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Kinetic Theory and First Law of Thermodynamics MCQs — Physics (page 7)

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.

151. A cup of hot tea left in a room eventually:

  1. reaches the room's temperature
  2. becomes hotter
  3. becomes colder than the room
  4. stays hot forever

Answer: A) reaches the room's temperature

Explanation: It comes to thermal equilibrium. SLO P-12-C-01 · Knowledge

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152. A body at 80 °C is placed in contact with a body at 20 °C (isolated system). Their final common temperature is:

  1. 80 °C
  2. 20 °C
  3. between 20 °C and 80 °C
  4. 100 °C

Answer: C) between 20 °C and 80 °C

Explanation: The hot one cools and the cold one warms. SLO P-12-C-01 · Knowledge

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153. A gas receives 600 J of heat and its internal energy rises by 600 J. The work done by the gas is:

  1. 0 J
  2. -600 J
  3. 1200 J
  4. 600 J

Answer: A) 0 J

Explanation: W = Q - Delta U = 0 (constant volume). SLO P-11-C-11 · Application

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154. The SI unit of thermodynamic temperature is:

  1. kelvin
  2. calorie
  3. joule
  4. degree Celsius

Answer: A) kelvin

Explanation: A base SI unit. SLO P-12-C-01 · Knowledge

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155. In the sign convention Delta U = Q + Won, a gas that is compressed has:

  1. Won = 0
  2. Won 0
  3. Won < 0
  4. Q = 0 always

Answer: B) Won 0

Explanation: Work done on the gas is positive in compression. SLO P-11-C-07 · Understanding

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156. 1 J of work equals:

  1. 1 Pa m
  2. 1 Pa m^3
  3. 1 Pa/m^3
  4. 1 N/m

Answer: B) 1 Pa m^3

Explanation: N m^-2 × m^3 = N m. SLO P-11-C-06 · Application

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157. 3 mol of an ideal gas at 350 K occupies 0.1 m^3. Its pressure is (R = 8.31 J mol^-1 K^-1):

  1. 8730 Pa
  2. 8.73 10^5 Pa
  3. 1.75 10^5 Pa
  4. 87,300 Pa

Answer: D) 87,300 Pa

Explanation: P = nRTV = 3 8.31 3500.1 = 87,300 Pa. SLO P-11-C-03 · Application

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158. Isothermal curves on a p–V diagram at higher temperatures lie:

  1. on the axes
  2. further from the origin
  3. on the same curve
  4. closer to the origin

Answer: B) further from the origin

Explanation: pV = nRT is larger. SLO P-11-C-10 · Application

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159. The internal energy of 1 mol of a monatomic ideal gas is U = 32nRT. When it is heated from 300 K to 400 K, its internal energy increases by:

  1. 1250 J
  2. 2490 J
  3. 831 J
  4. 623 J

Answer: A) 1250 J

Explanation: Delta U = 32nRDelta T = 1.5 1 8.31 100 = 1250 J. SLO P-11-C-12 · Understanding

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160. The area under an isothermal p–V curve between two volumes represents:

  1. the work done by the gas
  2. the temperature
  3. the heat capacity
  4. the pressure

Answer: A) the work done by the gas

Explanation: As for any p–V graph. SLO P-11-C-10 · Application

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161. When an ideal gas expands isothermally, the heat absorbed:

  1. is wasted
  2. is all converted into work done by the gas
  3. decreases internal energy
  4. increases internal energy

Answer: B) is all converted into work done by the gas

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

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162. For an adiabatic change:

  1. Q = 0, Delta U = -W
  2. Q = W
  3. Delta U = 0
  4. W = 0

Answer: A) Q = 0, Delta U = -W

Explanation: No heat exchanged. SLO P-11-C-08 · Application

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163. Thermal equilibrium is reached faster when:

  1. they are in a vacuum flask
  2. the bodies are good thermal conductors and in good contact
  3. they are insulated from each other
  4. they have no contact

Answer: B) the bodies are good thermal conductors and in good contact

Explanation: Energy transfer is faster. SLO P-12-C-01 · Knowledge

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164. In the compression stroke of an engine, the piston:

  1. receives work from the gas
  2. does work on the gas
  3. cools the gas
  4. does no work

Answer: B) does work on the gas

Explanation: Compression. SLO P-11-C-07 · Understanding

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165. For an isothermal change of an ideal gas:

  1. Q = 0
  2. W = 0
  3. Delta U = 0, Q = W
  4. Delta U = Q

Answer: C) Delta U = 0, Q = W

Explanation: Constant T ⇒ constant internal energy. SLO P-11-C-08 · Application

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166. The Boltzmann constant links:

  1. mass with energy
  2. the energy of a molecule with absolute temperature
  3. charge with current
  4. frequency with energy

Answer: B) the energy of a molecule with absolute temperature

Explanation: e.g. barEk = 32kT. SLO P-11-C-04 · Knowledge

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167. Why does a gas fill any container it is put in?

  1. it is attracted by the walls
  2. it has a fixed volume
  3. its molecules move randomly and freely in all directions
  4. it is heavier than air

Answer: C) its molecules move randomly and freely in all directions

Explanation: No strong intermolecular forces. SLO P-11-C-09 · Understanding

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168. Which quantity decides the direction of heat flow?

  1. temperature
  2. heat capacity
  3. mass
  4. internal energy

Answer: A) temperature

Explanation: A large cold lake can have more internal energy than a hot nail. SLO P-12-C-01 · Knowledge

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169. A bubble rising from the bottom of a lake (constant temperature) grows bigger because:

  1. the water pushes it outward
  2. the water pressure on it decreases
  3. the temperature rises
  4. gas is added

Answer: B) the water pressure on it decreases

Explanation: Boyle's law. SLO P-11-C-09 · Understanding

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170. The number of molecules in 2 mol of gas is:

  1. 2.76 10^-23
  2. 3.01 10^23
  3. 6.02 10^23
  4. 1.20 10^24

Answer: D) 1.20 10^24

Explanation: N = nNA = 2 6.02 10^23. SLO P-11-C-04 · Knowledge

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171. If the internal energy of a fixed mass of ideal gas increases, its temperature must:

  1. become zero
  2. stay the same
  3. decrease
  4. increase

Answer: D) increase

Explanation: U propto T for an ideal gas. SLO P-11-C-12 · Understanding

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172. The Boltzmann constant is related to the molar gas constant by:

  1. k = NAR
  2. k = R + NA
  3. k = RNA
  4. k = RNA

Answer: D) k = RNA

Explanation: The gas constant per molecule. SLO P-11-C-04 · Knowledge

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173. 2 mol of an ideal gas at 300 K occupies 0.05 m^3. Its pressure is (R = 8.31 J mol^-1 K^-1):

  1. 99,700 Pa
  2. 9970 Pa
  3. 1.99 10^5 Pa
  4. 9.97 10^5 Pa

Answer: A) 99,700 Pa

Explanation: P = nRTV = 2 8.31 3000.05 = 99,700 Pa. SLO P-11-C-03 · Application

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174. At high pressure a real gas departs from ideal behaviour because:

  1. the molecules stop moving
  2. the gas becomes a solid
  3. the molecules' own volume becomes significant
  4. temperature becomes zero

Answer: C) the molecules' own volume becomes significant

Explanation: The point-molecule assumption fails. SLO P-11-C-05 · Understanding

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175. Doubling the number of gas molecules in a fixed container at constant temperature:

  1. quadruples the pressure
  2. doubles the pressure
  3. halves the pressure
  4. does not change pressure

Answer: B) doubles the pressure

Explanation: Twice as many collisions. SLO P-11-C-09 · Understanding

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