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UPCET / UPSEE Thermodynamics & Kinetic Theory — practice questions

Explore 47 free Thermodynamics & Kinetic Theory questions in UPCET / UPSEE Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Thermodynamics & Kinetic Theory questions and exercises with answers

Showing questions 1–30 of 47.

The root-mean-square (rms) speed of gas molecules is given by:Read the answer and worked solution →The mean translational kinetic energy of one molecule of an ideal gas at temperature $T$ is:Read the answer and worked solution →If the absolute temperature of an ideal gas is doubled, the rms speed of its molecules:Read the answer and worked solution →The mean free path $\lambda$ of a gas molecule of diameter $d$ at number density $n$ is approximately:Read the answer and worked solution →For an ideal gas, the relationship between $C_p$ and $C_v$ is (Mayer's relation):Read the answer and worked solution →The ideal gas equation is:Read the answer and worked solution →The relation between universal gas constant $R$, Avogadro number $N_A$, and Boltzmann constant $k_B$ is:Read the answer and worked solution →The **mean free path** of a gas molecule is:Read the answer and worked solution →The pressure of an ideal gas in kinetic theory is given by:Read the answer and worked solution →The **rms speed** of gas molecules at absolute temperature $T$ is:Read the answer and worked solution →Average translational kinetic energy of a gas molecule at temperature $T$ is:Read the answer and worked solution →The **law of equipartition of energy** states that, at temperature $T$, each quadratic degree of freedom contributes:Read the answer and worked solution →Number of degrees of freedom (DOF) of a monatomic gas (e.g. helium) molecule:Read the answer and worked solution →Mayer's relation for an ideal gas is:Read the answer and worked solution →**Adiabatic ratio** $\gamma$ of a gas is defined as:Read the answer and worked solution →Number of degrees of freedom of a **rigid diatomic molecule** (e.g. O₂ at room temperature, vibrations not excited):Read the answer and worked solution →If vibrations are excited, a diatomic molecule has DOF:Read the answer and worked solution →For a monatomic ideal gas, $C_V$ and $C_P$ per mole are:Read the answer and worked solution →For a **rigid diatomic** ideal gas, $\gamma = C_P/C_V$ equals:Read the answer and worked solution →The rms speed of N₂ at 300 K (M = 28 g/mol) is approximately (use $R = 8.314$ J/(mol·K)):Read the answer and worked solution →If the temperature of a gas is doubled (at constant volume), the rms speed changes by a factor of:Read the answer and worked solution →Average translational KE of a He atom at 300 K is approximately ($k_B = 1.38 \times 10^{-23}$ J/K):Read the answer and worked solution →Mean free path of nitrogen at STP (d = 324 pm) is approximately:Read the answer and worked solution →If P is doubled at constant T (compress gas isothermally), the mean free path:Read the answer and worked solution →An ideal gas is held in a container at temperature T. The total internal energy of one mole of a monatomic gas is:Read the answer and worked solution →For a diatomic gas with all vibrational modes excited, the molar specific heat at constant volume is:Read the answer and worked solution →The ratio of average KE of a monatomic gas to that of a rigid diatomic gas, at the **same T**, is:Read the answer and worked solution →Velocity of sound in an ideal gas is $v_s = \sqrt{\gamma RT/M_0}$. Compared to rms speed, $v_s$:Read the answer and worked solution →The rms speed of helium atom at 300 K is approximately (M_He = 4 g/mol):Read the answer and worked solution →The internal energy of 2 moles of nitrogen gas (rigid diatomic) at 300 K is approximately ($R = 8.314$):Read the answer and worked solution →

Worked practice questions

The root-mean-square (rms) speed of gas molecules is given by:

  1. $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$
  2. $v_{\text{rms}} = \sqrt{\dfrac{RT}{M}}$
  3. $v_{\text{rms}} = \sqrt{\dfrac{2RT}{M}}$
  4. $v_{\text{rms}} = \sqrt{\dfrac{8RT}{\pi M}}$
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Correct answer: A. $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$

From kinetic theory, the average translational kinetic energy of a molecule is $\dfrac{1}{2} m \overline{v^2} = \dfrac{3}{2} kT$ giving $\overline{v^2} = \dfrac{3kT}{m} = \dfrac{3RT}{M}$, where $M$ is molar mass. The **rms** speed is the square root: $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$. Other useful speeds: most probable speed $v_p = \sqrt{\dfrac{2RT}{M}}$ (option C) average speed $\langle v\rangle = \sqrt{\dfrac{8RT}{\pi M}}$ (option D). Order: $v_p < \langle v\rangle < v_{\text{rms}}$.

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The mean translational kinetic energy of one molecule of an ideal gas at temperature $T$ is:

  1. $\dfrac{1}{2} kT$
  2. $kT$
  3. $\dfrac{3}{2} kT$
  4. $3 kT$
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Correct answer: C. $\dfrac{3}{2} kT$

Equipartition theorem: each translational degree of freedom contributes $\dfrac{1}{2} kT$ of energy. There are 3 translational degrees ($x, y, z$) so total mean translational KE per molecule is $\dfrac{3}{2} kT$. For a mole: $\dfrac{3}{2} RT$. If the molecule has internal degrees (rotation, vibration), more energy is stored — but those don't add to **translational** KE specifically.

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If the absolute temperature of an ideal gas is doubled, the rms speed of its molecules:

  1. Doubles
  2. Increases by a factor of $\sqrt{2}$
  3. Quadruples
  4. Halves
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Correct answer: B. Increases by a factor of $\sqrt{2}$

$v_{\text{rms}} \propto \sqrt{T}$ (in Kelvin). Doubling $T$ scales the rms speed by $\sqrt{2}$. Intuition: kinetic energy is proportional to $T$, and KE $\propto v^2$, so $v \propto \sqrt{T}$. Option A (doubles) would be true if $v \propto T$, but it's $\sqrt{T}$.

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