Kinetic Theory of Gases flash cards

1. 1.State the basic assumption of kinetic theory about the size of gas molecules.

   A gas consists of a very large number of identical molecules whose actual volume is negligible compared to the total volume of the container. Molecules are treated as point particles.

   Hint: Molecular size vs container size.

2. 2.According to kinetic theory, how do molecules of an ideal gas interact?

   Molecules exert no force on each other except during collisions. There is no intermolecular potential energy, so the total internal energy is purely kinetic.

   Hint: No forces between collisions.

3. 3.What is assumed about collisions of gas molecules in kinetic theory?

   Collisions (molecule–molecule and molecule–wall) are perfectly elastic and of negligible duration compared with the time between collisions. Kinetic energy and momentum are conserved.

   Hint: Elastic, instantaneous.

4. 4.How do molecules move between collisions according to kinetic theory?

   Between collisions molecules move in straight lines with constant velocity (no external force, gravity neglected). Directions are random and uniformly distributed.

   Hint: Straight-line free flight.

5. 5.What does 'molecular chaos' / random motion assumption mean?

   At any instant molecular velocities are distributed randomly in all directions; there is no preferred direction, so the average velocity vector is zero even though the average speed is not.

   Hint: $\langle \vec v\rangle = 0$, $\langle v\rangle \neq 0$.

6. 6.Why is $\langle v_x^2\rangle = \langle v_y^2\rangle = \langle v_z^2\rangle$ for a gas?

   By isotropy (no preferred direction), each component contributes equally. Since $\langle v^2\rangle = \langle v_x^2\rangle + \langle v_y^2\rangle + \langle v_z^2\rangle$, each equals $\frac13\langle v^2\rangle$.

   Hint: Equal sharing among 3 axes.

7. 7.Write the kinetic theory expression for pressure of an ideal gas in terms of density.

   $P = \dfrac13 \rho\, \overline{v^2}$, where $\rho$ is the gas density and $\overline{v^2}$ is the mean square speed of the molecules.

   Hint: One-third rho v-squared.

8. 8.Write the pressure expression in terms of number of molecules $N$, mass $m$, and volume $V$.

   $P = \dfrac{1}{3}\dfrac{N m}{V}\,\overline{v^2} = \dfrac13\, n\, m\, \overline{v^2}$, where $n=N/V$ is number density.

   Hint: $\rho = Nm/V$.

9. 9.In deriving $P=\frac13\rho\overline{v^2}$, what is the change in momentum when one molecule hits a wall elastically?

   For a wall perpendicular to $x$, momentum change per collision is $\Delta p = 2 m v_x$ (the $x$-component reverses; $y,z$ unchanged).

   Hint: Reverses $v_x$.

10. 10.Why does the factor $\frac13$ appear in the pressure formula?

    Only the velocity component perpendicular to the wall causes pressure. Because $\overline{v_x^2}=\frac13\overline{v^2}$ by isotropy, the $\frac13$ enters the final expression.

    Hint: One of three components hits the wall.

11. 11.Relate pressure to the average translational kinetic energy per unit volume.

    $P = \dfrac23\left(\dfrac{N}{V}\right)\overline{E}$ where $\overline{E}=\frac12 m\overline{v^2}$. So $P=\frac23 \times$ (translational KE per unit volume).

    Hint: $P=\frac23 n\overline E$.

12. 12.State the ideal gas equation in terms of moles.

    $PV = nRT$, where $n$ is number of moles, $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$ is the universal gas constant, and $T$ is absolute temperature.

    Hint: Moles form.

13. 13.State the ideal gas equation in terms of number of molecules.

    $PV = N k_B T$, where $N$ is the number of molecules and $k_B$ is the Boltzmann constant.

    Hint: Molecule form uses $k_B$.

14. 14.How are $R$, $k_B$ and Avogadro number $N_A$ related?

    $k_B = \dfrac{R}{N_A}$, so $k_B = \dfrac{8.314}{6.022\times10^{23}} \approx 1.38\times10^{-23}\,\text{J K}^{-1}$.

    Hint: Gas constant per molecule.

15. 15.What is the physical meaning of the Boltzmann constant $k_B$?

    $k_B$ is the gas constant per molecule; it links the average thermal energy of a particle to temperature: energy per degree of freedom $=\frac12 k_B T$.

    Hint: Energy scale of temperature.

16. 16.By combining $P=\frac13\frac{Nm}{V}\overline{v^2}$ with $PV=Nk_BT$, derive the kinetic interpretation of temperature.

    $\frac13 m\overline{v^2} = k_B T \Rightarrow \frac12 m\overline{v^2} = \frac32 k_B T$. Average translational KE per molecule $=\frac32 k_BT$.

    Hint: Equate the two expressions for $PV$.

17. 17.What is the average translational kinetic energy of a gas molecule?

    $\overline{E}_{trans} = \dfrac32 k_B T$. It depends only on temperature, not on the mass or nature of the gas.

    Hint: Three-halves $k_BT$.

18. 18.What is the total translational kinetic energy of one mole of an ideal gas?

    $E = \dfrac32 N_A k_B T = \dfrac32 R T$ per mole (translational only).

    Hint: Multiply per-molecule value by $N_A$.

19. 19.State the kinetic interpretation of temperature in words.

    Absolute temperature is a direct measure of the average translational kinetic energy of the molecules of a gas. Higher $T$ means faster mean-square molecular motion.

    Hint: $T \propto \overline{E}_{trans}$.

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