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Kinetic Theory and First Law of Thermodynamics MCQs — Physics (page 2)

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.

26. When ice is put in a drink, energy flows:

  1. from the ice to the drink
  2. in neither direction
  3. from the glass to the room only
  4. from the drink to the ice

Answer: D) from the drink to the ice

Explanation: Hot to cold. SLO P-12-C-01 · Knowledge

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27. The pressure of a gas, from kinetic theory, is:

  1. p = 13rhooverlinec^2
  2. p = 3rhooverlinec^2
  3. p = 12rho c
  4. p = rho gh

Answer: A) p = 13rhooverlinec^2

Explanation: overlinec^2 is the mean square speed. SLO P-11-C-05 · Understanding

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28. In an adiabatic expansion, the work done by the gas comes from:

  1. its internal energy (it cools)
  2. heat supplied
  3. the surroundings' heat
  4. nowhere

Answer: A) its internal energy (it cools)

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

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29. The first law of thermodynamics expresses conservation of energy because:

  1. it says entropy increases
  2. heat supplied = increase in internal energy + work done; no energy is created or destroyed
  3. it defines temperature
  4. it says heat always flows hot to cold

Answer: B) heat supplied = increase in internal energy + work done; no energy is created or destroyed

Explanation: It accounts for all energy. SLO P-11-C-11 · Application

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30. A gas at 2 10^5 Pa and 300 K occupies 0.01 m^3. The amount of gas is:

  1. 0.802 mol
  2. 8.02 mol
  3. 1.6 mol
  4. 0.401 mol

Answer: A) 0.802 mol

Explanation: n = PVRT = 0.802 mol. SLO P-11-C-03 · Application

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31. The internal energy of a gas can be increased by:

  1. heating it or doing work on it
  2. only heating it
  3. only cooling it
  4. only letting it expand

Answer: A) heating it or doing work on it

Explanation: First law. SLO P-11-C-02 · Understanding

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32. An ideal gas is heated from 300 K to 600 K. The average kinetic energy of its molecules is multiplied by:

  1. 2
  2. 0.5
  3. 4
  4. 1.41

Answer: A) 2

Explanation: barEk propto T, so the factor is 600300 = 2. SLO P-11-C-02 · Understanding

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33. When ice melts at 0 °C, its internal energy:

  1. increases although its temperature stays constant
  2. decreases
  3. stays the same
  4. becomes zero

Answer: A) increases although its temperature stays constant

Explanation: Energy increases molecular PE (breaking bonds). SLO P-11-C-02 · Understanding

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34. The formula W = pDelta V is valid when:

  1. the gas is a liquid
  2. the temperature is zero
  3. the volume is constant
  4. the pressure is constant

Answer: D) the pressure is constant

Explanation: Otherwise use the area under the p–V graph. SLO P-11-C-06 · Application

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35. 5 10^22 molecules of an ideal gas at 400 K fill 0.002 m^3. The pressure is (k = 1.38 10^-23 J/K):

  1. 1.38 10^5 Pa
  2. 1.38 10^6 Pa
  3. 2.76 10^5 Pa
  4. 13,800 Pa

Answer: A) 1.38 10^5 Pa

Explanation: PV = NkT Rightarrow P = NkTV = 1.38 10^5 Pa. SLO P-11-C-03 · Application

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36. According to kinetic theory, raising the temperature of a gas:

  1. increases the size of the molecules
  2. increases the mean speed of its molecules
  3. decreases the mean speed
  4. stops molecular motion

Answer: B) increases the mean speed of its molecules

Explanation: Mean KE ∝ T. SLO P-11-C-05 · Understanding

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37. When 1000 J of heat is supplied to a gas which does 400 J of work, the 'missing' 600 J:

  1. is stored as increased internal energy
  2. is destroyed
  3. becomes mass
  4. is lost as light

Answer: A) is stored as increased internal energy

Explanation: Energy is conserved. SLO P-11-C-11 · Application

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38. The first law treats heat and work as:

  1. the same as temperature
  2. forms of mass
  3. unrelated quantities
  4. two ways of transferring energy

Answer: D) two ways of transferring energy

Explanation: Both change internal energy. SLO P-11-C-11 · Application

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39. Heating a metal rod increases the amplitude of atomic vibrations. This shows:

  1. increased internal energy with temperature
  2. decreased internal energy
  3. loss of mass
  4. a change of state

Answer: A) increased internal energy with temperature

Explanation: Kinetic part. SLO P-11-C-12 · Understanding

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40. When a gas does work by expanding (with no heat flow), its temperature:

  1. doubles
  2. rises
  3. stays the same
  4. falls

Answer: D) falls

Explanation: Energy leaves as work. SLO P-11-C-07 · Understanding

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41. Points A (p = 100 kPa, V = 3 L) and B (p = 150 kPa, V = 2 L) lie on the same:

  1. adiabatic line
  2. isochore
  3. isobar
  4. isotherm (same pV)

Answer: D) isotherm (same pV)

Explanation: pV = 300 in both. SLO P-11-C-10 · Application

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42. The internal energy of 2 mol of a monatomic ideal gas is U = 32nRT. When it is heated from 250 K to 350 K, its internal energy increases by:

  1. 1250 J
  2. 4990 J
  3. 2490 J
  4. 1660 J

Answer: C) 2490 J

Explanation: Delta U = 32nRDelta T = 1.5 2 8.31 100 = 2490 J. SLO P-11-C-12 · Understanding

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43. Two bodies are in thermal equilibrium when:

  1. they are made of the same material
  2. they are at the same temperature, so there is no net heat flow between them
  3. they have the same mass
  4. they have the same internal energy

Answer: B) they are at the same temperature, so there is no net heat flow between them

Explanation: Equal temperature means no net energy transfer. SLO P-12-C-01 · Knowledge

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44. The value of the Boltzmann constant is approximately:

  1. 8.31 J mol^-1 K^-1
  2. 1.38 10^-23 J/K
  3. 6.63 10^-34 J s
  4. 6.02 10^23 mol^-1

Answer: B) 1.38 10^-23 J/K

Explanation: 8.316.02 10^23. SLO P-11-C-04 · Knowledge

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45. Which equation contains the Boltzmann constant?

  1. E = mc^2
  2. pV = NkT
  3. pV = nRT
  4. F = ma

Answer: B) pV = NkT

Explanation: Molecular form of the gas law. SLO P-11-C-04 · Knowledge

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46. A gas at constant pressure 2 10^5 Pa changes volume from 0.01 m^3 to 0.025 m^3. The work done by the gas is:

  1. -3000 J
  2. 30,000 J
  3. 3000 J
  4. 300 J

Answer: C) 3000 J

Explanation: W = pDelta V = 2 10^5 (0.025 - 0.01) = 3000 J. SLO P-11-C-06 · Application

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47. A gas is compressed at constant pressure. The work done by the gas is:

  1. positive
  2. infinite
  3. zero
  4. negative

Answer: D) negative

Explanation: Delta V < 0. SLO P-11-C-06 · Application

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48. In a Boyle's law experiment, the gas is compressed slowly so that:

  1. it heats up
  2. its volume stays constant
  3. its temperature stays constant
  4. it leaks out

Answer: C) its temperature stays constant

Explanation: Isothermal conditions. SLO P-11-C-10 · Application

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49. In terms of particles, the pressure law (p propto T) arises because at higher T:

  1. molecules collide less
  2. both the rate and force of collisions increase
  3. only the number of molecules increases
  4. the walls move outward

Answer: B) both the rate and force of collisions increase

Explanation: Momentum change per collision and collision rate both rise. SLO P-11-C-09 · Understanding

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50. When a gas expands against a piston, work is done:

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

Answer: B) by the gas on the surroundings

Explanation: The gas pushes the piston out. SLO P-11-C-07 · Understanding

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