Questions: Kinetic Theory

Q1. Derive the expression for the pressure exerted by an ideal gas on the walls of its container.

Consider a cubical container of side L holding N molecules, each of mass m. Take one molecule moving with velocity components vx, vy and vz. It bounces off the wall perpendicular to the x-axis, so its x-velocity reverses and the momentum change is 2m vx. The time between two collisions with the same wall is 2L/vx, so the average force from this molecule is 2m vx divided by 2L/vx, which is m vx^2/L. Adding the contributions of all molecules and dividing by the wall area L^2 gives P = (m/L^3)(sum vx^2). Since the motion is random, each axis carries one third of the mean square speed, so P = (1/3)(mN/V)<v^2>, where V = L^3. Hinglish: Ek molecule wall se takraata hai to momentum change 2m vx hota hai. Time 2L/vx me phir takraata hai, isliye average force m vx^2/L banti hai. Sab molecules ka contribution jodo aur area se divide karo, to P = (1/3)(mN/V) v_rms^2 milta hai.

Q2. Explain the kinetic interpretation of temperature and its consequences.

Comparing the kinetic expression for pressure with the ideal gas equation gives (1/2)m<v^2> = (3/2)k_BT. The mean translational kinetic energy of a molecule therefore depends only on the absolute temperature and not on the identity of the gas. Two consequences follow. At a given temperature, all gases have the same average translational kinetic energy per molecule, so a lighter gas has a larger rms speed. The rms speed is proportional to sqrt(T/M), which is why hydrogen has a much larger rms speed than oxygen at the same temperature.

Hinglish: Temperature average translational KE ka measure hai: (3/2)kT per molecule. Isliye same temperature par sab gases ki average KE same hoti hai, lekin halki gas ki rms speed zyada hoti hai. Absolute zero par yeh energy zero hoti hai.

Q3. State the law of equipartition of energy and use it to find the specific heats of a monatomic gas.

The law says that in thermal equilibrium each quadratic degree of freedom of a molecule carries the same average energy, (1/2)k_BT. A monatomic gas has only translational motion, with three degrees of freedom, so the internal energy of one mole is (3/2)RT. At constant volume all the heat goes into internal energy, so C_V = (3/2)R. At constant pressure the gas also does work, so C_P = C_V + R = (5/2)R. The ratio is gamma = C_P/C_V = 5/3, about 1.67. These values match the measured specific heats of helium and other noble gases at room temperature.

Hinglish: Equipartition kehta hai har quadratic degree ko (1/2)kT milti hai. Monatomic gas ke teen translational degrees hain, isliye U = (3/2)nRT, Cv = 3R/2, Cp = 5R/2 aur gamma = 5/3. Helium ka measured value isi se match karta hai.

Q4. Find the rms speed and the average translational kinetic energy of oxygen molecules at 300 K. Take M = 32 g mol^-1.

The rms speed is sqrt(3RT/M) = sqrt(3 x 8.31 x 300/0.032) = sqrt(2.34 x 10^5) = 483 m s^-1. The average translational kinetic energy per molecule is (3/2)k_BT = 1.5 x 1.38 x 10^-23 x 300 = 6.2 x 10^-21 J. Per mole the same energy is (3/2)RT = 3740 J. Hinglish: rms speed sqrt(3RT/M) se 483 m/s aati hai. Average translational KE per molecule (3/2)kT = 6.2 x 10^-21 J. Same temperature par nitrogen ki KE same hogi lekin speed zyada hogi kyunki nitrogen halka hai.

Q5. Why do the specific heats of a diatomic gas differ from those of a monatomic gas?

A diatomic molecule can rotate about two axes perpendicular to its bond in addition to translating, so at ordinary temperatures it has five quadratic degrees of freedom. The extra rotational degrees of freedom carry energy, so the internal energy rises faster with temperature and the molar specific heats are larger: C_V = (5/2)R and C_P = (7/2)R. The ratio gamma is 7/5, about 1.40. A monatomic gas has three translational degrees only, so C_V = (3/2)R and gamma = 5/3. Hinglish: Diatomic molecule translate ke saath do axes par rotate bhi karta hai, isliye f = 5. Zyada degrees energy store karte hain, isliye Cv aur Cp bade hote hain aur gamma 7/5 hota hai. High temperature par vibration bhi add hota hai.

Q6. What is the mean free path? How does it change with pressure and temperature?

The mean free path is the average distance a molecule travels between two successive collisions. For molecules of diameter d and number density n, it is lambda = 1/(sqrt(2) pi d^2 n). Because n = P/k_BT at fixed temperature, the mean free path falls as the pressure rises. At fixed pressure, raising the temperature reduces the number density, so the mean free path increases. In air at room temperature and pressure, lambda is about 10^-7 m, which is about a thousand molecular diameters; this is why the ideal-gas model, which ignores collisions except with walls, works well at ordinary conditions.

Hinglish: Mean free path collisions ke beech ki average distance hai: lambda = 1/(sqrt(2) pi d^2 n). Pressure badhne par n badhta hai to lambda ghatta hai; temperature badhne par lambda badhta hai. Air me room conditions par lambda 10^-7 m hota hai.

Q7. Explain how the ideal gas equation leads to Boyle's law and Charles's law.

The ideal gas equation is PV = Nk_BT. If the temperature and the amount of gas are fixed, the right side is constant, so PV is constant: doubling the pressure halves the volume. This is Boyle's law. If the pressure and the amount of gas are fixed, then V/T = Nk_B/P is constant, so the volume is proportional to the absolute temperature. This is Charles's law, and it requires absolute temperature, because a Celsius temperature would make the ratio meaningless. Avogadro's law follows by fixing T and P: equal volumes then hold equal numbers of molecules.

Hinglish: PV = NkT me agar T fixed hai to PV constant, yeh Boyle's law hai. Agar P fixed hai to V/T constant, yeh Charles's law hai. Volume fixed ho to P/T constant hota hai. In sab me T absolute temperature hona chahiye.

Q8. A gas has C_V = 5R/2. Find gamma and the change in internal energy when 2 moles are heated through 20 K at constant volume.

The specific heat at constant pressure is C_P = C_V + R = 7R/2, so gamma = C_P/C_V = 7/5 = 1.4, which is the value for a rigid diatomic gas. The change in internal energy at constant volume is dU = nC_V delta T = 2 x (5R/2) x 20 = 100R = 831 J. No work is done at constant volume, so the heat supplied is also 831 J. Hinglish: Cv = 5R/2 matlab diatomic gas, Cp = 7R/2 aur gamma = 1.4. Constant volume par work zero hoti hai, isliye dU = nCv delta T = 831 J aur Q bhi 831 J. Constant pressure par Q = 1163 J hota hai aur extra energy work me jaati hai.

Q9. Two gases at the same temperature have different rms speeds. Explain why.

At the same temperature the average translational kinetic energy per molecule is the same for both gases, because it is (3/2)k_BT and does not involve the mass. The rms speed follows from (1/2)m<v^2> = (3/2)k_BT, so v_rms = sqrt(3k_BT/m). A lighter molecule therefore has a larger rms speed. For example, at 300 K the rms speed of hydrogen is about four times that of oxygen because the mass of an oxygen molecule is about sixteen times larger and the speed scales as the square root of the inverse mass. This is why the composition of the upper atmosphere is different from the lower atmosphere.

Hinglish: Same temperature par average KE same hoti hai, lekin rms speed mass par depend karti hai: v_rms = sqrt(3kT/m). Halki molecule ki speed zyada hoti hai, isliye hydrogen ki speed oxygen se lagbhag chaar guna hoti hai.

Q10. Why does a real gas deviate from ideal behaviour at high pressure and low temperature?

The ideal model treats molecules as points with no forces between them. At high pressure the molecules are packed closer, so the finite volume they occupy becomes a noticeable part of the container volume, and the free space available to them is smaller than V. At low temperature the attractive forces between molecules matter, because the thermal energy is no longer large enough to overcome them, so the molecules pull together and the pressure is lower than the ideal prediction. The measured isotherms of steam and carbon dioxide show these effects as flattened curves near condensation. The ideal equation recovers its accuracy when the gas is rarefied and hot.

Hinglish: High pressure par molecules ka apna volume matter karta hai, aur low temperature par attractive forces important ho jaati hain, isliye real gas ideal se hat jaati hai. Steam aur CO2 ke isotherms mein yeh deviation dikhta hai. Low pressure aur high temperature par ideal equation theek kaam karti hai.

Quiz below: five questions on pressure from molecular motion, kinetic temperature, equipartition, specific heats and mean free path.