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P' = P + \(\frac{0.4}{100}P\)

T' = T + 1

By Gay Lussac’s law,

\(\frac{P}{T}=\frac{P'}{T'}\)

\(\frac{P}{T}=\frac{[P+{\frac{0.4}{100}P'}]}{T+1}\) 

On solving, 

T = 250 K

Because the molecules of ammonia move at random and continuously collide with one another. As a result of which they are not able to advance in one particular direction speedly.

According to kinetic theory, the molecules of a gas are in a state of continuous random motion. They collide with one another and also with the walls of the vessel.

Whenever a molecule collides with the wall, it returns with a changed momentum and an equal momentum is transferred to the wall.

The air molecule travel along a zig-zag path due to frequency collisions. As a result, their displacement per unit time is very small. Hence the small of scent spreads very slowly.

(c) This is a constant pressure (p = Mg/ A) arrangement

On pulling the piston out, the volume of the cylinder for the given gas increases due to which the molecules of gas more space to move about As a result of which less molecules will collide with the wall of cylinder per second and hence less momentum is transferred to the wall pre second. In addition of it, now these collisions take place on a larger area of the walls. Due to both these reasons the pressure decreases.

When a gas is heated, the root mean square velocity of its molecules increases. 

As, \(v_{rms}∝\sqrt{T}\). 

So, the temperature of the gas increases

On raising the temperature the average velocity of the gas molecule increases. As a result of which more molecules collide with the wall of the cylinder per second and hence greater momentum is transferred to the wall per second. Due to both these reason the pressure increases.

(∵ P=\(\frac{1}{3}\frac{mnc^2}{V}\))

As, 

PV = nRT

for vessel B, 

n_B = \(\frac{PV}{RT}\)

and for vessel A, 

n_A = \(\frac{2P\times2V}{R\times2T}\)

= \(2\frac{PV}{RT}\)

and n_A: n_B = 2 : 1

At S.T.P., actual volume occupied by 1 mole of oxygen is

V = 22400 ml

= 22400 cm³

Also, molecular volume,

V' = 4/3 πR³N

where R = D/2 = 3/2 Å

= 3/2 x 10^-8 = 1.5 x 10^-8 cm

and N = 6.023 x 10²³

Hence, V' = 4/3 x 3.143 x (1.5 x 10^-8)³ x 6.023 x 10²³

= 8.52 cm³

Fraction of molecular volume to actual volume = V'/V

= 8.52/22400 = 3.8 x 10^-4 = 4 x 10^-4

Real gases obey Boyle’s law at low density.

(i) Yes, because the mean K.E. per molecule depends only upon the temperature.

(ii) No, because for different gases r.m.s. velocity depends upon the mass of the gas molecules.

(iii) No definite idea about pressure because masses of the gases are not given.

\(\frac{(v_{rms})_{kr}}{(v_{rms})_{H_2}}=\sqrt\frac{M_{H_2}}{M_{kr}}\)

\(=\sqrt\frac{1}{2.25}\)

\(=\sqrt\frac{4}{9}\)

= 2: 3

(i) Yes, because the mean K.E. per molecule (= 3/2 KT) depends only on the temperature.

(ii) No, because for different gases r.m.s. velocity depends upon the mass of the gas molecules.

(iii) No definite idea about pressure because masses of the gases are not given

O₂ has 5 degree of freedom. 

Therefore, energy per mole = \(\frac{5}{2}\)RT

∴ For 2 moles of O₂, energy = 5RT

Neon has 3 degrees of freedom,

∴ Energy per mole = \(\frac{3}{2}\)RT

∴ For 4 mole of neon, energy = 4 × \(\frac{3}{2}\) RT 

= 6RT

∴ Total energy = 11RT.

As, 

P = \(\frac{1}{3}\frac{M}{V}\)c² 

but c² ∝ T; 

therefore, if V and T are constant, 

the P ∝ M. as M becomes 2M, 

P becomes 2P.

Since all vessels are of same capacity and are at same pressure and temperature,

P = P₁ = P₂ = P₃ ……..(1)

V = V₁ = V₂ = V₃ ……..(2)

T = T₁= T₂ = T₃ ……..(3)

where P_i V_i and T_i are pressure, volume and temperature of i^th vessel respectively.

From ideal gas law

PV = nRT

∴ P_i V_i = n_i RT_i i = 1,2, 3

From equation (1), (2) and (3).

PV = n_i RT

or n_i = \(\frac{PV}{RT}\) = constant for all the three vessels.

∴ All three vessels have same no of moles of respective gases. From Avagadro’s law, equal mols of gases contain equal no of molecules. The root mean square (RMS) speed of a molecule is given by

V_rms = \(\sqrt \frac{3kg}{m}\)

where kB = Boltzmann constant

T = temperature

m = molecular mass of the gas.

As Vrms α 1/√m the rms speed of gas molecules in the three vessels are different. Since neon has the least molecular mass of the three gases, it has the highest rms speed.

As,

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

so P ∝ nc²

(i) Here, M and V are constant.

Since c ∝ \(\sqrt{T}\)

But, T is constant

So, c₁: c₂ = 1 : 1

(ii) When temperature is same for two vessels then, 

P ∝ n.

i.e., \(\frac{n_1}{n_2}=\frac{P_1}{P_2}=\frac{1}{2}\)

V₁ = 2.0 litre, 

V₂ = 3.0 litre

μ₁ = 4.0 moles, 

μ₂ = 5.0 moles

P₁ = 1.00 atm, 

P₂ = 2.00 atm

P₁V₁ =μ₁RT₁, 

P₂V₂ = μ₂RT₂

μ = μ₁ + μ₂ 

V = V₁ + V₂

For 1 mole, PV =  \(\frac{2}{3}\)E

For μ₁ moles, P₁V₁ = \(\frac{2}{3}\)μ₁E₁

For μ₂ moles, P₂V₂ = \(\frac{2}{3}\)μ₂E₂

Total energy is (μ₁E₁ + μ₂E₂) 

= \(\frac{3}{2}\)(P₁V₁ + P₂V₂)

PV = \(\frac{2}{3}\)E_total = \(\frac{2}{3}\) μE_{per mole}

P(V₁ + V₂) = \(\frac{2}{3}\) × \(\frac{3}{2}\) (P₁V₁ + P₂V₂)

\(P=\frac{P_1V_1+P_2V_2}{V_1+V_2}\)

\(=(\frac{1.00\times 2.0+2.00\times 3.0}{2.0+3.0})atm\)

\(=\frac{8.0}{5.0}\)

= 1.60 atm.

Comment : This form of Ideal gas law represented by Equation marked* becomes very useful for adiabatic changes

Let us find the product of atomic mass and specific heat i.e., molar sp. heat from the given data

(i) For Helium, M x c_v = 4 x 0.748 = 2.99 Cal/K

(ii) For Neon, M x c_v = 20.18 x 0.147 = 2.97 Cal/K

(iii) For Argon, M x c_v = 39.94 x 0.076 = 3.04 Cal/K

(iv) For Crypton, M x c_v = 83.83 x 0.358 = 3.00 Cal/K

(v) For Xenon, M x c_v = 131.3 x 0.226 = 2.97 Cal/K

and 3/2 R = 3/2 x 1.98 = 2.97 Cal/K

So, from the above data, it is clear that molar specific heat i.e., product of a atomic mass and specific heat is nearly constant for all the gases and is  equal to 3/2 R, in case of all monoatomic gases as given by the kinetic theory of gas

There are two types of molar specific heats of gases.

(a) At constant volume (C₁): It is the amount of heat energy required to raise the temperature of one mole gas at constant volume by 1°C.

(b) At constant pressure (C_p)- It is the amount of heat required to raise the temperature of one mole gas at constant pressure by 1°C.

∴ Molar mass of gold is 197 g mole^-1, the number of atoms = 6.0×10²³ 

∴ No. of atoms in 39.4g = (6.0 x 10²³ x 39.4/197) = 1.2 x 10²³ 

In evaporation, because escaping molecules have higher kinetic energy, hence the average kinetic energy of the molecules left behind decreases. As average kinetic energy is directly related with temperature, hence evaporation causes cooling.

According to kinetic theory of gases,

P = \(\frac{1}{3}ρv^2\)

= \(\frac{2}{3}\frac{ρ}{m}.\frac{1}{2}mv^2\)

P = \(\frac{2}{3}\frac{ρ}{m}.\frac{3}{2}k_BT\)

∴ Density  \(ρ=\frac{Pm}{k_BT}\)

Zero, the molecules cannot hit this portion on the wall, so no pressure is experienced.

It states that volume (V) of a given mass of a gas changes by a constant fraction of its volume (V₀) at 0°C for every changes of 1°C when pressure remains constant.

Let y_p be a constant fraction called coefficient of thermal expansion at constant pressure, then

V = V₀₍1 + y_pθ),

Where θ is change in temperature.

According to kinetic theory, pressure, P exerted by a gas is

\(P=\frac{1}{3}ρv^2\)

\(=\frac{1}{3}(\frac{M}{V})v^2\)

∴ PV = \(\frac{1}{3}Mv^2\)

But at constant temperature total K.E. of gas \(\frac{1}{3}\) (M/V)v² or v² will be constant.

∴ At a constant temperature, PV = constant.

According to kinetic theory of gases, pressure of a gas,

\(P=\frac{1}{3}.ρv^2\)

\(=\frac{1}{3}\frac{M}{V}v^2\)

\(P=\frac{1}{3}.\frac{M}{V}v^2\)

For a given mass of a gas at constant pressure P, we have

V ∝ v²

According to kinetic theory,

V² ∝ T

so, V ∝ T

This proves Charles’s law.

C ∝ √T. Therefore, the r.m.s. velocity becomes √3 C. Hence, increase in r.m.s. velocity = √3 C - C = 0.732 C.

According to Avogadro’s hypothesis, both will have equal no. of molecules.

An ideal gas is a gas that obeys all the gas laws at all temperatures and is given by

PV = µRT

where P = pressure of the gas

V = Volume of the gas

µ = number of mols of the gas

R = Universal gas constant

T = absolute temperature .

(d)  20 times the pressure initially.

Comment for discussion: The usual statement for the perfect gas law somehow emphasizes molecules. If a gas exists in atomic form (perfectly possible) or a combination of atomic and molecular form, the law is not clearly stated.

One, because the kinetic energy per molecule of the gas depends only upon the temperature.

Pressure exerted by one mole of gas,

\(P=\frac{1}{3}ρv^2_{rms}\) 

\(=\frac{1}{3}\frac{M}{V}v^2_{rms}\) 

or, \(PV=\frac{1}{3}Mv^2_{rms}\) 

But  PV = RT

∴ \(\frac{1}{3}Mv^2_{rms}\) = RT

\(Mv^2_{rms}\) = 3RT

Now, average KE

\(\frac{1}{3}Mv^2_{rms}\) = \(\frac{3}{2}\)RT

\(\frac{1}{2}(Nm)v^2\) = \(\frac{3}{2}\)RT (∵ M = Nm)

(KE)_Avg for one mole

\(\frac{1}{3}Mv^2_{rms}\) = \(\frac{3}{2}\frac{R}{N}T\)= \(\frac{3}{2}\)k_BT

∴ Total random K.E. for one mole =\(\frac{3}{2}\)RT

and average K.E. per molecule = \(\frac{3}{2}\)k_BT

The change in gravitational potential energy are negligibly small as compared to the mean kinetic energy all molecules.

Here, 

ρ = 0.09 kg/m³

P = 1.01 × 10⁵ Pa

(i) According to kinetic theory of gases,

\(P=\frac{1}{3}ρc^2\) 

or  \(c=\sqrt\frac{3P}{ρ}\)

\(=\sqrt\frac{3\times1.01\times10^5}{0.09}\) 

= 1837.5 m/s

(ii) Volume occupied by one mole of hydrogen at S.T.P. 

= 22.4 litres

= 22.4 × 10^-3m³

Mass of hydrogen, 

M = Volume × Density

= 22.4 × 10^-3 × 0.09 kg

= 2.016 × 10^-3 kg

Mean K.E. of one-gram molecule of hydrogen at S.T.P.

= 1/2 MV²

= 1/2 × 2.016 × 10^-3 × (1837.5)²

= 3.4 × 10³ J

1 : 1

Because kinetic energy per molecule of the gas only depends upon the temperature.

Here, p = 1 atm = 1 x 1.01 x 10⁵ Pa

V = 25 m³

T = 27°C = 27 + 273 = 300 K

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

Now pV = uRT

⇒ u = pV/RT = {1.01 x 10⁵ x 25}/{831 x 300} = 1.013 x 10³ moles

Now, one mole of a gas contains = 6.023 x 10²³ molecules

∴ 1.013 x 10³ moles would contain

= 6.023 x 10²³ 1.013 x 10³ = 6.10 x 10²⁶ molecules

c₁/c₂ = √{M₂/M₁} = √{ρ₂/ρ₁} = √{9/8}

= 3 : 2√r

\(\frac{c_1}{c_2}=\sqrt\frac{m_2}{m_1}=\sqrt\frac{ρ_2}{ρ_1}\)

=\(\sqrt\frac{9}{8}\)

= c₁: c₂

= \(3:2\sqrt2\)

When the two gases have exactly the same temperature the average kinetic energy per molecule for each gas is the same. But as the different gases may have molecules of different masses, the r.m.s. speed (c) of molecules of different gases shall be different.

Yes. It is equal to 3/2 k_BT.

Atomicity of nitrogen = 2

Atomicity of oxygen = 2

Atomicity of hydrogen = 2

No, it does not increase monoatomaically with the increase in atomic mass. It is because of the fact that the size of atom is due to the electron shells around the nucleus. When the mass of atom increases, it means the number of nucleons in the nucleus of the atom has increased. As a result atomic number may increase. However, the increase in the number of electrons (atomic number) may be only into a shell of the atom, Thus, the size of the atom may not increase.

The law of equipartition of energy states that in thermal equilibrium, the total energy of a particle is equally distributed in all possible modes of energy, with each mode having an average energy equal to \(\frac{1}{2}\) kB T.

−273.15°C is the lowest temperature attainable according to Charle’s law.

Charles’ law states that at constant pressure, the volume of a given mass of an ideal gas is directly proportional to its temperature.

consider a gas at pressure ‘P’, temperature ‘T’ having volume ‘V’, Then Charles’ law states that

v ∝ T

or V = kT where k is the proportionality constant.

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

For one mole of an ideal gas, we have

pV = RT

Hence, V = RT/p

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

T = 273 K

and p = 1 atmosphere = 1.013 x 10⁵ Nm^-2

Thus, V = {8.31 x 273}/{1.013 x 10⁵} 

= 0.024 m³

= 0.0224 x 10⁶ cm³

= 22400 ml  (1 cm³ = 1 ml)

(a) Brownian motion decreases with increase in the size of the particle.

(b) Brownian motion increases with decrease in the density of the medium.

(c) Brownian motion increases with the increase in temperature of the medium.

(d) Brownian motion decreases with the increase in the value of viscosity of the medium.