is a good conductor of heat. Ona cold morning, a metal surface on touching is felt colder than wood because metal is a good conductor of heat (high thermal conductivity) hence, when we touch metal, the metal rapidly conducts heat from our hand and gives a cold feeling. Wood on the conducts heat is a bad conductor of heat, it conducts heat slowly from our hand and appears less cold.

55. The processes or systems that do not involve heat are called

a. Isothermal processes

b. Equilibrium processes

c. Thermal processes

d. Adiabatic processes

Solution: Adiabatic processes

56. In an open system, for maximum work, the process must be entirely

a. Irreversible

b. Reversible

c. Adiabatic

d. None of the mentioned

Solution: Reversible

A reversible process gives the maximum work.

57. During the adiabatic compression of a gas, its temperature will

a. Falls

b. Remains constant

c. Rises

d. Becomes zero

Solution: Rises

The work done on the gas during adiabatic process increases its internal energy and hence its temperature rises.

58. The work done on the gas during adiabatic process increases its internal energy and hence its temperature rises.

c. PV

Solution:

∆U =

59. A Carnot engine having an efficiency of as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is … [NEET 2017]

a) 1 J

b) 90 J

c) 99 J

d) 100 J

Solution: 90 J

Efficiency of the Carnot Engine η =

Q_input=100J

Q_output=100J = 90J

This Q_input is the amount of energy that will be absorbed when this Carnot engine is used as a refrigerator.

60. Thermodynamic processes are indicated in the following diagram. Match the following

Column-1	Column-2
Process I	Isochoric
Process II	Adiabatic
Process III	Isothermal
Process IV	Isobaric

a. P → a, Q → c, R → d, S → b

b. P → c, Q → a, R → d, S → b

c. P → c, Q → d, R → b, S → a

d. P → d, Q → b, R → a, S → c

Solution: P → c, Q → a, R → d, S → b

Process I = Isochoric

Process II = Adiabatic

Process III = Isothermal

Process IV = Isobaric

61. A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas? [NEET II – 2016]

b. mkT

Solution:

Density = =

62. In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas released 20 J of heat and 8 J of work has done on the gas. If the initial internal energy of the gas was 30 J, then the final internal energy will be

a. 58J

b. 2J

c. 42J

d. 18J

Solution: 18J

= +

= 30 -20 +8

= 18J

63. One mole of an ideal monatomic gas undergoes a process described by the equation PV₃ = constant. The heat capacity of the gas during this process is [NEET II – 2016]

a. 2 R

b. R

c. 3

d. 5

Solution: R

PV₃ = constant.

For a poly tropic process, PVα = constant

𝐶=𝐶𝑣+

= +

64. A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then [AIPMT 2015]

a. Compressing the gas isothermally will require more work to be done

b. Compressing the gas through adiabatic process will require more work to be done.

c. Compressing the gas isothermally or adiabatically will require the same amount of work.

d. Which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend up on the atomicity of the gas.

Solution: Compressing the gas through adiabatic process will require more work to be done

65. An ideal gas is compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas?

a. Isothermal

b. Adiabatic

c. Isobaric

d. Isochoric

Solution: Adiabatic

Work done on the gas

W_isochoric= 0, then

W_adiabatic> W_isothermal> W_isobaric

66. The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is -20°C, the temperature of the surroundings to which it rejects heat is: …[ReAIPMT 2015]

a) 21°C

b) 31°C

c) 41°C

d) 11°C

Solution: 31°C

Coefficient of performance of refrigerator

COP =

Where T_L→ lower temperature

T_H→higher temperature

T_H= T_L =(253)

= 303.6k

67. Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is [REAIPMT 2015]

c. 2

Solution:

According to ideal gas equation

= (1.5) ()

68. The ratio of the specific heats γ= in terms of degrees of freedom (n) is given by: … [AIPMT 2015]

Solution:

The specific heat of gas at constant volume in terms of degree of freedom 'n' is:
C_v =R
C_p - C_v = R

C_p = R + R
=R

= =
=

69. One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure, then the change in internal energy of the gas during the transition is: [ NEET 2015]

a. -12 kJ

b. 20 kJ

c. 20 kJ

d. 20 J

Solution: 20 kJ

The charge in internal energy of gas in the transition from A to B is
=nCvdT

= nR

=(2 10³6 - 5 10³4)

= -

= - 20 KJ

70. A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and then adiabatically to a volume 16V.The final pressure of the gas is …. (Take γ = 5/3)[AIPMT 2014]

b. 16P

c. 64P

d. 32P

Solution:

P₁V₁=P₂V₂

P₂ = P₁=

P₃=P₂

P₃ =

71. In the given (V – T) diagram, what is the relation between pressures P₁ and P₂?

a. P₂ = P₁

b. P₂ > P₁

c. P₂ < P₁

d. Cannot be predicted

Solution: P₂ < P₁

The slope of the graph directly proportional to

72. The efficiency of carnot’s engine operating between two reservoirs, maintained at temperature 27° C and -123° C is

a. 0.5

b. 0.4

c.0.6

d.0.25

Solution: 0.5

η = 1 -

= 1 –

=0.4229

=0.5

73. A gas is taken through the cycle A → B → C → A, as shown. What is the net work done by the gas?

a. 2000 J

b. 1000 J

c. Zero

d. –2000 J

Solution: 1000 J

Work done = area enclosed under the curve

= J

= 1000J

74. The molar specific heats of an ideal gas at constant pressure and volume are denoted by C_p and C_v respectively. If γ= and R is the universal gas constant, then C_v is equal to … [NEET 2013]

Solution:

C_v(γ-1) =R

C_p-C_v = R

From given C_p = γC_v

γC_v – C_v = R

C_v(γ-1) =R

C_v =

74. An ideal gas exerts a pressure P. The mean kinetic energy per unit volume is E. Which of the following relations is correct.

a. P=E

b. P=

c. P=E

d. P=E

Solution: P=E

K.E=

75. The change in internal energy, when a gas is cooled from 927°C to 27°C

a. 200%

b. 100%

c. 300%

d. 400%

Solution: 300%

We know that Energy of gas E ∝ T

Let T₁=927 + 273=1200 K

T₂=27 + 273=300 K

76. Four molecules of gas are having speeds of 1, 4, 8 and 16 ms^-1. The root mean square velocity of the gas molecules is

a. 7.25 ms^-1

b. 52.56 ms^-1

c. 84.25 ms^-1

d. 9.2 ms^-1

Solution: 9.2 ms^-1

C = ms^-1

= ms^-1

=9.2 ms^-1

77. At room temperature the r.m.s speed of the molecules of certain diatomic gas is found to be 1933 ms^-1. The gas is

a) H₂

b) F₂

c) O₂

d) Cl₂

Solution: H₂

M =

= 0.002Kg

M = 2gms

∴ H₂

78. By what percentage should the pressure of a given mass of gas be increased so as to decrease its volume by 10% at constant temperature?

a. 8.1%

b. 9.1%

c. 10.1%

d. 11.1%

Solution: 11.1%

Using Boyle's law

PV = P'× (0.9V)

P'= (10/9)P=1.111P

Thus Increase in P=11.1%

79. A sample of Oxygen gas and a sample of Hydrogen gas both have the same mass, the same volume and the same pressure. The ratio of their absolute temperature is

a. 1:4

b. 4:1

c. 1:16

d. 16:1

Solution: 16:1

PV = nRT

PV = RT

PV = RT₁ and PV= RT₂

Thus

=16:1

80. At what temperature is the root mean square velocity of gaseous hydrogen molecules equal to that of oxygen molecules at 47°C?

a. 20 K

b. 80 K

c. -73 K

d. 3 K

Solution: 20 K

T = 273+47 = 320K

M₀= 32

T =

= 20K

81. One kg of diatomic gas is at a pressure of 8×104N/m². The density of the gas is 4kg/m3. What is the energy of the gas due to thermal motion? [AIEEE 2009]

a. 5×104J

b. 6×104J

c. 7×104J

d. 3×104J

Solution: 5×104J

Volume=

= m³

Degree of freedom of diatomic gas=5

K.E.= PV

K.E.= ×8×10⁴ ×

=5×10⁴ J

82. If C_p and C_v denotes the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively then. [AIEEE 2007]

a. C_p - C_v=28R

b. C_p - C_v=

c. C_p - C_v=

d. C_p - C_v=R

Solution: C_p - C_v=

C_vC_p =

C_p – C_v = R

C_p – C_v =

83. The work of 146kJ is performed in order to compresses one kilo mole of gas adiabatically and in this process the temperature of the gas increased by 7°C. The gas is (R=8.3 J mol-1K-1) [AIEEE 2006]

a. diatomic

b. triatomic

c. a mixture of monoatomic and diatomic

d. monoatomic

Solution: diatomic

For adiabatic process,

dQ = 0

dU =

nC_vdT= +146×10³J

×7 =+146×10³

=+146×10³

f= 5.02=5

So, it is a diatomic gas.

85. During an adiabatic process, the pressure of a gas is found to be proportional to the absolute temperature. The ratio for the gas is

Solution:

P T⁵, an adiabatic process P T

86. For an isothermal expansion of a perfect gas, the value of is equal to

Solution:

PV = Constant

PV- VP = 0

87. If r denotes the ratio of adiabatic of two specific heats of a gas. Then what is the ratio of slope of adiabatic and isothermal P→V curves at their point of intersection?

b. γ – 1

c. γ

d. γ + 1

Solution: γ

For isothermal process, PV = constant

∴ P ΔV – VΔP = 0

i.e. = – ----- (1)

For adiabatic process PVγ = constant

PγV⁽^γ–1) ΔV + Vγ ΔP = 0

∴ γV⁽^γ–1) ΔV =

∴ =

(ΔP / P) = [(– γ ΔV) / V]

i.e. (ΔP / ΔV) = – γ (P/V) ----- (2)

At point of intersection, sloper’s ratio is

= γ

88. A container that suits the occurrence of an isothermal process should be made of

a. wood

b. coppers

c. glass

d. cloth

Solution: coppers

An isothermal process that takes place at constant temperature, must be carried out in a vessel with conducting wall so that heat generated should go out at once.

89. A thermodynamic process in which temperature T of the system remains constant throughout variable P and V may change is called

a. Isothermal process

b. Isochoric process

c. Isobaric process

d. None of this

Solution: Isothermal process

90. The volume of an ideal gas is 1 liter column and its pressure is equal to 72 cm of Hg. The volume of gas is made 900 cm³ by compressing it isothermally. The stress of the gas will be

a. 4cm

b. 6cm

c. 7cm

d. 8cm

Solution: 8cm

Given:

Initial pressure on gas P₁=72cm of Hg

Initial volume of gas =V₁=1L

Final volume of gas=V₂=9L

Let P₂ be the final pressure on Gas

From Boyle's law: P₁V₁=P₂V₂

P₂=

=72x

=8cm of Hg.

91. In adiabatic expansion the initial value will be

a. ∆U = 0

b. ∆U = Positive

c. ∆U = Negative

d. ∆W = 0

Solution: ∆U = Negative.

92. The pressure and density of a diatomic gas γ = then, Change adiabatically from (p, d) to (p¹, d¹) if = 32 then should be

a. 128

c. 32

d. None of this

Solution: 128

= = (32)^7/5

= 128

93. An engine is supposed to operate between two reservoirs at temperature 727° C and 227° C. The maximum possible efficiency of such an engine is

d. 1

Solution:

n = 1 -

=1-

94. A gas expands 0.25m³at constant pressure 10³the work done is

a. 250 J

b. 2.5 erg

c. 250 W

d. 250 N

Solution: 250 J

=PV = 10³× 0.25

=250 J

95. The volume of air increases by 5% in an adiabatic expansion. The percentange decrease in its pressure will be

a. 5%

b. 6%

c. 7%

d. 8%

Solution: 7%

PV^γ= K or p^γV^γ-1dV+dP.V^γ= 0

×100= -γ

= -1.4×5

=7%

96. For adiabatic process which relation is true mentioned below? γ =

a. P^γV = Constant

b. T^γV = Constant

c. TV^γ= Constant

d. TV^γ-1= Constant

Solution: TV^γ-1= Constant.

PV = RT

P =

= Constant

TV ^(γ-1)= Constant.

97. For adiabatic process which one is wrong statement?

a. dQ = 0

b. entropy is not constant

c. du = - dw

d. Q = Constant

Solution: Q = Constant

Internal energy of the system can be constant, that is ∆U = Constant but the quantity (Q) cannot be constant.

98. Carnot engine working between a source temperature of T₂ and sink temperature of T₁ has efficiency of 25%. If the sink temperature is reduced by 20°C, the efficiency is increased to 30%. Then the source and the sink temperature is

a. 400°C and 300°C

b. 300°C and 400°C

c. 200°C and 300°C

d. 300°C and 200°C

Solution: 400°C and 300°C

Efficiency, η = (T₂ – T₁) / T₂= 25% = 0.25

Where T₂ is the source temperature and T₁ is the sink temperature.

T₂ – T₁ = 0.25 T₂

T₁ = 0.75 T₂

In the second case, if the sink is reduced by 20o C, then

η= T₂ – (T₁ – 20) / T₂ = 30% = 0.3

T₂ – T₁ + 20 = 0.3 T₂ ………………………………. (1)

Putting the value of T1 in equation (1), we will get:

T2 – 0.75 T2 + 20 = 0.3 T2

0.25 T2 + 20 = 0.3 T2

20 = 0.3 T2 – 0.25 T2 = 0.05 T2

T2 = 20 / 0.05 = 400°C

Hence, T1 = 0.75 T2

T1 = 0.75 X 400 = 300°C

Hence, temperature of the source and the sink is 400o C and 300o C respectively.

99. A diatomic gas is used in a carnot engine as the working substance, if during the adiabatic expansion part of the cycle the volume of the gas increases from V to 32V_,the efficiency of the engine is

a. 0.99

b. 0.75

c. 0.50

d. 0.25

Solution: 0.75

TV ^(γ-1)= Constant.

T_bV_b⁽^γ-1)= T_cV_c^(γ-1)= [32]^(7/5)-1

= [2⁵]^2/5

T_b= 4T_c= 4

1- = 1 - =

100. A carnot’ s engine whose sink is at a temperature of 300 K has an efficiency of 40% by space should the temperature of the source be increase the efficiency to 50% of original efficiency?

a. a 275 K

b. 325 K

c. 300 K

d. 250 K

Solution: 250 K

Temperature of sink (T_c) = 300K

Temperature of source = T_h

Initial efficiency = 40%

Efficiency n₁ = 1 -