Thermodynamics All subtopics

The wall has effectively infinite mass, so its velocity remains zero. The atom’s perpendicular momentum component reverses direction while its parallel component stays unchanged. The total momentum of the atom–wall system is constant because the wall provides the external surface but the impulse pair (atom on wall, wall on atom) are equal and opposite by Newton’s third law.

The total momentum before the collision is zero because the two momenta are equal in magnitude and opposite in direction. Conservation of momentum requires the total momentum after the collision to also be zero. Therefore the atoms rebound with equal speeds in opposite directions, preserving zero net momentum.

Reducing the volume means atoms travel shorter distances between wall collisions, so they strike the walls more frequently. Each collision still delivers the same average impulse because the temperature (and therefore average speed) is unchanged. More collisions per unit time per unit area means a greater total perpendicular force per unit area, which is a greater pressure.

Converting radius to SI: $$r=5.0\text{ cm}=0.050\text{ m}$$ Finding the area: $$A=\pi {r}^{2}=\pi \times (0.050{)}^{2}=7.854\times {10}^{-3}{\text{ m}}^{2}$$ Equation used — rearranged pressure definition: $${F}_{⟂}=PA$$ Substitution: $${F}_{⟂}=2.0\times {10}^{5}\times 7.854\times {10}^{-3}$$ $${F}_{⟂}=1.57\times {10}^{3}\text{ N}$$ The gas pushes on the window with roughly 1570 N of perpendicular force.

The pressure is equal. Pressure exists throughout the gas, not only at the boundary. At any imaginary surface inside the gas, atoms on both sides bombard that surface equally, producing the same force per unit area as at the wall. For a gas in equilibrium at uniform density and temperature, pressure is the same everywhere.

The rms speed increases by a factor of $\sqrt{2}$. Because ${v}_{\text{rms}}=\sqrt{\frac{3{k}_{B}T}{m}}$, the speed depends on the square root of $T$. Doubling $T$ multiplies the quantity under the square root by 2, so ${v}_{\text{rms}}$ increases by $\sqrt{2}$, not by 2.

Equation used: $${K}_{\text{avg}}=\frac{3}{2}{k}_{B}T$$ Given: $${k}_{B}=1.38\times {10}^{-23}\text{ J/K}$$ $$T=295\text{ K}$$ Working — substitution: $${K}_{\text{avg}}=\frac{3}{2}\times 1.38\times {10}^{-23}\times 295$$ $${K}_{\text{avg}}=2.07\times {10}^{-23}\times 295$$ $${K}_{\text{avg}}=6.11\times {10}^{-21}\text{ J}$$ The average translational kinetic energy is $6.11\times {10}^{-21}$ J.

Starting point — the rms speed equation from kinetic theory: $${v}_{\text{rms}}=\sqrt{\frac{3{k}_{B}T}{m}}$$ Writing the rms speed for each gas: $${v}_{\text{rms, A}}=\sqrt{\frac{3{k}_{B}T}{m}}$$ $${v}_{\text{rms, B}}=\sqrt{\frac{3{k}_{B}T}{4m}}$$ Forming the ratio: $$\frac{{v}_{\text{rms, A}}}{{v}_{\text{rms, B}}}=\frac{\sqrt{\frac{3{k}_{B}T}{m}}}{\sqrt{\frac{3{k}_{B}T}{4m}}}$$ Combining under one square root: $$\frac{{v}_{\text{rms, A}}}{{v}_{\text{rms, B}}}=\sqrt{\frac{3{k}_{B}T}{m}\times \frac{4m}{3{k}_{B}T}}$$ Cancelling common factors: $$\frac{{v}_{\text{rms, A}}}{{v}_{\text{rms, B}}}=\sqrt{4}=2$$ The lighter gas atoms move at twice the rms speed of the heavier gas atoms at the same temperature.

The peak shifts to a higher speed because the average kinetic energy increases. The peak height decreases and the curve becomes broader because a wider range of speeds is populated. The total area under the curve remains unchanged because the number of atoms is the same.

The peak for Gas X is to the left of the peak for Gas Y. At the same temperature, both gases have the same average kinetic energy (${K}_{\text{avg}}=\frac{3}{2}{k}_{B}T$). Because Gas X atoms are more massive, they must move more slowly to have that same kinetic energy. Therefore the most probable speed and rms speed for Gas X are both lower, shifting the distribution peak to the left.

Increasing temperature shifts the speed distribution to the right, meaning atoms move faster on average. Faster atoms strike the container walls with greater momentum, so each collision delivers a larger impulse. The force per collision increases, and the frequency of collisions also increases because atoms cross the container more quickly. The combined effect is a greater total perpendicular force per unit area on the walls, which is a greater pressure.

The atoms have random instantaneous velocities. The volumes of individual atoms are negligible compared to the container volume. Collisions between atoms are elastic. The only appreciable forces on atoms occur during collisions, so no long-range forces act between them.

The ideal gas model assumes no forces act between atoms except during brief collisions, so there is no inter-particle potential energy. All internal energy is the kinetic energy of the atoms. Because kinetic energy depends only on temperature, internal energy depends only on temperature.

A real gas at very high pressure is less likely to behave ideally. At high pressure the atoms are closer together, so the volume of the atoms is no longer negligible compared to the container volume. Inter-molecular forces between nearby atoms also become significant, violating two assumptions of the ideal gas model.

Equation used: $$PV=N{k}_{B}T ⇒ N=\frac{PV}{{k}_{B}T}$$ Converting temperature to kelvin: $$T=20+273=293 \text{K}$$ Given: $$P=1.01\times {10}^{5} \text{Pa}$$ $$V=45 {\text{m}}^{3}$$ $${k}_{B}=1.38\times {10}^{-23} \text{J/K}$$ Substituting: $$N=\frac{1.01\times {10}^{5}\times 45}{1.38\times {10}^{-23}\times 293}$$ $$N=\frac{4.545\times {10}^{6}}{4.0434\times {10}^{-21}}$$ $$N=1.12\times {10}^{27} \text{molecules}$$ This is the total number of gas molecules. Oxygen makes up about 21% of air, so the number of oxygen molecules is approximately $0.21\times 1.12\times {10}^{27}\approx 2.36\times {10}^{26}$.

At constant volume and constant amount of gas: $$\frac{{P}_{1}}{{T}_{1}}=\frac{{P}_{2}}{{T}_{2}} ⇒ {P}_{2}={P}_{1}\times \frac{{T}_{2}}{{T}_{1}}$$ Given: $${P}_{1}=2.0\times {10}^{5} \text{Pa}, {T}_{1}=400 \text{K}, {T}_{2}=200 \text{K}$$ Substituting: $${P}_{2}=2.0\times {10}^{5}\times \frac{200}{400}$$ $${P}_{2}=1.0\times {10}^{5} \text{Pa}$$ Halving the absolute temperature at constant volume halves the pressure, because pressure is directly proportional to temperature.

The pressure exactly doubles. From $P=\frac{nRT}{V}$, pressure is directly proportional to $n$ when $T$ and $V$ are constant. Doubling $n$ therefore doubles $P$.

Plot measured pressure values against temperature in °C at constant volume. The data form a straight line. Extrapolate the line to the temperature axis where pressure equals zero. The temperature at that intercept is absolute zero, approximately $-273 ^{\circ}\text{C}$.

The pressure triples. From $P=\frac{nRT}{V}$, pressure is directly proportional to absolute temperature at constant $n$ and $V$. Tripling $T$ therefore triples $P$.

The graph is a smooth curve (hyperbola) in the first quadrant that decreases as volume increases. It is not linear because $P=\frac{nRT}{V}$, so $P$ is inversely proportional to $V$. A linear graph would require $P∝V$, which contradicts the ideal gas equation at constant $T$.