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IGNOU BPHCT-135 Solved Assignment 2024 | B.Sc (G) CBCS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU BPHCT-135 Assignment Question Paper 2024

bphct-135-solved-assignment-2024-qp-6d9e9760-7dd6-4cec-a975-b50451cfdbad

bphct-135-solved-assignment-2024-qp-6d9e9760-7dd6-4cec-a975-b50451cfdbad

BPHCT-135 Solved Assignment 2024 QP
PART A
  1. a) Write the assumptions of kinetic theory of gases. Derive the following expression of the pressure exerted by an ideal gas:
p = 1 3 m n v ¯ 2 p = 1 3 m n v ¯ 2 p=(1)/(3)mn bar(v)^(2)p=\frac{1}{3} m n \bar{v}^2p=13mnv¯2
Also, using this expression deduce Avogadro’s law. What is the kinetic interpretation temperature?
b) The expression for the number of molecules in a Maxwellian gas having speeds in the range v v vvv to v + d v v + d v v+dvv+d vv+dv is given by
d N v = 4 π N ( m 2 π k B T ) 3 / 2 v 2 exp ⁡ [ − ( m v 2 2 k B T ) ] d v d N v = 4 π N m 2 π k B T 3 / 2 v 2 exp ⁡ − m v 2 2 k B T d v dN_(v)=4pi N((m)/(2pik_(B)T))^(3//2)v^(2)exp[-((mv^(2))/(2k_(B)T))]dvd N_v=4 \pi N\left(\frac{m}{2 \pi k_{\mathrm{B}} T}\right)^{3 / 2} v^2 \exp \left[-\left(\frac{m v^2}{2 k_{\mathrm{B}} T}\right)\right] d vdNv=4πN(m2πkBT)3/2v2exp⁡[−(mv22kBT)]dv
Obtain an expression of average speed and root mean square speed of a molecule.
c) Define mean free path of the molecules of a gas. Show that it is equal to 1 n π d 2 1 n π d 2 (1)/(n pid^(2))\frac{1}{n \pi d^2}1nπd2 under Zeroth order approximation.
d) What is Brownian motion? Write any four characteristics of Brownian motion.
2. a) What are Intensive and extensive variables. Write two examples of each. List the intensive and extensive variables required to specify the thermodynamic systems (i) paramagnetic solid and (ii) stretched wire.
b) State Zeroth law of thermodynamics. Discuss how this law introduces the concept of temperature. Write parametric as well as exact equation of state for one mole of a real gas and paramagnetic substance.
c) For a p V T p V T pVTp V TpVT – system, show that:
d V V = α d T − β T d p d V V = α d T − β T d p (dV)/(V)=alpha dT-beta _(T)dp\frac{d V}{V}=\alpha d T-\beta_T d pdVV=αdT−βTdp
where β T β T beta _(T)\beta_TβT is the isothermal compressibility and α α alpha\alphaα is isobaric coefficient of volume expansion.
d) What is an adiabatic index? Using the first law of thermodynamics, show that T V − 1 = K T V − 1 = K TV^(-1)=KT V^{-1}=KTV−1=K, when one mole of an ideal gas is made to undergo quasi-static adiabatic expansion.
e) Derive an expression for the work done in an isothermal process of an ideal gas.
PART B
3. a) Define efficiency of a Carnot engine. A Carnot engine has an efficiency of 50 % 50 % 50%50 \%50% when its sink temperature is at 27 ∘ C 27 ∘ C 27^(@)C27^{\circ} \mathrm{C}27∘C. Calculate the source temperature for increasing its efficiency to 60 % 60 % 60%60 \%60%.
b) State third law of thermodynamics. Write its mathematical expression. Discuss some important consequences of third law.
c) Write Maxwell’s relations and using these relations derive first and second energy equations.
d) Derive Clausius-Clapeyron equation for two phases to coexist in equilibrium.
e) What is Joule-Thomson effect? Write the mathematical expression of Joule Thomson Coefficient for van der Waals’ gas. What will be the effect on gas if the intermolecular forces are strong?
4. a) Obtain an expression of single particle partition function. Hence, using this expression obtain expressions for entropy and pressure.
b) Define phase space of the system. Draw the phase space for a linear harmonic oscillator.
c) Establish the Boltzmann relation S = k B ln ⁡ W S = k B ln ⁡ W S=k_(B)ln WS=k_{\mathrm{B}} \ln WS=kBln⁡W.
d) Show that Bose derivation of Planck’s law for energy density is given by
U v d v = 8 π h c 3 v 3 d v exp ⁡ [ h v k B T − 1 ] U v d v = 8 π h c 3 v 3 d v exp ⁡ h v k B T − 1 U_(v)dv=(8pi h)/(c^(3))(v^(3)dv)/(exp[(hv)/(k_(B)T)-1])U_v d v=\frac{8 \pi h}{c^3} \frac{v^3 d v}{\exp \left[\frac{h v}{k_{\mathrm{B}} T}-1\right]}Uvdv=8πhc3v3dvexp⁡[hvkBT−1]
\(sin\left(2\theta \right)=2\:sin\:\theta \:cos\:\theta \)

BPHCT-135 Sample Solution 2024

bphct-135-solved-assignment-2024-ss-b0a87631-f2d0-400e-939b-0b99078c2d98

bphct-135-solved-assignment-2024-ss-b0a87631-f2d0-400e-939b-0b99078c2d98

BPHCT-135 Solved Assignment 2024
PART A
  1. a) Write the assumptions of kinetic theory of gases. Derive the following expression of the pressure exerted by an ideal gas:
p = 1 3 m n v ¯ 2 p = 1 3 m n v ¯ 2 p=(1)/(3)mn bar(v)^(2)p=\frac{1}{3} m n \bar{v}^2p=13mnv¯2
Also, using this expression deduce Avogadro’s law. What is the kinetic interpretation temperature?
Answer:
The kinetic theory of gases is a fundamental theory that explains the behavior of gases and the properties of gases at the molecular level. It is based on several key assumptions that simplify the complex interactions of gas particles, allowing for the derivation of important gas laws and principles. Here are the assumptions of the kinetic theory of gases:
  1. Gas consists of a large number of small particles (atoms or molecules), which are in constant, random motion.
  2. The volume of the individual particles is negligible compared to the total volume of the gas. This means that the actual volume of the gas molecules is so small relative to the space between them that it can be assumed to be zero.
  3. There are no forces of attraction or repulsion between the particles of the gas. This assumption implies that the gas particles do not interact with each other except during collisions.
  4. The collisions between gas particles and between particles and the walls of the container are perfectly elastic. This means that there is no net loss of kinetic energy in the system of particles after the collisions.
  5. The average kinetic energy of the gas particles is directly proportional to the absolute temperature of the gas. This implies that as the temperature increases, the average kinetic energy of the gas particles increases, and vice versa.
  6. The gas particles obey Newton’s laws of motion, and their motion can be described by the laws of mechanics (classical mechanics).
  7. The time of collision between the particles is much shorter than the time between successive collisions, meaning that the actual duration of a collision is negligible compared to the time particles spend flying freely.
These assumptions allow for the derivation of the ideal gas law and the explanation of gas properties such as pressure, temperature, and volume. It’s important to note that these assumptions are idealizations; real gases may deviate from these assumptions under certain conditions, such as high pressure or low temperature.
To derive the expression for the pressure exerted by an ideal gas, p = 1 3 m n v ¯ 2 p = 1 3 m n v ¯ 2 p=(1)/(3)mn bar(v)^(2)p = \frac{1}{3} m n \bar{v}^2p=13mnv¯2, we’ll follow the kinetic theory of gases and its assumptions. Here, p p ppp represents the pressure of the gas, m m mmm is the mass of a single molecule, n n nnn is the number density of the gas molecules (number of molecules per unit volume), and v ¯ 2 v ¯ 2 bar(v)^(2)\bar{v}^2v¯2 is the mean square velocity of the gas molecules.

Step 1: Consider a Gas in a Cubic Container

Imagine a cubic container of volume V V VVV with sides of length L L LLL. Gas molecules are moving randomly in all directions. We’ll focus on the molecules moving perpendicular to one of the walls of the cube.

Step 2: Calculate the Change in Momentum for a Single Molecule

When a molecule with velocity v x v x v_(x)v_xvx in the x-direction collides elastically with a wall, it rebounds with the same speed but in the opposite direction. The change in momentum ( Δ p Δ p Delta p\Delta pΔp) for this molecule during a collision with the wall is:
Δ p = 2 m v x Δ p = 2 m v x Delta p=2mv_(x)\Delta p = 2mv_xΔp=2mvx

Step 3: Determine the Rate of Collisions Against a Wall

The time ( t t ttt) it takes for a molecule to hit the wall and return to the same spot (after hitting the opposite wall and coming back) is 2 L v x 2 L v x (2L)/(v_(x))\frac{2L}{v_x}2Lvx. The rate of collisions of one molecule against a wall is the inverse of this time:
Rate of collisions = 1 t = v x 2 L Rate of collisions = 1 t = v x 2 L “Rate of collisions”=(1)/(t)=(v_(x))/(2L)\text{Rate of collisions} = \frac{1}{t} = \frac{v_x}{2L}Rate of collisions=1t=vx2L

Step 4: Calculate the Force Exerted by a Single Molecule

The force ( F F FFF) exerted by a single molecule on the wall is the rate of change of momentum:
F = Δ p t = 2 m v x ⋅ v x 2 L = m v x 2 L F = Δ p t = 2 m v x ⋅ v x 2 L = m v x 2 L F=(Delta p)/(t)=(2mv_(x)*v_(x))/(2L)=(mv_(x)^(2))/(L)F = \frac{\Delta p}{t} = \frac{2mv_x \cdot v_x}{2L} = \frac{mv_x^2}{L}F=Δpt=2mvx⋅vx2L=mvx2L

Step 5: Extend to All Molecules and Directions

Considering all molecules and averaging over all directions (since the container is cubic and the gas molecules’ movements are isotropic), the average force due to all molecules is proportional to the mean square velocity ( v ¯ 2 v ¯ 2 bar(v)^(2)\bar{v}^2v¯2) and the number of molecules ( N N NNN). The total force ( F total F total F_(“total”)F_{\text{total}}Ftotal) exerted by N N NNN molecules is:
F total = N ⋅ m v ¯ 2 3 L F total = N ⋅ m v ¯ 2 3 L F_(“total”)=N*(m bar(v)^(2))/(3L)F_{\text{total}} = N \cdot \frac{m\bar{v}^2}{3L}Ftotal=N⋅mv¯23L
Here, we divide by 3 to account for the three-dimensional motion of the molecules (x, y, and z directions).

Step 6: Relate Force to Pressure

Pressure ( p p ppp) is defined as force per unit area. For one face of the cube, the area ( A A AAA) is L 2 L 2 L^(2)L^2L2, so:
p = F total A = N ⋅ m v ¯ 2 3 L L 2 = N m v ¯ 2 3 L 3 p = F total A = N ⋅ m v ¯ 2 3 L L 2 = N m v ¯ 2 3 L 3 p=(F_(“total”))/(A)=(N*(m bar(v)^(2))/(3L))/(L^(2))=(Nm bar(v)^(2))/(3L^(3))p = \frac{F_{\text{total}}}{A} = \frac{N \cdot \frac{m\bar{v}^2}{3L}}{L^2} = \frac{N m \bar{v}^2}{3L^3}p=FtotalA=N⋅mv¯23LL2=Nmv¯23L3
Since L 3 L 3 L^(3)L^3L3 is the volume ( V V VVV) of the container and n = N V n = N V n=(N)/(V)n = \frac{N}{V}n=NV is the number density of the molecules, we can rewrite the pressure as:
p = 1 3 m n v ¯ 2 p = 1 3 m n v ¯ 2 p=(1)/(3)mn bar(v)^(2)p = \frac{1}{3} m n \bar{v}^2p=13mnv¯2
This derivation shows how the pressure exerted by an ideal gas can be related to the mass of the molecules, their mean square velocity, and their number density, under the assumptions of the kinetic theory of gases.

Deduction of Avogadro’s Law

Avogadro’s law states that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules. From the derived pressure formula, we see that the pressure is directly proportional to the number of molecules ( n n nnn) and their average kinetic energy ( 1 2 m v ¯ 2 1 2 m v ¯ 2 (1)/(2)m bar(v)^(2)\frac{1}{2}m\bar{v}^212mv¯2), which is a measure of temperature.
For a given p p ppp and T T TTT (temperature), if two gases have the same volume V V VVV, the number of molecules n n nnn must be the same, according to the formula, regardless of the type of gas. This is because the kinetic energy component ( 1 2 m v ¯ 2 1 2 m v ¯ 2 (1)/(2)m bar(v)^(2)\frac{1}{2}m\bar{v}^212mv¯2) reflects the temperature, which is the same for both gases. Thus, Avogadro’s law is a direct consequence of the kinetic theory of gases.

Kinetic Interpretation of Temperature

The kinetic interpretation of temperature is that it is a measure of the average kinetic energy of the molecules in a substance. From the kinetic theory, the average kinetic energy of a molecule is given by:
Average kinetic energy = 1 2 m v ¯ 2 Average kinetic energy = 1 2 m v ¯ 2 “Average kinetic energy”=(1)/(2)m bar(v)^(2)\text{Average kinetic energy} = \frac{1}{2}m\bar{v}^2Average kinetic energy=12mv¯2
And since the average kinetic energy is directly proportional to the absolute temperature ( T T TTT) of the gas, we can write:
Average kinetic energy ∝ T Average kinetic energy ∝ T “Average kinetic energy”prop T\text{Average kinetic energy} \propto TAverage kinetic energy∝T
This relationship indicates that the temperature of a gas is a measure of the average kinetic energy of its particles. Higher temperatures correspond to higher average kinetic energies and faster-moving particles.
b) The expression for the number of molecules in a Maxwellian gas having speeds in the range v v vvv to v + d v v + d v v+dvv+d vv+dv is given by
d N v = 4 π N ( m 2 π k B T ) 3 / 2 v 2 exp ⁡ [ − ( m v 2 2 k B T ) ] d v d N v = 4 π N m 2 π k B T 3 / 2 v 2 exp ⁡ − m v 2 2 k B T d v dN_(v)=4pi N((m)/(2pik_(B)T))^(3//2)v^(2)exp[-((mv^(2))/(2k_(B)T))]dvd N_v=4 \pi N\left(\frac{m}{2 \pi k_{\mathrm{B}} T}\right)^{3 / 2} v^2 \exp \left[-\left(\frac{m v^2}{2 k_{\mathrm{B}} T}\right)\right] d vdNv=4πN(m2πkBT)3/2v2exp⁡[−(mv22kBT)]dv
Obtain an expression of average speed and root mean square speed of a molecule.
Answer:
To find the average speed and root mean square (rms) speed of a molecule in a Maxwellian gas, we’ll use the distribution function given:
d N v = 4 π N ( m 2 π k B T ) 3 / 2 v 2 exp ⁡ [ − ( m v 2 2 k B T ) ] d v d N v = 4 π N m 2 π k B T 3 / 2 v 2 exp ⁡ − m v 2 2 k B T d v dN_(v)=4pi N((m)/(2pik_(B)T))^(3//2)v^(2)exp[-((mv^(2))/(2k_(B)T))]dvd N_v = 4 \pi N \left(\frac{m}{2 \pi k_B T}\right)^{3/2} v^2 \exp\left[-\left(\frac{m v^2}{2 k_B T}\right)\right] dvdNv=4πN(m2πkBT)3/2v2exp⁡[−(mv22kBT)]dv
where:
  • N N NNN is the total number of molecules,
  • m m mmm is the mass of a molecule,
  • k B k B k_(B)k_BkB is the Boltzmann constant,
  • T T TTT is the temperature in Kelvin,
  • v v vvv is the speed of a molecule.

Average Speed ( v ¯ v ¯ bar(v)\bar{v}v¯)

The average speed is given by the first moment of the speed distribution:
v ¯ = 1 N ∫ 0 ∞ v d N v v ¯ = 1 N ∫ 0 ∞   v d N v bar(v)=(1)/(N)int_(0)^(oo)vdN_(v)\bar{v} = \frac{1}{N} \int_{0}^{\infty} v \, d N_vv¯=1N∫0∞vdNv
Substituting the expression for d N v d N v dN_(v)d N_vdNv:
v ¯ = 1 N ∫ 0 ∞ 4 π N ( m 2 π k B T ) 3 / 2 v 3 exp ⁡ [ − ( m v 2 2 k B T ) ] d v v ¯ = 1 N ∫ 0 ∞   4 π N m 2 π k B T 3 / 2 v 3 exp ⁡ − m v 2 2 k B T d v bar(v)=(1)/(N)int_(0)^(oo)4pi N((m)/(2pik_(B)T))^(3//2)v^(3)exp[-((mv^(2))/(2k_(B)T))]dv\bar{v} = \frac{1}{N} \int_{0}^{\infty} 4 \pi N \left(\frac{m}{2 \pi k_B T}\right)^{3/2} v^3 \exp\left[-\left(\frac{m v^2}{2 k_B T}\right)\right] dvv¯=1N∫0∞4πN(m2πkBT)3/2v3exp⁡[−(mv22kBT)]dv
Simplifying and integrating, we get:
v ¯ = 8 k B T π m v ¯ = 8 k B T π m bar(v)=sqrt((8k_(B)T)/(pi m))\bar{v} = \sqrt{\frac{8 k_B T}{\pi m}}v¯=8kBTπm

Root Mean Square Speed ( v rms v rms v_(“rms”)v_{\text{rms}}vrms)

The root mean square speed is given by the square root of the second moment of the speed distribution:
v rms = 1 N ∫ 0 ∞ v 2 d N v v rms = 1 N ∫ 0 ∞   v 2 d N v v_(“rms”)=sqrt((1)/(N)int_(0)^(oo)v^(2)dN_(v))v_{\text{rms}} = \sqrt{\frac{1}{N} \int_{0}^{\infty} v^2 \, d N_v}vrms=1N∫0∞v2dNv
Substituting the expression for d N v d N v dN_(v)d N_vdNv:
v rms = 1 N ∫ 0 ∞ 4 π N ( m 2 π k B T ) 3 / 2 v 4 exp ⁡ [ − ( m v 2 2 k B T ) ] d v v rms = 1 N ∫ 0 ∞   4 π N m 2 π k B T 3 / 2 v 4 exp ⁡ − m v 2 2 k B T d v v_(“rms”)=sqrt((1)/(N)int_(0)^(oo)4pi N((m)/(2pik_(B)T))^(3//2)v^(4)exp[-((mv^(2))/(2k_(B)T))]dv)v_{\text{rms}} = \sqrt{\frac{1}{N} \int_{0}^{\infty} 4 \pi N \left(\frac{m}{2 \pi k_B T}\right)^{3/2} v^4 \exp\left[-\left(\frac{m v^2}{2 k_B T}\right)\right] dv}vrms=1N∫0∞4πN(m2πkBT)3/2v4exp⁡[−(mv22kBT)]dv
Simplifying and integrating, we get:
v rms = 3 k B T m v rms = 3 k B T m v_(“rms”)=sqrt((3k_(B)T)/(m))v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}vrms=3kBTm
So, the average speed and root mean square speed of a molecule in a Maxwellian gas are given by 8 k B T π m 8 k B T π m sqrt((8k_(B)T)/(pi m))\sqrt{\frac{8 k_B T}{\pi m}}8kBTπm and 3 k B T m 3 k B T m sqrt((3k_(B)T)/(m))\sqrt{\frac{3 k_B T}{m}}3kBTm, respectively.
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