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Here, torsion constant, C = 1.8 x 10^-17 J

<θ²> = 2.6 x 10^-4 rad², and T = 300 K

Using the relation: 1/2 k_BT = 1/2 C <θ²>

⇒ k_B = {c<θ²>}/{T}

= {1.8 x 10^-7 x 2.6 x 10^-4}/{300}

= 1.56 x 10^-23 JK^-1

This value is different from the actual value of Boltzmann's constant, i.e., 1.38 x 10^-23 J K^-1.

(a) For Brownian motion, the typical size of the suspended particles should be about 10^-6 m to 10^-5 m. If the size of the particle is too small or too large, no Brownian motion takes place. When particle is too small (=10^-10 m) then chances of bombardment of the particle by the molecules of the liquid becomes negligible. On the other hand, when particle is too large (= 1 m) the suspended particle is bombarded equally from all sides and in absence of a net unbalanced force, Brownian motion does not occur.

(b) Brownian motion is observed due to fluctuations in the number of molecules actually striking a suspended particle from the average number of molecules striking the particle.

(c) Yes, the assembly of suspended particles can be treated as a gas. Since, the suspended particles are in thermal equilibrium with the liquid at temperature T, the temperature of the suspended particles will also be T.

Since, Avogadro's number is finite, each suspended particle experiences a net unbalanced force resulting in Brownian motion. In case, Avogadro number was infinite, each suspended particle would be bombarded by such a large number of atoms from all the sides that, there will be no net unbalanced force and thus, no Brownian motion will be observed.

Boyle’s law states that under constant temperature, the pressure of given mass of a gas is inversely proportional to its volume.

Loss in K.E of the gas = ΔE = (1/2)(mn)v_o² 

where n = no: of moles.

If its temperature changes by ΔT, then

(n)(3/2)RΔT = (1/2)mn v_o² . 

ΔT = (mv_o²/3R)

\(\frac{P_1}{T_1}=\frac{P_2}{T_2}\)

or  \(T_2=\frac{P_2T_1}{P_1}\)

= \(\frac{2\times(273-173)}{1}\)

= 200K

= −73°C

No, temperature can not be changed without changing V or P.

According to Dalton’s law of partial pressure,

P = P₁ + P₂ 

= (2.5 + 1.5)atm

= 4.0 atmosphere.

According to Dalton's law of partial pressure,

p = p₁ + p₂ = 2.5 + 1.5 = 4 atmosphere

∴ Number of moles in 16 g of O₂

_{\(= \frac{1}{32}\times16=\frac{1}{2}\)} 

As, PV = nRT and 

n = \(\frac{1}{2}\),

so, PV = \(\frac{1}{2}\)RT.

Boltzmann constant, 

k_B = R/N, 

Where R is universal gas constant and N is the Avogadro’s number.

k_{B = \(\frac{8.31\,J\,mol^{-1}K^{-1}}{6.023\times10^{23}\,mole^{-1}}\)} 

= 1.38 × 10^-23 JK^-1

R = 8.31 J mol^-1K^-1

or, R = 1.98 cal mol-1K^-1

Numerical value R : 

Consider one mole of a gas at STP, then 

R = \(\frac{P_0V_0}{T}\)

Standard pressure

P₀ = 0.76 m of Hg column

= 0.76 × 13.6 × 10³ × 9.8 N/m2

Standard temperature = T₀ = 273.15K

Volume of one mole of gas at

= 22.4 × 10^-3m³

R = \(\frac{0.76\times13.6\times10^3\times9.8\times22.4\times10^{-3}}{273.15}\)

= 8.31 J mol^-1K^-1

In the C.G.S. system,

R = \(\frac{8.31}{4.2}\) cal mol^-1°C^-1

= 1.98 cal mol^-1°C^-1

Numerical value of k_B:

We know that,

k_B = \(\frac{R}{NA}\)

k_B = \(\frac{8.31\,J\,mole^{-1}K^{-1}}{6.02\times10^{23}mole^{-1}}\)

=1.38 × 10^-23 J/K.

Suppose a polyatomic gas molecule has n degree of freedom.

Total energy associated with one gram molecule of the gas, i.e.,

E = n × \(\frac{1}{2}\)RT × 1

=\(\frac{n}{2}\)RT

As,

C_v = \(\frac{d}{dT}\)€

= \(\frac{d}{dT}\)(\(\frac{n}{2}\)RT)

= \(\frac{n}{2}\)R

C_p = C_v + R

C_p = \(\frac{n}{2}\)R + R

= (\(\frac{n}{2}\)+1)R

γ= \(\frac{C_p}{C_V}\)

γ= \(\frac{(\frac{n}{2}+1)R}{\frac{n}{2}R}\)

∴ = \(\frac{2}{n}\)(\(\frac{n}{2}\)+1)

γ= 1+\(\frac{2}{n}\)

\(R=\frac{PV}{T}\)

= \(\frac{[ML^{-1}T^{-2}][L^3]}{[K]}\)

= [ML²T^-2K^-1]

Gas C is ideal, because pV is constant for this gas. It means the gas C obeys Boyle's law at all pressures.

The correct value of 0°C on the Kelvin scale is 273.15 K.

It is a straight line passing through the origin.

The number of collisions per unit time can be increased by:

(i) Increasing the temperature of the gas.

(ii) Increasing the number of molecules, and

(iii) Decreasing the volume of the gas.

At zero kelvin temperature all molecular motion cease.

The phenomenon of diffusion of gases and Brownian motion provide direct experimental evidence in support of molecular motion.

lα(1/d²)

d₁ = 1Å  α₂ = 2Å

l₁ : l₂ = 4 : 1 

O₂ has 5 degrees of freedom. 

Therefore, energy per mole = (5/2)RT

∴ For 2 moles of O₂, energy = 5RT

Neon has 3 degrees of freedom 

∴ Energy per mole = (3/2)RT

∴ For 4 mole of neon, energy = 4 x (3/2) RT = 6RT

∴ Total energy = 11RT.

Since ∝ = 1/3 mn/v c², i.e., p∝ n, therefore, pressure will be doubled when number of molecules n becomes 2n. As kinetic energy of the gas = 1/2 x mn c², therefore K.E. of the gas also becomes double when n changes to 2n. Since, r.m.s. velocity of molecules of a gas only depends upon the temperature, hence r.m.s. velocity of the molecules will remain the same when n changes to 2n.

When p becomes double, then temperature becomes double. As C ∝ √T, therefore r.m.s velocity becomes √2 C.

Since,

P = \(\frac{1}{3}\frac{mnc^2}{V}\),

i.e., P ∝ n, 

Therefore, pressure will be double when number of molecules n becomes 2n. As kinetic energy of the gas = \(\frac{1}{2}mnc^2\) , 

Therefore, K.E. of the gas also becomes double when n changes to 2n. Since, r.m.s. velocity of molecules of a gas only depends upon the temperature, hence, r.m.s. velocity of the molecules will remain the same when n changes to 2n.

An ideal gas is that which obeys the gas laws i.e., Charles’s law, Boyle’s law etc. at all values off temperature and pressure. Molecules of such a gas should be free intermolecular attraction and have negligible volume.

Apply Boyle’s law

V ∝ \(\frac{1}{P}\)   …(i)

Apply Charles’s law,

V ∝ T   …(ii)

Combining eq (i) & (ii)

V ∝\(\frac{T}{P}\)

or, V = \(\frac{KT}{P}\)

or, \(\frac{PV}{T}\)=K

\(\frac{PV}{T}\)= constant

= R (i.e., universal gas constant)

PV = RT

for n moles; 

PV = nRT

This equation give the relation between pressure P, volume V and absolute temperature T of a gas.

PV = nRT

Derivation.

According to Boyle’s law,

V ∝ \(\frac{1}{P}\) …(1)

According to Charle’s law,

V ∝ T …(2)

Comparing (1) and (2), we get

\(\frac{PV}{T}\)= constant

As PV = RT

For n moles of gas PV = nRT

This is perfect or ideal gas equation.

With the expansion, the molecules get more space to move, which decrease the number of collisions per second. Less number of collisions on the walls and larger area of the walls available for collisions result in decrease in pressure.

IN an ideal gas, the molecules can be considered as point masses with no intermolecular forces between them. So there can neither be internal potential energy nor the internal energy due to rotation or vibration. The molecules can have only translations K.E.

All molecular motion ceases at absolute zero or at 0 K.

Thus at 0 K, the velocity of molecules becomes zero.

S.I. unit of heat is Joule and C.G.S. unit of heat is calorie. 1 cal = 4.18 J

This is because in a harbor town, the relative humidity is more than in a desert town. Hence the climate of a harbor town is without extremes of hot and cold.

Sp. heat of water is 4180 J kg^-1 K^-1. Yes, it does vary-very little with temperature.

It states that volume (V) remaining constant, pressure (P) of a given mass of a gas is directly proportional to its absolute temperature (θ), i.e.,

P ∝ θ

P = P₀(1 + y_pθ)

Where y_p is the coefficient of expansion of a gas at constant pressure.

According to kinetic theory of gases, the pressure p exerted by one mole of an ideal gas is given by:

p = 1/3 M/v C²

Or, pV = 1/3 MC²     ...(i)

(pV = RT)

or, C² = 3RT/M or C² ∝ T

(R an M are constants)

or, C² ∝ √T or √T ∝ C

Thus, the square root of the absolute temperature of an ideal gas is directly proportional to root mean square velocity of its molecules:

From (i), we have wnen T = 0, C = 0

Hence, absolute zero of the temperature may be defined as that temperature at which the root mean square velocity of the gas molecules reduces to zero.

It means molecular motion ceases at absolute zero.

It is important to note that this definition holds in case of an ideal or a perfect gas. The gases used in actual practice are far from being perfect, particularly at very low temperatures.

(d) between p and 1.1.

Comment for discussion: In this chapter, one has discussed constant pressure and constant volume situations but in real life there are many situations where both change. If the surfaces were rigid, p would rise to 1.1 p. However, as the pressure rises, V also rises such that pv finally is 1.1 RT with p_final > p and V_final > V. Hence (d).

(b)  f₁ (v), f₂ (v) will obey the Maxwell’s distribution law separately.

Comment: In a mixture, the average kinetic energy are equating. Hence, distribution in velocity are quite different.

On reducing the volume the space for the given number of molecules of the gas decreases i.e., number of molecules per unit volume increases. As a result of which more molecules collide with the walls of the vessel per second and hence a large momentum is transferred to the walls per second due to which the pressure of the gas increases.

On reducing the volume, the number of molecules per unit volume increases. Hence a large number of molecules collide with the walls of the vessel per second and a larger momentum is transferred to the wall per second. This increase the pressure of the gas.

The average velocity of the molecules of an ideal gas is zero, because the molecules possess all sorts of velocities in all possible direction so their vector sum is zero and hence average is zero.

2.078 x 10⁷ ergs g^-1°C^-1

Rise in temperature of both the balls will be the same.

Let us consider two gases A and B diffusing into one another. 

Let ρ₁ and ρ₂ be their densities and v₁ and v₂ be their respective r.ms. velocities.

Pressure exerted by gas A,

\(P_1= \frac{1}{3}ρ_1V_1^2\) 

and pressure exerted by gas B,

\(P_2= \frac{1}{3}ρ_2v_2^2\)

When steady state of diffusion is reached

P₁ = P₂

\( \frac{1}{3}ρ_1v_1^2\) = \( \frac{1}{3}ρ_2v_2^2\) 

\( \frac{v_1^2}{v_2^2}=\frac{ρ_2}{ρ_1}\)

or \( \frac{v_1}{v_2}=\,\sqrt\frac{ρ_2}{ρ_1}\) 

If r₁ and r₂ be the rates of diffusion of gases A and respectively.

\(\frac{r_1}{r_2}=\frac{v_1}{v_2}=\sqrt\frac{ρ_2}{ρ_1}\)

Thus this law states that the rate of diffusion of a gas is inversely proportional to the square root of its density.

r ∝ \(\frac{1}{\sqrtρ}\)

K_i = p one atmosphere = 1.01 x 10⁵ Nm^-2

Avagadro’s hypothesis states that under fixed conditions of temperature and pressure, equal volumes of all gases contain equal number of molecules.

No, the statement does not hold good in case, the word 'molecule' is replaced by 'atoms'. It would then lead to the conclusion that fractions of atoms can take part in a chemical reaction and it is totally against Dalton's atomic theory.

The Boltzmann's constant

k_B = R/N, where R = Real Gas Constant

Zero; because ΔQ = 0; use C = ΔQ/mΔT

Average speed is the arithmetic mean of the speeds of the molecules

∴ Average speed = \(\frac{v_1+v_2+v_3}{3}\)

r.m.s. speed is the root mean square speed and is defined as the square root of the mean of the squares of random speeds of the individual molecules.

v_{rms =} \(\sqrt\frac{v_1^2+v_2^2+v_3^2}{3}\)