Thermodynamics and Kinetic Theory — Practice

Question 1

How much heat is needed to raise the temperature of 2 kg of water by 10 °C? (specific heat of water = 4200 J kg⁻¹ K⁻¹)
(a) 84 000 J (b) 8400 J (c) 42 000 J (d) 21 000 J

Accepted Answer

(a) 84 000 J.

Q = mcΔT = 2 × 4200 × 10 = 84 000 J.

Question 2

The heat required to melt 0.1 kg of ice at 0 °C is (latent heat of fusion = 3.36 × 10⁵ J/kg):
(a) 33 600 J (b) 3360 J (c) 336 000 J (d) 336 J

Accepted Answer

(a) 33 600 J.

Q = mL = 0.1 × 3.36 × 10⁵ = 33 600 J, absorbed at a constant 0 °C with no temperature rise.

Question 3

100 J of heat is supplied to a gas, which does 40 J of work. The increase in its internal energy is:
(a) 60 J (b) 140 J (c) 40 J (d) 100 J

Accepted Answer

(a) 60 J.

ΔU = ΔQ − ΔW = 100 − 40 = 60 J.

Question 4

In an isothermal expansion of an ideal gas, the change in internal energy is:
(a) zero (b) positive (c) negative (d) equal to the heat supplied

Accepted Answer

(a) zero.

Internal energy of an ideal gas depends only on temperature; in an isothermal process temperature is constant, so ΔU = 0 (and all the heat becomes work).

Question 5

In an adiabatic process, which quantity is zero?
(a) Heat exchanged ΔQ (b) Work done (c) Change in internal energy (d) Change in temperature

Accepted Answer

(a) Heat exchanged ΔQ.

An adiabatic process exchanges no heat, so ΔQ = 0. Then ΔU = −ΔW, and the gas cools on expansion (PVγ = constant).

Question 6

At constant pressure, the absolute temperature of a fixed mass of ideal gas is doubled. Its volume:
(a) doubles (b) halves (c) is unchanged (d) quadruples

Accepted Answer

(a) doubles.

At constant P and n, V ∝ T (Charles's law). Doubling the absolute temperature doubles the volume.

Question 7

The average translational kinetic energy of a gas molecule depends on:
(a) only the absolute temperature (b) only the pressure (c) the type of gas (d) the volume

Accepted Answer

(a) only the absolute temperature.

Average KE = (3/2)kT, set by temperature alone — the same at a given T for any ideal gas, regardless of molecular mass.

Question 8

If the absolute temperature of a gas is increased to 4 times its value, the rms speed of its molecules becomes:
(a) 2 times (b) 4 times (c) 16 times (d) √2 times

Accepted Answer

(a) 2 times.

v_rms ∝ √T, so multiplying T by 4 multiplies the rms speed by √4 = 2.

Question 9

At the same temperature, the rms speed of hydrogen molecules (M = 2) compared with oxygen molecules (M = 32) is:
(a) 4 times greater (b) equal (c) 16 times greater (d) 4 times smaller

Accepted Answer

(a) 4 times greater.

v_rms ∝ 1/√M at a fixed temperature, so the ratio is √(32/2) = √16 = 4. Lighter molecules move faster.

Question 10

For one mole of an ideal gas, the relation between the molar specific heats is:
(a) Cp − Cv = R (b) Cv − Cp = R (c) Cp + Cv = R (d) Cp = Cv

Accepted Answer

(a) Cp − Cv = R.

At constant pressure, extra heat also does expansion work, so Cp exceeds Cv by exactly the gas constant R.

Question 11

The ratio of specific heats γ = Cp/Cv for a monatomic ideal gas is:
(a) 5/3 (b) 7/5 (c) 4/3 (d) 1

Accepted Answer

(a) 5/3.

A monatomic gas has 3 degrees of freedom, so Cv = (3/2)R and Cp = (5/2)R, giving γ = 5/3 ≈ 1.67.

Question 12

At ordinary temperatures, the number of degrees of freedom of a diatomic gas molecule is:
(a) 5 (b) 3 (c) 6 (d) 7

Accepted Answer

(a) 5.

A diatomic molecule has 3 translational plus 2 rotational degrees of freedom at ordinary temperatures (vibration is not excited), giving f = 5 and γ = 7/5.

Question 13

At constant temperature, if the pressure on a fixed mass of gas is doubled, its volume:
(a) halves (b) doubles (c) is unchanged (d) quadruples

Accepted Answer

(a) halves.

Boyle's law: PV = constant at constant T. Doubling P halves V.

Question 14

A Carnot engine works between 400 K and 300 K. Its maximum efficiency is:
(a) 25% (b) 75% (c) 33% (d) 100%

Accepted Answer

(a) 25%.

η = 1 − T₂/T₁ = 1 − 300/400 = 1 − 0.75 = 0.25 = 25%. Both temperatures must be in kelvin.

Question 15

A Carnot refrigerator operates between 300 K (hot) and 250 K (cold). Its coefficient of performance is:
(a) 5 (b) 6 (c) 0.83 (d) 1.2

Accepted Answer

(a) 5.

COP = T₂/(T₁ − T₂) = 250/(300 − 250) = 250/50 = 5. A refrigerator's COP can exceed 1, unlike an engine's efficiency.

Practice — Thermodynamics and Kinetic Theory

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