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Kinetic Theory and First Law of Thermodynamics MCQs — Physics (page 5)

180 multiple-choice questions on Kinetic Theory and First Law of Thermodynamics from Physics, each with the correct answer and, where available, a teacher's explanation. Read through them to revise the chapter, then practise the same questions one by one to test yourself.

101. When a gas is compressed by a piston, work is done:

  1. by the gas
  2. only by heat
  3. by nobody
  4. on the gas by the surroundings

Answer: D) on the gas by the surroundings

Explanation: The piston pushes inward. SLO P-11-C-07 · Understanding

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102. A syringe of air is sealed and the plunger is pushed in slowly. The molecules:

  1. become smaller
  2. move faster
  3. hit the walls less frequently
  4. hit the walls more frequently

Answer: D) hit the walls more frequently

Explanation: Same speed, smaller space. SLO P-11-C-09 · Understanding

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103. An ideal gas is heated from 200 K to 800 K. The average kinetic energy of its molecules is multiplied by:

  1. 2
  2. 0.25
  3. 16
  4. 4

Answer: D) 4

Explanation: barEk propto T, so the factor is 800200 = 4. SLO P-11-C-02 · Understanding

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104. A gas at constant pressure 3 10^5 Pa changes volume from 0.5 m^3 to 0.8 m^3. The work done by the gas is:

  1. 9000 J
  2. 90,000 J
  3. -90,000 J
  4. 9 10^5 J

Answer: B) 90,000 J

Explanation: W = pDelta V = 3 10^5 (0.8 - 0.5) = 90,000 J. SLO P-11-C-06 · Application

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105. In pV = NkT, N stands for:

  1. the number of moles
  2. the number of molecules
  3. Avogadro's constant
  4. the number of collisions

Answer: B) the number of molecules

Explanation: n = moles; N = molecules. SLO P-11-C-03 · Application

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106. A gas exchanges no heat with its surroundings and does 150 J of work. The change in its internal energy is: (Use Q = Delta U + W, where Q = heat supplied to the gas and W = work done by the gas.)

  1. -150 J
  2. -300 J
  3. -1500 J
  4. -75 J

Answer: A) -150 J

Explanation: Delta U = Q - W = 0 - (150) = -150 J. SLO P-11-C-08 · Application

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107. 0.5 mol of an ideal gas at 400 K occupies 0.01 m^3. Its pressure is (R = 8.31 J mol^-1 K^-1):

  1. 3.32 10^5 Pa
  2. 1.66 10^5 Pa
  3. 16,600 Pa
  4. 1.66 10^6 Pa

Answer: B) 1.66 10^5 Pa

Explanation: P = nRTV = 0.5 8.31 4000.01 = 1.66 10^5 Pa. SLO P-11-C-03 · Application

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108. A hammer hitting a nail repeatedly makes the nail warm. The nail's internal energy:

  1. becomes zero
  2. increases due to work done on it
  3. decreases
  4. stays constant

Answer: B) increases due to work done on it

Explanation: Work → internal energy. SLO P-11-C-12 · Understanding

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109. The internal energy of a body is:

  1. the gravitational energy of the body
  2. the sum of the random kinetic and potential energies of its molecules
  3. the heat it contains
  4. only the kinetic energy of the whole body

Answer: B) the sum of the random kinetic and potential energies of its molecules

Explanation: Internal energy is microscopic. SLO P-11-C-02 · Understanding

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110. The average translational kinetic energy of a gas molecule is:

  1. 32kT
  2. 12kT
  3. 3kT
  4. kT^2

Answer: A) 32kT

Explanation: k = Boltzmann constant. SLO P-11-C-02 · Understanding

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111. At constant volume, the pressure of a gas rises when it is heated because the molecules:

  1. move faster, hitting the walls harder and more often
  2. stop moving
  3. attract each other
  4. become bigger

Answer: A) move faster, hitting the walls harder and more often

Explanation: Pressure ∝ T at constant V. SLO P-11-C-09 · Understanding

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112. For an ideal gas, internal energy depends only on:

  1. its temperature (for a fixed amount of gas)
  2. its pressure only
  3. its volume only
  4. the container's shape

Answer: A) its temperature (for a fixed amount of gas)

Explanation: No intermolecular PE in an ideal gas. SLO P-11-C-02 · Understanding

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113. A graph of pV against p for an ideal gas at constant temperature is:

  1. a straight line through the origin
  2. a horizontal line
  3. a vertical line
  4. a hyperbola

Answer: B) a horizontal line

Explanation: pV does not change. SLO P-11-C-10 · Application

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114. The Boltzmann constant can be thought of as:

  1. the speed of a molecule
  2. the charge on an electron
  3. the gas constant for one mole
  4. the gas constant for one molecule

Answer: D) the gas constant for one molecule

Explanation: k = R/NA. SLO P-11-C-04 · Knowledge

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115. At absolute zero, the kinetic energy of the molecules of an ideal gas would be:

  1. maximum
  2. minimum (zero for an ideal gas)
  3. infinite
  4. the same as at 0 °C

Answer: B) minimum (zero for an ideal gas)

Explanation: Molecular motion is minimal. SLO P-11-C-02 · Understanding

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116. In a car engine's power stroke, the hot gas:

  1. does no work
  2. has work done on it
  3. does work on the piston
  4. absorbs work from the crankshaft

Answer: C) does work on the piston

Explanation: Expansion pushes the piston. SLO P-11-C-07 · Understanding

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117. Which statement is correct?

  1. Work done on the gas is always positive
  2. Both are always zero
  3. Work done by the gas = −(work done on the gas)
  4. Work done by the gas = work done on the gas

Answer: C) Work done by the gas = −(work done on the gas)

Explanation: Same energy transfer, opposite viewpoints. SLO P-11-C-07 · Understanding

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118. When a gas is heated at constant volume, all the heat supplied:

  1. is converted to work
  2. is lost
  3. increases the internal energy
  4. goes into the container

Answer: C) increases the internal energy

Explanation: W = 0 Rightarrow Q = Delta U. SLO P-11-C-11 · Application

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119. A gas at 1 10^5 Pa, 3 L and 300 K is changed to 1.5 10^5 Pa and 450 K. Its new volume is:

  1. 6 L
  2. 1.5 L
  3. 3 L
  4. 12 L

Answer: C) 3 L

Explanation: P1V1T1 = P2V2T2 Rightarrow V2 = 3 L. SLO P-11-C-03 · Application

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120. The zeroth law of thermodynamics states that if A and B are each in thermal equilibrium with C, then:

  1. A is hotter than B
  2. A and B are in thermal equilibrium with each other
  3. C is hotter than both
  4. A and B have equal masses

Answer: B) A and B are in thermal equilibrium with each other

Explanation: This allows thermometers to work. SLO P-12-C-01 · Knowledge

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121. A large iceberg at −10 °C can have more internal energy than a cup of boiling water because:

  1. ice is hotter
  2. temperature measures total energy
  3. boiling water has no energy
  4. internal energy also depends on the amount of matter

Answer: D) internal energy also depends on the amount of matter

Explanation: Temperature is not the same as internal energy. SLO P-11-C-12 · Understanding

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122. The internal energy of a gas increases by 150 J while it does 100 J of work. The heat supplied to it is: (Use Q = Delta U + W, where Q = heat supplied to the gas and W = work done by the gas.)

  1. 2500 J
  2. 500 J
  3. 250 J
  4. 125 J

Answer: C) 250 J

Explanation: Q = Delta U + W = 150 + 100 = 250 J. SLO P-11-C-08 · Application

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123. A fixed mass of gas at 2 10^5 Pa occupies 3 L. At constant temperature it is changed to 12 L. The new pressure is:

  1. 8 10^5 Pa
  2. 50,000 Pa
  3. 1 10^5 Pa
  4. 25,000 Pa

Answer: B) 50,000 Pa

Explanation: p1V1 = p2V2 Rightarrow p2 = 2 10^5 312 = 50,000 Pa. SLO P-11-C-10 · Application

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124. In a Boyle's law experiment, which quantity must be kept constant (besides temperature)?

  1. the mass (amount) of gas
  2. the pressure
  3. the gradient
  4. the volume

Answer: A) the mass (amount) of gas

Explanation: No gas should leak. SLO P-11-C-10 · Application

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125. A gas at constant pressure 1 10^5 Pa changes volume from 0.001 m^3 to 0.003 m^3. The work done by the gas is:

  1. 20 J
  2. 2000 J
  3. -200 J
  4. 200 J

Answer: D) 200 J

Explanation: W = pDelta V = 1 10^5 (0.003 - 0.001) = 200 J. SLO P-11-C-06 · Application

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