Physics · Thermodynamics · chapter 11 of 14

Class 11 Physics Chapter 11: Thermodynamics MCQs with answers

10 multiple-choice questions on Thermodynamics, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
Three bodies P, Q and R are tested in pairs. P is found to be in thermal equilibrium with R, and Q is also found to be in thermal equilibrium with R. P and Q are now placed in contact through a conducting wall. Then
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Answer: (c) no net heat flows between P and Q, since they are already in thermal equilibrium

By the Zeroth Law, two systems each in thermal equilibrium with a third are in thermal equilibrium with each other: TP = TR and TQ = TR give TP = TQ, so nothing changes when P and Q touch. Masses do not enter the condition for equilibrium.

Question 2
Which one of the following quantities, for a given mass of gas, depends only on the state of the gas and not on how the gas reached that state?
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Answer: (a) its internal energy

Internal energy is a state variable. Heat and work are modes of energy transfer whose values depend on the path; the heat given out on the way back depends on the return path chosen.

Question 3
A sealed flask of air of total mass M is carried in an aircraft flying steadily at a speed v. The air in the flask is at the same temperature as when the flask stood on the ground. Compared with the flask on the ground, the internal energy of the air in the flask is
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Answer: (b) the same

Internal energy is the energy of the random molecular motion measured in the frame in which the centre of mass of the gas is at rest; the kinetic energy of the flask as a whole is not part of it. At the same temperature the internal energy is unchanged.

Question 4
A gas gives out 150 J of heat to its surroundings while its internal energy rises by 250 J. The work done on the gas by the piston is
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Answer: (c) 400 J

ΔQ = −150 J (heat removed) and ΔU = +250 J. The work done by the gas is ΔW = ΔQ − ΔU = −150 − 250 = −400 J, so 400 J of work is done on the gas. −400 J is the work done by the gas, not on it; 100 J and −100 J drop the sign of the heat.

Question 5
A gas is taken from state i to state f along path 1, absorbing 80 J of heat and doing 50 J of work. It is then brought back from f to i along a different path, on which 30 J of work is done on the gas. The heat absorbed by the gas on the return path is
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Answer: (a) −60 J

ΔU(i → f) = 80 − 50 = 30 J, so ΔU(f → i) = −30 J whatever the path. On the return, ΔW = −30 J, so ΔQ = ΔU + ΔW = −30 − 30 = −60 J: the gas gives out 60 J. Over the whole cycle ΔU = 0 and net heat 80 − 60 = 20 J equals net work 50 − 30 = 20 J, as it must.

Question 6
A gas in a cylinder expands against a constant pressure of 1.5 × 10⁵ Pa, its volume increasing from 2.0 L to 6.0 L. The work done by the gas is
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Answer: (a) 600 J

ΔW = PΔV = 1.5 × 10⁵ Pa × (6.0 − 2.0) × 10⁻³ m³ = 600 J. 600000 J forgets to convert litres to m³; 900 J uses the final volume instead of the change.

Question 7
A quantity of heat raises the temperature of a 0.50 kg block of aluminium by 20 K. The same quantity of heat is given to a 0.50 kg block of copper. Using the specific heats 900 J kg⁻¹ K⁻¹ for aluminium and 386.4 J kg⁻¹ K⁻¹ for copper, the rise in temperature of the copper block is about
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Answer: (b) 46.6 K

ΔQ = msΔT is the same for both blocks of equal mass, so ΔT ∝ 1/s: ΔT(Cu) = 20 × 900/386.4 ≈ 46.6 K. 8.6 K inverts the ratio; 20 K ignores the difference in specific heat.

Question 8
The law of equipartition predicts a molar specific heat of about 3R (≈ 24.9 J mol⁻¹ K⁻¹) for solids. Among the solids listed in NCERT's Table 11.1, the one whose measured molar specific heat at room temperature departs most from this prediction is
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Answer: (c) carbon

Table 11.1 gives 24.4 (aluminium), 24.5 (copper), 26.5 (lead) but only 6.1 J mol⁻¹ K⁻¹ for carbon, far below 3R. The other three lie within about 2 J mol⁻¹ K⁻¹ of 24.9.

Question 9
One calorie is now defined as the amount of heat required to raise the temperature of 1 g of water
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Answer: (b) from 14.5 °C to 15.5 °C

Because the specific heat of water varies slightly with temperature, the interval has to be fixed; the calorie is defined for the rise from 14.5 °C to 15.5 °C. The older definition, 'by 1 °C' at any temperature, was given up for this reason.

Question 10
According to NCERT's graph of the specific heat capacity of water against temperature between 0 °C and 100 °C, the specific heat is least
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Answer: (c) somewhere between 30 °C and 50 °C

The curve falls from the left, reaches its lowest value (below 1.000 cal g⁻¹ °C⁻¹) near 40 °C and rises again towards 100 °C. So the least value lies between 30 °C and 50 °C, and the specific heat is not constant.

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