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Kinetic Theory and First Law of Thermodynamics MCQs — Physics (page 6)

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.

126. If k is 1.38 10^-23 J/K and NA is 6.02 10^23 mol^-1, then R is:

  1. 1.0
  2. 8.31 J mol^-1 K^-1
  3. 4.36 10^46
  4. 2.29 10^-47

Answer: B) 8.31 J mol^-1 K^-1

Explanation: R = kNA. SLO P-11-C-04 · Knowledge

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127. A graph of p against 1V at constant temperature is:

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

Answer: A) a straight line through the origin

Explanation: p = const 1V. SLO P-11-C-10 · Application

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128. A gas in a cylinder is compressed by 400 J of work while losing no heat. Its internal energy:

  1. decreases by 400 J
  2. increases by 800 J
  3. increases by 400 J
  4. is unchanged

Answer: C) increases by 400 J

Explanation: Adiabatic compression. SLO P-11-C-07 · Understanding

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129. A gas expands from 1 L to 3 L at 100 kPa. The work done by the gas is:

  1. 2 10^5 J
  2. 300 J
  3. 200 J
  4. 100 J

Answer: C) 200 J

Explanation: Delta V = 2 10^-3 m^3; W = 10^5 2 10^-3 = 200 J. SLO P-11-C-06 · Application

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130. 2 mol of an ideal gas is heated at constant volume and its internal energy rises by 500 J. The heat supplied is:

  1. 250 J
  2. 0 J
  3. 1000 J
  4. 500 J

Answer: D) 500 J

Explanation: W = 0, so Q = Delta U. SLO P-11-C-08 · Application

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131. In an expansion at constant pressure, the p–V graph is:

  1. a circle
  2. a hyperbola
  3. a vertical line
  4. a horizontal straight line

Answer: D) a horizontal straight line

Explanation: p constant. SLO P-11-C-06 · Application

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132. The time a molecule spends in a collision is assumed to be:

  1. infinite
  2. equal to the time between collisions
  3. negligible compared with the time between collisions
  4. longer than the time between collisions

Answer: C) negligible compared with the time between collisions

Explanation: Collisions are instantaneous. SLO P-11-C-05 · Understanding

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133. A gas at constant pressure 1 10^5 Pa changes volume from 0.002 m^3 to 0.005 m^3. The work done by the gas is:

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

Answer: C) 300 J

Explanation: W = pDelta V = 1 10^5 (0.005 - 0.002) = 300 J. SLO P-11-C-06 · Application

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134. A machine that produces work without any energy input would violate:

  1. the zeroth law
  2. Newton's first law
  3. the second law only
  4. the first law of thermodynamics

Answer: D) the first law of thermodynamics

Explanation: It would create energy (perpetual motion of the first kind). SLO P-11-C-11 · Application

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135. When water at 100 °C turns to steam at 100 °C, the molecules' average kinetic energy:

  1. becomes zero
  2. stays the same, while their potential energy increases
  3. decreases
  4. increases

Answer: B) stays the same, while their potential energy increases

Explanation: Constant temperature. SLO P-11-C-12 · Understanding

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136. A 2 kg body of specific heat capacity 4200 J kg^-1 K^-1 is warmed by 10 K (no work done). The increase in its internal energy is:

  1. 84,000 J
  2. 42,000 J
  3. 1.68 10^5 J
  4. 8400 J

Answer: A) 84,000 J

Explanation: Delta U = mcDeltatheta = 2 4200 10 = 84,000 J. SLO P-11-C-12 · Understanding

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137. At constant temperature, a gas at 200 kPa expands until its pressure is 50 kPa. Its volume:

  1. becomes one quarter
  2. halves
  3. becomes 4 times as large
  4. doubles

Answer: C) becomes 4 times as large

Explanation: V propto 1/p. SLO P-11-C-10 · Application

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138. Molecules of an ideal gas have no potential energy because:

  1. they have no mass
  2. there are no intermolecular forces (except during collisions)
  3. they are at rest
  4. they are very hot

Answer: B) there are no intermolecular forces (except during collisions)

Explanation: So internal energy = total KE. SLO P-11-C-05 · Understanding

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139. The pressure law states that at constant volume:

  1. p propto T
  2. V propto T
  3. p propto 1T
  4. p propto V

Answer: A) p propto T

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

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140. NA is the number of particles in:

  1. one gram of any substance
  2. one mole of a substance
  3. one kilogram
  4. one cubic metre

Answer: B) one mole of a substance

Explanation: Definition. SLO P-11-C-04 · Knowledge

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141. On a p–V graph, the work done by a gas is:

  1. zero always
  2. the area under the curve
  3. the gradient
  4. the intercept

Answer: B) the area under the curve

Explanation: W = int p,dV. SLO P-11-C-06 · Application

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142. The Avogadro constant NA is about:

  1. 8.31 mol^-1
  2. 1.38 10^-23 mol^-1
  3. 6.02 10^23 mol^-1
  4. 3 10^8 mol^-1

Answer: C) 6.02 10^23 mol^-1

Explanation: Particles per mole. SLO P-11-C-04 · Knowledge

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143. 2 mol of an ideal gas is heated at constant pressure so its temperature rises by 50 K. The work done by the gas is:

  1. 831 J
  2. 416 J
  3. 1250 J
  4. 1660 J

Answer: A) 831 J

Explanation: W = pDelta V = nRDelta T = 2 8.31 50 = 831 J. SLO P-11-C-06 · Application

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144. k is a fundamental constant whose value is:

  1. zero for ideal gases
  2. the same for all gases
  3. different for each gas
  4. dependent on pressure

Answer: B) the same for all gases

Explanation: It's universal. SLO P-11-C-04 · Knowledge

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145. A gas expands from 2 L to 5 L at 1 atm (10^5 Pa). The work done ON the gas is:

  1. +300 J
  2. -300 J
  3. +700 J
  4. -500 J

Answer: B) -300 J

Explanation: Work by gas = +300 J, so work on gas = -300 J. SLO P-11-C-07 · Understanding

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146. Temperature is the property that determines whether a body:

  1. is in thermal equilibrium with other bodies
  2. is a conductor
  3. will float
  4. has mass

Answer: A) is in thermal equilibrium with other bodies

Explanation: Basis of the zeroth law. SLO P-12-C-01 · Knowledge

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147. When the temperature of a gas rises, its internal energy:

  1. increases, because the molecules move faster
  2. stays the same
  3. decreases
  4. becomes zero

Answer: A) increases, because the molecules move faster

Explanation: Mean KE ∝ T. SLO P-11-C-02 · Understanding

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148. A sealed can explodes when thrown into a fire because:

  1. the can melts first
  2. the gas pressure rises as its temperature rises at constant volume
  3. the gas cools
  4. the volume increases

Answer: B) the gas pressure rises as its temperature rises at constant volume

Explanation: Pressure law. SLO P-11-C-09 · Understanding

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149. A real gas behaves most like an ideal gas at:

  1. high temperature and low pressure
  2. high pressure and high density
  3. low temperature and high pressure
  4. low temperature and low pressure only

Answer: A) high temperature and low pressure

Explanation: Molecules are far apart and fast. SLO P-11-C-05 · Understanding

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150. A gas absorbs 500 J of heat and does 200 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. 600 J
  3. 300 J
  4. 3000 J

Answer: C) 300 J

Explanation: Delta U = Q - W = 500 - (200) = 300 J. SLO P-11-C-08 · Application

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