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## Monday, March 17, 2008

### AP Physics B – Answers to Practice Questions (MCQ) on Kinetic Theory

Some typical multiple choice questions for practice were given to you in the post dated 15th March 2008. As promised, I give below the answers with explanation.

(1) If a diatomic gas molecule has an additional vibrational mode which contributes to both kinetic and potential energies, the sum of which is equal to kT where k is Boltzmann’s constant and T is the absolute temperature, what is the ratio of specific heats of the gas?

(a) 3/2

(b) 4/3

(c) 5/3

(d) 7/5

(e) 9/7

A diatomic molecule without vibrational mode has average energy equal to (5/2)kT since it has 5 degrees of freedom. Since the vibrational mode contributes an extra energy of kT, the total average energy per molecule is (5/2)kT+kT = (7/2)kT.

The molar specific heat at constant volume (Cv) is therefore equal to (7/2)kT×N where N is Avogadro’s number.

Therefore, Cv = (7/2)R and the molar specific heat at constant pressure (Cp) is [(7/2)R+R] = (9/2)R where R is universal gas constant.

Therefore, the ratio of specific heats, γ = Cp/Cv = 9/7

(2) Four moles of helium (a mono atomic gas) is contained in a barrel provided with a light, frictionless piston (fig.). The quantity of heat to be supplied to the gas for raising its temperature by 5 K is (universal gas constant, R = 8.31 J mol–1 K–1)

(a) 415.5 J

(b) 249.3 J

(c) 136.5 J

(d) 166.2 J

(e) 581.7 J

When heat is supplied to the gas, the internal energy of the gas is increased, raising the temperature of the gas. The pressure of the gas is unchanged since the piston moves outwards, doing work against the atmospheric pressure. It is the molar specific heat at constant pressure (Cp) that is involved here so that the quantity of heat (Q) to be supplied is given by

Q = no. of moles × molar specific heat at constant pressure × rise in temperature.

Helium being mono atomic, molar specific heat at constant pressure (Cp) is (5/2)R

Therefore, Q = 4×(5/2)R×5 = 4×(5/2)×8.31×5 = 415.5 J

(3) The intermolecular attraction is an important reason why real gases behave differently compared to ideal gas. But real gases too can behave like ideal gas at

(a) low temperature and low pressure

(b) low temperature and high pressure

(c) high temperature and high pressure

(d) high temperature and low pressure

(e) absolute zero (0 K) temperature

At high temperatures and low pressures the molecules are far apart and molecular interactions are negligible. Under these conditions even a real gas behaves like an ideal gas [Option (d)].

(4) A fixed mass of an ideal gas at pressure P is contained in a closed vessel of volume V. It is heated so that the root mean square velocity of the gas molecules is doubled. The thermal expansion of the vessel is negligible. Then the increase in pressure (P) of the gas is

(a) P

(b) 2P

(c) 3P

(d) 4P

(e) √2 P

The root mean square velocity of the gas molecules is given by c = √(3kT/m) where k is Boltzmann’s constant, T is the absolute temperature and m is the mass of the molecule. The r.m.s. velocity is therefore directly proportional to the square root of absolute temperature. Since the r.m.s. velocity is doubled on heating the gas, the temperature must be quadrupled.

By Charle’s law, we have PV/T = constant for a given mass of gas. When the temperature is quadrupled at constant volume, the pressure must be quadrupled. The final pressure is thus 4P and the increase in pressure is 3P.

(5) A cubical vessel of side contains N molecules of an ideal gas. If the mass of a molecule is m and the root mean square velocity is c, the pressure of the gas is

(a) mNc2 /33

(b) mNc2 /3

(c) mc2 /33

(d) 3mNc2 /3

(e) (1/3)mNc2

The pressure exerted by a gas is given by P = (1/3) ρc2 where ρ is the density of the gas and c is the root mean square velocity of the gas molecules. The density of the gas here is

ρ = Total mass/Total volume = mN/3

Therefore, pressure P = mNc2 /33

(6) Four moles of oxygen (diatomic) is mixed with eight moles of neon (mono atomic). If they are treated as ideal gases, the effective molar specific heat at constant volume of the mixture is (R = universal gas constant)

(a) (5/3) R

(b) (7/5) R

(c) (9/5) R

(d) (11/6) R

(e) (13/8) R

The molar specific heat at constant volume (Cv) of a diatomic gas is (5/2)R and that of a mono atomic gas is (3/2)R where R is universal gas constant.

Molar specific heat is the heat required to raise the temperature of one mole of the gas by 1 K. The quantity of heat required to raise the temperature of four moles of oxygen and eight moles of neon by 1 K is [4×(5/2)R + 8×(3/2)R] = 22 R and the total number of moles in the mixture is 12.

Therefore, molar specific heat of the mixture = 22 R/12 = (11/6) R

(7) The temperature at which the root mean square velocity of hydrogen gas molecules is twice that at 0º C is

(a) 20º C

(b) 40º C

(c) 819º C

(d) 859º C

(e) 1132º C

Since the root mean square velocity is directly proportional to the square root of absolute temperature, it will be doubled at 4 times the absolute temperature corresponding to 10º C. Therefore, the answer is 4×283 K = 1132 K = (1132 – 273) K = 859º C.

(8) Equal number of oxygen molecules (molar mass m1) and helium molecules (molar mass m2) are kept in two identical vessels. If they are at the same temperature, their pressures will be in the ratio

(a) m1/m2

(b) m2/m1

(c) √(m1/m2)

(d) √(m2/m1)

(e) 1

Most of you know that equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules. If equal numbers of molecules of different gases have the same temperature and volume their pressure must therefore be the same. The correct option is (e).

The expression for pressure in the form, P = nkT also is handy in working out similar problems. Since k is Boltzmann’s constant, the pressure depends only on the temperature and the number density (number per unit volume). You can easily obtain this expression by substituting for the r.m.s. velocity [c = √(3kT/m)] in the expression for pressure, P = (1/3) nmc2.