Degrees of Freedom & Specific Heat — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Degrees of Freedom & Specific Heat MCQs with step-by-step solutions (31 questions). Part of Kinetic Theory of Gases & Radiation. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Degrees of Freedom & Specific Heat · easy · theory
The number of degrees of freedom of a monatomic gas molecule is:
A. $6$
B. $3$ ✓ Correct
C. $2$
D. $5$
Solution: A monatomic molecule is a point mass, so it can only translate along three independent directions. It has no rotational or vibrational degrees of freedom.
Q2 — Degrees of Freedom & Specific Heat · easy · theory
The number of degrees of freedom of a rigid diatomic molecule at ordinary temperatures is:
A. $6$
B. $5$ ✓ Correct
C. $7$
D. $3$
Solution: Three translational plus two rotational (about the two axes perpendicular to the bond) gives $f = 5$. Rotation about the bond axis contributes negligibly.
Q3 — Degrees of Freedom & Specific Heat · easy · theory
The ratio of specific heats $\gamma = \dfrac{C_p}{C_v}$ for a rigid diatomic gas is:
A. $\dfrac{4}{3}$
B. $\dfrac{9}{7}$
C. $\dfrac{5}{3}$
D. $\dfrac{7}{5}$ ✓ Correct
Solution: With $f = 5$, $\gamma = 1 + \dfrac{2}{f} = 1 + \dfrac{2}{5} = \dfrac{7}{5} = 1.40$.
Q4 — Degrees of Freedom & Specific Heat · easy · theory
The ratio of specific heats $\gamma$ for a monatomic ideal gas is:
A. $\dfrac{7}{5}$
B. $\dfrac{3}{2}$
C. $\dfrac{4}{3}$
D. $\dfrac{5}{3}$ ✓ Correct
Solution: With $f = 3$, $\gamma = 1 + \dfrac{2}{3} = \dfrac{5}{3} \approx 1.67$.
Q5 — Degrees of Freedom & Specific Heat · medium · theory
The general relation between the ratio of specific heats $\gamma$ and the number of degrees of freedom $f$ is:
A. $\gamma = 1 - \dfrac{2}{f}$
B. $\gamma = 1 + \dfrac{2}{f}$ ✓ Correct
C. $\gamma = \dfrac{2}{f}$
D. $\gamma = 1 + \dfrac{f}{2}$
Solution: Equipartition gives $C_v = \dfrac{f}{2}R$ and $C_p = C_v + R$, so $\gamma = \dfrac{C_p}{C_v} = 1 + \dfrac{2}{f}$.
Q6 — Degrees of Freedom & Specific Heat · easy · theory
Mayer's relation connecting the molar specific heats of an ideal gas is:
A. $C_p - C_v = R$ ✓ Correct
B. $C_p C_v = R$
C. $C_p + C_v = R$
D. $\dfrac{C_p}{C_v} = R$
Solution: At constant pressure the gas must additionally do external work $R\,\Delta T$ per mole, so $C_p$ exceeds $C_v$ by exactly $R$.
Q7 — Degrees of Freedom & Specific Heat · medium · numerical
The molar specific heat at constant volume of an ideal gas is $C_v = \dfrac{5}{2}R$. Its adiabatic index $\gamma$ is:
A. $1.67$
B. $1.33$
C. $1.28$
D. $1.40$ ✓ Correct
Solution: By Mayer's relation $C_p = C_v + R = \dfrac{7}{2}R$, so $\gamma = \dfrac{7/2}{5/2} = \dfrac{7}{5} = 1.40$.
Q8 — Degrees of Freedom & Specific Heat · easy · theory
The molar specific heat at constant volume of a monatomic ideal gas is:
A. $R$
B. $\dfrac{5}{2}R$
C. $\dfrac{7}{2}R$
D. $\dfrac{3}{2}R$ ✓ Correct
Solution: $C_v = \dfrac{f}{2}R$ with $f = 3$ gives $C_v = \dfrac{3}{2}R$, and correspondingly $C_p = \dfrac{5}{2}R$.
Q9 — Degrees of Freedom & Specific Heat · easy · theory
The law of equipartition of energy states that, in thermal equilibrium, the energy of a system is shared:
A. Equally among all its degrees of freedom, each receiving $\dfrac{1}{2}k_B T$ ✓ Correct
B. Equally among all the molecules regardless of temperature
C. In proportion to the mass of each molecule
D. Entirely among the translational degrees of freedom
Solution: Every independent quadratic term in the energy — translational, rotational or vibrational — carries an average of $\dfrac{1}{2}k_B T$ per molecule.
Q10 — Degrees of Freedom & Specific Heat · medium · theory
The internal energy of $n$ moles of a monatomic ideal gas at absolute temperature $T$ is:
A. $\dfrac{1}{2}nRT$
B. $\dfrac{3}{2}nRT$ ✓ Correct
C. $nRT$
D. $\dfrac{5}{2}nRT$
Solution: A monatomic gas stores energy only as translation, so $U = n C_v T = \dfrac{3}{2}nRT$.
Q11 — Degrees of Freedom & Specific Heat · medium · numerical
The internal energy of $2\text{ moles}$ of a monatomic ideal gas at $300\text{ K}$ is ($R = 8.314\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$):
A. $7483\text{ J}$ ✓ Correct
B. $2494\text{ J}$
C. $12472\text{ J}$
D. $4988\text{ J}$
Solution: $U = \dfrac{3}{2}nRT = \dfrac{3}{2}(2)(8.314)(300) \approx 7483\text{ J}$.
Q12 — Degrees of Freedom & Specific Heat · hard · theory
For a diatomic gas in which vibrational modes are also excited, the number of degrees of freedom becomes:
A. $5$
B. $7$ ✓ Correct
C. $3$
D. $6$
Solution: At high temperature a vibrating diatomic molecule gains two more degrees of freedom (one kinetic and one potential), giving $f = 5 + 2 = 7$ and $\gamma = \dfrac{9}{7}$.
Q13 — Degrees of Freedom & Specific Heat · medium · theory
A non-linear triatomic molecule has how many degrees of freedom at ordinary temperatures?
A. $7$
B. $3$
C. $6$ ✓ Correct
D. $5$
Solution: Three translational plus three rotational degrees of freedom give $f = 6$, so $\gamma = 1 + \dfrac{2}{6} = \dfrac{4}{3}$.
Q14 — Degrees of Freedom & Specific Heat · hard · numerical
During the isobaric heating of an ideal diatomic gas, the fraction of the heat supplied that is converted into external work is:
A. $\dfrac{2}{7}$ ✓ Correct
B. $\dfrac{3}{5}$
C. $\dfrac{2}{5}$
D. $\dfrac{5}{7}$
Solution: At constant pressure $W = nR\,\Delta T$ and $Q = nC_p\,\Delta T$, so $\dfrac{W}{Q} = \dfrac{R}{C_p} = \dfrac{R}{\frac{7}{2}R} = \dfrac{2}{7}$.
Q15 — Degrees of Freedom & Specific Heat · medium · numerical
During the isobaric heating of a monatomic ideal gas, the fraction of the heat supplied that goes into external work is:
A. $\dfrac{3}{5}$
B. $\dfrac{2}{7}$
C. $\dfrac{2}{5}$ ✓ Correct
D. $\dfrac{5}{7}$
Solution: $\dfrac{W}{Q} = \dfrac{R}{C_p} = \dfrac{R}{\frac{5}{2}R} = \dfrac{2}{5}$; the remaining $\dfrac{3}{5}$ raises the internal energy.
Q16 — Degrees of Freedom & Specific Heat · easy · theory
The molar specific heat at constant pressure is greater than that at constant volume because:
A. The molecules move faster at constant pressure
B. At constant pressure part of the heat supplied is spent doing external work ✓ Correct
C. The gas becomes denser at constant pressure
D. Heat is lost to the surroundings at constant pressure
Solution: At constant volume all the heat raises the internal energy. At constant pressure the gas also expands, so extra heat $R\,\Delta T$ per mole is needed for the work done.
Q17 — Degrees of Freedom & Specific Heat · easy · theory
The SI unit of molar specific heat capacity is:
A. $\text{J}\cdot\text{kg}^{-1}\cdot\text{K}^{-1}$
B. $\text{J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ ✓ Correct
C. $\text{J}\cdot\text{K}^{-1}$
D. $\text{J}\cdot\text{mol}^{-1}$
Solution: Molar specific heat is the heat required to raise one mole through one kelvin, giving $\text{J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.
Q18 — Degrees of Freedom & Specific Heat · medium · numerical
The molar specific heat at constant volume of a monatomic ideal gas is ($R = 8.314$):
A. $29.10\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
B. $20.79\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
C. $12.47\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ ✓ Correct
D. $8.314\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
Solution: $C_v = \dfrac{3}{2}R = 1.5 \times 8.314 = 12.47\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.
Q19 — Degrees of Freedom & Specific Heat · medium · numerical
The molar specific heat at constant pressure of a rigid diatomic gas is ($R = 8.314$):
A. $20.79\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
B. $29.10\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ ✓ Correct
C. $12.47\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
D. $24.94\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
Solution: $C_p = \dfrac{7}{2}R = 3.5 \times 8.314 = 29.10\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.
Q20 — Degrees of Freedom & Specific Heat · hard · numerical
The heat required to raise the temperature of $2\text{ moles}$ of a monatomic ideal gas by $100\text{ K}$ at constant volume is ($R = 8.314$):
A. $2494\text{ J}$ ✓ Correct
B. $1247\text{ J}$
C. $1663\text{ J}$
D. $4157\text{ J}$
Solution: $Q = nC_v\Delta T = 2 \times \dfrac{3}{2}(8.314) \times 100 = 2 \times 12.47 \times 100 \approx 2494\text{ J}$.
Q21 — Degrees of Freedom & Specific Heat · medium · numerical
A gas molecule has $6$ degrees of freedom. Its ratio of specific heats $\gamma$ is:
A. $1.33$ ✓ Correct
B. $1.29$
C. $1.67$
D. $1.40$
Solution: $\gamma = 1 + \dfrac{2}{f} = 1 + \dfrac{2}{6} = \dfrac{4}{3} \approx 1.33$.
Q22 — Degrees of Freedom & Specific Heat · medium · numerical
The molar specific heat at constant volume of a rigid diatomic gas is ($R = 8.314$):
A. $29.10\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
B. $12.47\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
C. $8.31\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
D. $20.79\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ ✓ Correct
Solution: $C_v = \dfrac{5}{2}R = 2.5 \times 8.314 = 20.79\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.
Q23 — Degrees of Freedom & Specific Heat · medium · numerical
The molar specific heat at constant pressure of a monatomic ideal gas is ($R = 8.314$):
A. $12.47\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
B. $16.63\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
C. $20.79\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ ✓ Correct
D. $29.10\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
Solution: $C_p = C_v + R = \dfrac{3}{2}R + R = \dfrac{5}{2}R = 20.79\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.
Q24 — Degrees of Freedom & Specific Heat · hard · numerical
The heat needed to raise $3\text{ moles}$ of a rigid diatomic gas through $50\text{ K}$ at constant volume is ($R = 8.314$):
A. $4365\text{ J}$
B. $6236\text{ J}$
C. $3118\text{ J}$ ✓ Correct
D. $1871\text{ J}$
Solution: $Q = nC_v\Delta T = 3 \times 20.79 \times 50 \approx 3118\text{ J}$.
Q25 — Degrees of Freedom & Specific Heat · hard · numerical
The heat needed to raise $2\text{ moles}$ of a monatomic ideal gas through $100\text{ K}$ at constant pressure is ($R = 8.314$):
A. $2494\text{ J}$
B. $1663\text{ J}$
C. $4157\text{ J}$ ✓ Correct
D. $5820\text{ J}$
Solution: $Q = nC_p\Delta T = 2 \times 20.79 \times 100 \approx 4157\text{ J}$.
Q26 — Degrees of Freedom & Specific Heat · medium · numerical
A gas molecule has $7$ degrees of freedom. Its ratio of specific heats is:
A. $1.67$
B. $1.33$
C. $1.29$ ✓ Correct
D. $1.40$
Solution: $\gamma = 1 + \dfrac{2}{f} = 1 + \dfrac{2}{7} = \dfrac{9}{7} \approx 1.29$.
Q27 — Degrees of Freedom & Specific Heat · easy · numerical
A gas molecule has $3$ degrees of freedom. Its ratio of specific heats is:
A. $1.33$
B. $1.40$
C. $1.29$
D. $1.67$ ✓ Correct
Solution: $\gamma = 1 + \dfrac{2}{3} = \dfrac{5}{3} \approx 1.67$ — the monatomic value.
Q28 — Degrees of Freedom & Specific Heat · hard · numerical
The internal energy of $4\text{ moles}$ of a rigid diatomic gas at $300\text{ K}$ is ($R = 8.314$):
A. $9.98 \times 10^3\text{ J}$
B. $3.49 \times 10^4\text{ J}$
C. $2.49 \times 10^4\text{ J}$ ✓ Correct
D. $1.50 \times 10^4\text{ J}$
Solution: $U = nC_vT = 4 \times 20.79 \times 300 \approx 2.494 \times 10^4\text{ J}$.
Q29 — Degrees of Freedom & Specific Heat · hard · numerical
During isobaric heating of a non-linear triatomic gas ($f = 6$), the fraction of heat converted into external work is:
A. $\dfrac{2}{5}$
B. $\dfrac{1}{3}$
C. $\dfrac{1}{4}$ ✓ Correct
D. $\dfrac{2}{7}$
Solution: With $f = 6$, $C_v = 3R$ and $C_p = 4R$. So $\dfrac{W}{Q} = \dfrac{R}{C_p} = \dfrac{1}{4}$.
Q30 — Degrees of Freedom & Specific Heat · easy · numerical
For any ideal gas, the difference $C_p - C_v$ has the value:
A. $12.47\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
B. $1.99\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
C. $8.314\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ ✓ Correct
D. $4.157\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
Solution: By Mayer's relation $C_p - C_v = R = 8.314\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$, whatever the atomicity of the gas.