MCQs / Physics / Kinetic Theory and First Law of Thermodynamics
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: 1.0 8.31 J mol^-1 K^-1 4.36 10^46 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: a straight line through the origin a parabola a hyperbola 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: decreases by 400 J increases by 800 J increases by 400 J 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: 2 10^5 J 300 J 200 J 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: 250 J 0 J 1000 J 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: a circle a hyperbola a vertical line 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: infinite equal to the time between collisions negligible compared with the time between collisions 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: 30 J 3000 J 300 J -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: the zeroth law Newton's first law the second law only 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: becomes zero stays the same, while their potential energy increases decreases 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: 84,000 J 42,000 J 1.68 10^5 J 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: becomes one quarter halves becomes 4 times as large 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: they have no mass there are no intermolecular forces (except during collisions) they are at rest 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: p propto T V propto T p propto 1T 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: one gram of any substance one mole of a substance one kilogram 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: zero always the area under the curve the gradient 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: 8.31 mol^-1 1.38 10^-23 mol^-1 6.02 10^23 mol^-1 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: 831 J 416 J 1250 J 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: zero for ideal gases the same for all gases different for each gas 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: +300 J -300 J +700 J -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: is in thermal equilibrium with other bodies is a conductor will float 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: increases, because the molecules move faster stays the same decreases 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: the can melts first the gas pressure rises as its temperature rises at constant volume the gas cools 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: high temperature and low pressure high pressure and high density low temperature and high pressure 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.) 150 J 600 J 300 J 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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