Heat and Thermodynamics

To determine the frequency of collisions for oxygen molecules, we use the formula for frequency:

$ \text{Frequency} = \frac{1}{T} = \frac{V_{\text{avg}}}{\lambda} $

where $ V_{\text{avg}} $ is the average speed of the molecules and $ \lambda $ is the mean free path.

Given:

Average speed ($ V_{\text{avg}} $) = 600 m/s

Mean free path ($ \lambda $) = $ 3 \times 10^{-7} $ m

Substitute these values into the formula:

$ \text{Frequency} = \frac{600}{3 \times 10^{-7}} = 2 \times 10^9 \, \text{s}^{-1} $

Therefore, the frequency of collisions is $ 2 \times 10^9 \, \text{collisions per second} $.

$\begin{aligned} & \omega_1=\mathrm{P}_0 \mathrm{v}_0 \ell \mathrm{n} 4 \\ & \omega_2=\frac{\mathrm{P}_0}{4}\left(-3 \mathrm{v}_0\right)=-\frac{3 \mathrm{P}_0 \mathrm{v}_0}{4} \\ & \omega_3=0 \\ & \mathrm{Q}_{\mathrm{T}}=\Delta \mathrm{U}_{\text {cyclic }}+\omega \\ & \mathrm{Q}_{\mathrm{T}}=\omega \quad\left(\Delta \mathrm{U}_{\text {cyclic }}=0\right) \\ & \mathrm{Q}_{\mathrm{T}}=\mathrm{P}_0 \mathrm{v}_0\left(\ln 4-\frac{3}{4}\right) \\ & =\mathrm{P}_0 \mathrm{v}_0(2 \ln 2-0.75) \end{aligned}$

In this scenario, we're dealing with an isochoric process, meaning the volume of the gas remains constant. For an ideal gas in an isochoric process, the pressure ($P$) is directly proportional to the temperature ($T$) in Kelvin.

The relationship can be expressed as:

$ P \propto T $

Given the percentage increase in pressure is $0.4\%$ when the temperature is increased by $1^{\circ} \text{C}$, we can use the relationship:

$ \frac{\Delta P}{P} = \frac{\Delta T}{T} $

Substituting the given values into the equation, we have:

$ \frac{0.4}{100} = \frac{1}{T} $

Solving for $T$, we find:

$ T = 250 \, \text{K} $

Therefore, the initial temperature of the gas in the vessel must be $250 \, \text{K}$.

To determine the change in temperature when a gas is adiabatically compressed, we are given the following initial conditions:

Initial volume, $ V_1 = 800 \, \text{cm}^3 $

Final volume, $ V_2 = 200 \, \text{cm}^3 $

Initial temperature, $ T_1 = 27^\circ \text{C} = 300 \, \text{K} $

Adiabatic index (ratio of specific heats) $ \gamma = 1.5 $

In an adiabatic process, the relationship between temperature and volume is given by the equation:

$ TV^{\gamma-1} = \text{constant} $

Applying this to the initial and final states:

$ 300 \times (800)^{0.5} = T_2 \times (200)^{0.5} $

Solving for the final temperature $ T_2 $:

$ T_2 = 300 \times \left( \frac{800}{200} \right)^{0.5} = 300 \times (4)^{0.5} $

$ T_2 = 300 \times 2 = 600 \, \text{K} $

Thus, the change in temperature is:

$ \Delta T = T_2 - T_1 = 600 \, \text{K} - 300 \, \text{K} = 300 \, \text{K} $

Therefore, when the gas is adiabatically compressed, the temperature increases by 300 K.

During the melting of a slab of ice at 273 K under atmospheric pressure, the following occurs:

As the ice melts, its volume decreases. Because of this volume reduction, the atmosphere exerts a positive work on the ice-water system. When heat is absorbed by the ice-water system, we denote the absorbed heat as $ \Delta \mathrm{Q} $, which is positive. The work done by the ice-water system is negative because it is being compressed by the external pressure.

According to the first law of thermodynamics:

$ \Delta \mathrm{U} = \Delta \mathrm{Q} + \Delta \mathrm{W} $

In this equation, $ \Delta \mathrm{U} $ represents the change in internal energy, $ \Delta \mathrm{Q} $ is the heat absorbed by the system, and $ \Delta \mathrm{W} $ is the work done by the system. Since the work done is negative and the absorbed heat is positive, the overall internal energy ($ \Delta \mathrm{U} $) increases.

Using WET

Total energy supplied $=$ gravitational potential energy + spring potential energy + work done by gas

$ \begin{aligned} & \mathrm{Mg}\left(\mathrm{~L}_1-\mathrm{L}_0\right)+\int_{\mathrm{L}_0}^{\mathrm{L}_1} \mathrm{kx}^3 \mathrm{dx}+\mathrm{nRT} \ell \mathrm{n} \\ & {\left[\frac{\mathrm{~L}_1 \mathrm{~A}}{\mathrm{~L}_0 \mathrm{~A}}\right]+\mathrm{W}_{\text {ext }}=0} \\ & \frac{\mathrm{~K}}{4}\left[\mathrm{x}^4\right]_{\mathrm{L}_0}^{\mathrm{L}_1}+\mathrm{Mg}\left(\mathrm{~L}_1-\mathrm{L}_0\right)+\int_{\mathrm{L}_0}^{\mathrm{L}_1} k x^3 \mathrm{dx}+\mathrm{nRT} \ell \mathrm{n} \\ & {\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_0}\right]+\mathrm{W}_{\text {ext }}=0} \\ & \frac{\mathrm{k}}{4}\left(\mathrm{~L}_1^4-\mathrm{L}_0^4\right)+\mathrm{Mg}\left(\mathrm{~L}_1-\mathrm{L}_0\right)+\mathrm{nRT} \ell \mathrm{n} \\ & {\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_0}\right]+\mathrm{W}_{\text {ext }}=0} \\ & \mathrm{~W}_{\text {ext }}=\frac{\mathrm{k}}{4}\left(\mathrm{~L}_1^4-\mathrm{L}_0^4\right)+\mathrm{Mg}\left(\mathrm{~L}_1-\mathrm{L}_0\right)+\mathrm{nRT} \ell \mathrm{n}\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_0}\right] \end{aligned}$

To match the entries in List-I with their corresponding units in List-II, we can use the definitions and formulas related to each concept in thermal physics:

Heat Capacity of a Body (C') is defined as:

$ C' = \frac{\Delta Q}{\Delta T} $

The unit for heat capacity (C') is Joules per Kelvin (J K$^{-1}$).

Specific Heat Capacity of a Body (S) indicates how much heat energy is required to change the temperature of a unit mass of a substance by one degree Kelvin:

$ S = \frac{\Delta Q}{m \Delta T} $

The unit for specific heat capacity (S) is Joules per kilogram per Kelvin (J kg$^{-1}$ K$^{-1}$).

Latent Heat (L) is the heat energy absorbed or released during a phase change of a substance at a constant temperature:

$ L = \frac{\Delta Q}{m} $

The unit for latent heat (L) is Joules per kilogram (J kg$^{-1}$).

Thermal Conductivity (K) measures the ability of a material to conduct heat:

$ \Delta Q = \frac{KA \Delta T}{L} \quad \Rightarrow \quad K = \frac{\Delta Q \cdot L}{A \Delta T} $

The unit for thermal conductivity (K) is Joules per meter per Kelvin per second (J m$^{-1}$ K$^{-1}$ s$^{-1}$).

Using these formulas and their corresponding units, the correct matches are:

(A) Heat capacity of body $\rightarrow$ (II) J K$^{-1}$

(B) Specific heat capacity of body $\rightarrow$ (III) J kg$^{-1}$ K$^{-1}$

(C) Latent heat $\rightarrow$ (I) J kg$^{-1}$

(D) Thermal conductivity $\rightarrow$ (IV) J m$^{-1}$ K$^{-1}$ s$^{-1}$

In an adiabatic process, the key characteristics for a monoatomic gas are:

No Heat Exchange: The adiabatic process occurs without heat transfer, meaning $ Q = 0 $.

Work Done and Internal Energy Change: The work done on or by the system is equal to the change in internal energy. Mathematically, this is expressed as:

$ \Delta U = -W $

where $\Delta U$ is the change in internal energy and $W$ is the work done.

Work and Temperature Relationship: The work done when changing the temperature from $ T_1 $ to $ T_2 $ is proportional to the difference between these temperatures. This relationship can be expressed as:

$ |\text{Work Done}| = nC_v \Delta T \propto (T_2 - T_1) $

Here, $ n $ is the number of moles, $ C_v $ is the molar heat capacity at constant volume, and $\Delta T$ is the change in temperature.

From these characteristics, the correct features of an adiabatic process in a monoatomic gas relate to the relationships involving work done and temperature change, specifically options (B) and (E).

Option B is correct.

• In an adiabatic process no heat is exchanged, so

$\delta Q = 0.$

• The (molar) heat capacity for any process is defined by

$C = \frac{\delta Q}{n\,dT}\,. $

Since $\delta Q=0$ even when $dT\neq0$, it follows that

$C_{\rm adiabatic}=0\,. $

All other statements are therefore false.

The equation for a real gas is given by:

$ \left(\mathrm{P} + \frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V} - \mathrm{b}) = \mathrm{RT} $

Here, $ \mathrm{P} $, $ \mathrm{V} $, $ \mathrm{T} $, and $ R $ are the pressure, volume, temperature, and gas constant, respectively.

To find the dimension of $ ab^{-2} $, we proceed with the following steps:

Rearrange the equation for dimensions:

$ [\mathrm{a}] = [\mathrm{P}][\mathrm{V}^2] $

Since $[\mathrm{P}] = \mathrm{ML}^{-1}\mathrm{T}^{-2}$ and $[\mathrm{V}] = \mathrm{L}^3$, we have:

$ [\mathrm{a}] = \mathrm{ML}^{-1}\mathrm{T}^{-2}(\mathrm{L}^6) = \mathrm{ML}^5\mathrm{T}^{-2} $

As derived from the equation:

$[\mathrm{b}] = [\mathrm{V}] = \mathrm{L}^3$

Now, to find $[\mathrm{ab}^{-2}]$:

$[\mathrm{ab}^{-2}] = \frac{[\mathrm{a}]}{[\mathrm{b}]^2} = \mathrm{ML}^5\mathrm{T}^{-2} \cdot \mathrm{L}^{-6} = \mathrm{ML}^{-1}\mathrm{T}^{-2}$

This dimension, $\mathrm{ML}^{-1}\mathrm{T}^{-2}$, is equivalent to the dimension of energy density.

To convert a temperature difference into electrical energy effectively, the material in question should have low thermal conductivity and high electrical conductivity. Such materials are classified as thermoelectric materials.

A low thermal conductivity ensures that the material does not easily transfer heat, which is crucial for maintaining a significant temperature difference between its hot and cold sides. This helps in generating a greater voltage. Meanwhile, high electrical conductivity is essential for facilitating the efficient movement of electrons across the material when there is a temperature difference, leading to maximal energy conversion.

Therefore, a material that optimally harvests heat energy into electrical energy should embody these characteristics.

We know, ideal gas equation,

$\mathrm{Pv = nRT}$

For isothermal process, T = constant

$\mathrm{\Rightarrow PV = const.}$

$\mathrm{\Rightarrow P \propto \frac{1}{V}}$

For an adiabatic process,

$\mathrm{PV^r=constant}$

We can see, for the same pressure incresement, ${P_2} - {P_1}$, $\left( {{v_4} - {v_3}} \right) > \left( {{v_2} - {v_1}} \right)$

i.e. the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process. Hence, option 4 is corret.

Here, we will use the Newton's law of cooling.

We know, ${{{T_1} - {T_2}} \over {\Delta t}} = k\left( {{{{T_1} + {T_2}} \over 2} - {T_s}} \right)$ .... (1)

where, T$_1$ and T$_2$ are initial and final temperatures respectively.

$\mathrm{T_s}$ = surrounding temperature

$\Delta t$ = time taken

$k$ = positive constant

Given,

Ist case :

${T_1} = 90^\circ C$

${T_2} = 80^\circ C$

$\Delta t = t$, ${T_s} = 20^\circ C$

IInd case :

${T_1} = 80^\circ C$

${T_2} = 60^\circ C$

${T_s} = 20^\circ C$, $\Delta t = ?$

Now, for Ist case,

${{90 - 80} \over t} = k\left( {{{90 + 80} \over 2} - 20} \right)$ ...... (from (1))

$ \Rightarrow {{10} \over t} = k\left( {85 - 20} \right)$

$ \Rightarrow {{10} \over t} = 65k$ .... (2)

For IInd case,

${{80 - 60} \over {\Delta t}} = k\left( {{{80 + 60} \over 2} - 20} \right)$ .... (From (1))

$ \Rightarrow {{20} \over {\Delta t}} = k\left( {70 - 20} \right)$

$ \Rightarrow {{20} \over {\Delta t}} = 50k$ .... (3)

Now, by (2) $\div$ (3),

${{{{10} \over t}} \over {{{20} \over {\Delta t}}}} = {{13} \over {10}}$

$ \Rightarrow {{\Delta t} \over t}\left( {{1 \over 2}} \right) = {{13} \over 5}$

$ \Rightarrow \Delta t = {{13} \over 5}t$

As A to B is can adiabatic path,

So, $\Delta {Q_{A \to B}} = o$ .... (1)

So, now we need to find the heat absorbed for the isothermal path (B to C)

We know, change in internal energy

$\Delta u = m{c_v}\Delta T$

$ \Rightarrow \Delta u = o$ (for isothermal as $\Delta T = o$)

Now, using Ist law of thermodynamics

$\Delta Q = \Delta u + w$

$ \Rightarrow \Delta Q = o + w \Rightarrow \Delta Q = w$

So, $\Delta {Q_{B \to C}} = {W_{B \to C}}$

$ = nRT\ln \left( {{{{v_C}} \over {{v_B}}}} \right)$

$ \Rightarrow \Delta {Q_{B \to C}} = nR(450)\ln \left( {{4 \over 3}} \right)$

$ \Rightarrow \Delta {Q_{B \to C}} = 450nR\ln \left( {{4 \over 3}} \right)$ .... (2)

So net heat absorbed,

$Q = \Delta {Q_{A \to B}} + \Delta {Q_{B \to C}}$

$ \Rightarrow Q = 450nR\ln \left( {{4 \over 3}} \right)$ ... (from 1 and 2)

Net heat absorbed per mole,

$ \Rightarrow {Q \over n} = 450R(\ln 4 - \ln 3)$.

Option 2 is correct.

$\Delta \mathrm{W}=-\Delta \mathrm{U}=-\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$

We are given that the ratio of the vapour densities of the two gases is

$\frac{4}{25}.$

Since vapour density is proportional to the molecular mass, we can write

$\frac{M_1}{M_2} = \frac{4}{25},$

where $M_1$ and $M_2$ are the molecular masses of the gases.

The root mean square (r.m.s.) velocity of a gas is given by

$v_{\text{rms}} = \sqrt{\frac{3RT}{M}},$

where:

$R$ is the universal gas constant,

$T$ is the temperature,

$M$ is the molecular mass of the gas.

Since both gases are at the same temperature, the ratio of their r.m.s. velocities is

$\frac{v_{\text{rms,1}}}{v_{\text{rms,2}}} = \sqrt{\frac{M_2}{M_1}}.$

Substitute the ratio of molecular masses:

$\frac{v_{\text{rms,1}}}{v_{\text{rms,2}}} = \sqrt{\frac{25}{4}} = \frac{5}{2}.$

Thus, the ratio of their r.m.s. velocities is

$\frac{5}{2}.$

The correct answer is Option A: $\frac{5}{2}.$

$\begin{aligned} & (\mathrm{KE})_{\text {Transsational }}=\left[\frac{3}{2} \mathrm{KT}\right] \times \text { no. of molecule } \\ & \text { No. of molecule }=\left[\frac{50}{44} \times 6.023 \times 10^{23}\right] \\ & (\mathrm{KE})_{\text {Transsational }}=4108.644 \mathrm{~J} \end{aligned}$

We know, for an ideal gas,

${V_{rms}} = \sqrt {{{3RT} \over M}} $

$ \Rightarrow {V_{ms}} = V_{rms}^2 = {{3RT} \over M}$

$ \Rightarrow {V_{ms}} \propto T$

$ \Rightarrow {V_{ms}} \propto T$

i.e. mean squared velocity is directly proportional to temperature.

$ \Rightarrow {V_{ms}} = {\left( {{{3R} \over M}} \right)^T}$

by comparing it with $y=mx$ (straight line)

Hence, option 3 is correct.

We know, efficiency of a carnot engine, $\eta$ or $e = 1 - {{{T_C}} \over {{T_H}}}$

where, T$_C$ = temperature of cold sink

T$_H$ = temperature of hot source

So, ${\eta _{12}} = 1 - {{273} \over {473}} = {{200} \over {473}} = 0.423$

${\eta _1} = 1 - {{373} \over {473}} = {{100} \over {473}} = 0.211$

${\eta _2} = 1 - {{273} \over {373}} = {{100} \over {373}} = 0.268$

Here, ${\eta _1} + {\eta _2} = 0.211 + 0.268 = 0.479 > 0.423$

So, ${\eta _{12}} < {\eta _1} + {\eta _2}$

$\begin{aligned} & \frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{t}}=\mathrm{K}\left[\mathrm{~T}_{\text {avg }}-\mathrm{T}_{\mathrm{s}}\right] \\ & \mathrm{T}_1=24^{\circ} \mathrm{C} ; \mathrm{T}_2=40^{\circ} \mathrm{C}, \mathrm{t}=4, \mathrm{~T}_{\mathrm{s}}=16^{\circ} \mathrm{C} \\ & \frac{40-24}{4}=\mathrm{K}[32-16] \\ & \mathrm{K}=\frac{4}{16}=\frac{1}{4} \\ & \text { Now } \frac{24-\mathrm{T}}{4}=\mathrm{K}\left[\frac{\mathrm{~T}+24}{2}-16\right] \\ & 24-\mathrm{T}=\frac{\mathrm{T}-16}{2}+16 \\ & \frac{3 \mathrm{~T}}{2}=28 \\ & \mathrm{~T}=\frac{56}{3} \mathrm{C} \end{aligned}$

Step 1: Find the Work Done ($\mathrm{W}$) in the Cycle

The work done in a cyclic process shown in a pressure-volume (P-V) graph is equal to the area enclosed by the cycle. For a circle, this area is $\frac{1}{2} \pi R^2$, where $R$ is the radius of the circle.

Step 2: Put in the Given Values

Radius $R$ on the graph is $ R = \frac{200}{2} \times 10^3 \text{ (for Pressure, P)} \\ R = \frac{200}{2} \times 10^{-6} \text{ (for Volume, V)} $

Step 3: Calculate the Area (Work Done)

$ \mathrm{W} = \frac{1}{2} \times \pi \times \left(\frac{200}{2} \times 10^3\right) \times \left(\frac{200}{2} \times 10^{-6}\right) $

First, calculate $\frac{200}{2} = 100$, so:

$ \mathrm{W} = \frac{1}{2} \times \pi \times 100 \times 10^3 \times 100 \times 10^{-6} $

Multiplying $100 \times 100 = 10,000$, and $10^3 \times 10^{-6} = 10^{-3}$, so:

$ \mathrm{W} = \frac{1}{2} \times \pi \times 10,000 \times 10^{-3} = \frac{1}{2} \times \pi \times 10 = 5\pi~\mathrm{J} $

Step 4: Relate to Heat Exchanged

In a complete cycle, the change in internal energy is zero, so total heat exchanged is equal to the total work done.

So, the magnitude of heat exchanged is $5\pi~\mathrm{J}$.

Let's analyze both statements step by step.

Assertion (A): "In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases."

 • In an adiabatic process, no heat is exchanged with the surroundings.

 • For a reversible adiabatic (compression) process, the relationship between temperature and volume is given by

  $T V^{\gamma-1} = \text{constant},$

  where $\gamma$ is the heat capacity ratio (greater than 1).

 • If the volume is reduced from $V$ to $V/2$, then

  $T_{\text{final}} = T_{\text{initial}} \left(\frac{V}{V/2}\right)^{\gamma-1} = T_{\text{initial}}\,(2)^{\gamma-1}.$

 • Since $(2)^{\gamma-1} > 1,$ the final temperature is higher than the initial temperature.

 • Therefore, the assertion that the temperature decreases is incorrect.

Reason (R): "Free expansion of an ideal gas is an irreversible and an adiabatic process."

 • In a free expansion, the gas expands into a vacuum without doing any work on the surroundings.

 • Since the container is insulated, no heat is exchanged, making it adiabatic.

 • However, the process is inherently irreversible due to the spontaneous expansion and mixing of states.

 • Thus, this statement is true.

Since Assertion (A) is false and Reason (R) is true (but the reason does not explain the false assertion about temperature change during compression), the correct answer is:

Option C

(A) is false but (R) is true.

$\frac{C}{5}=\frac{F-32}{9} \Rightarrow C=\frac{5 F}{9}-\frac{160}{9}$

For an ideal gas, the equation of state is given by

$ pV = nRT. $

If during a process the pressure increases linearly with temperature (i.e., $p \propto T$), then from the ideal gas law (with a fixed amount of gas, $n$, and with the gas constant, $R$), it follows that

$ \frac{p}{T} = \frac{nR}{V}. $

Since $nR$ is constant, a linear relationship between $p$ and $T$ implies that $\frac{p}{T}$ remains constant throughout the process. This can only be true if the volume $V$ is constant. Hence, the process is isochoric.

For an isochoric (constant volume) process:

The work done by the gas is given by

$ W = \int p\, dV, $

and since $dV = 0$, we have $W = 0.$ (Statement A is true.)

The internal energy change for an ideal gas depends only on temperature:

$ \Delta U = nC_V\Delta T. $

If the temperature increases, then $\Delta U > 0.$ (Statement D is true.)

For an isochoric process, no expansion work is done, so any heat added goes entirely into increasing the internal energy. This means the heat added is not different from the change in internal energy (in fact, $Q = \Delta U$). (Statement B is false.)

Since the process is isochoric, the volume does not change (and certainly is not increased). (Statement C is false, and Statement E stating it is isochoric is true.)

Thus, the correct true statements are A, D, and E.

To determine the change in entropy when water is heated from temperature $T_1$ to $T_2$ (here, $T_\gamma$ is equivalent to $T_2$), we use the definition of entropy change for a reversible process:

$ \Delta S = \int_{T_1}^{T_2} \frac{dQ}{T} $

Since the water is being heated slowly, the process is reversible, and the heat added is given by:

$ dQ = mc\,dT $

where:

• $m$ is the mass of the water,

• $c$ is the specific heat capacity (given as $1\,\mathrm{J\,kg^{-1}\,K^{-1}}$).

Substitute $dQ$ into the integral:

$ \Delta S = \int_{T_1}^{T_2} \frac{mc\,dT}{T} = mc\, \int_{T_1}^{T_2} \frac{dT}{T}. $

Since $c = 1$, the equation simplifies to:

$ \Delta S = m \int_{T_1}^{T_2} \frac{dT}{T}. $

Evaluating the integral, we have:

$ \int_{T_1}^{T_2} \frac{dT}{T} = \ln{T}\Big|_{T_1}^{T_2} = \ln\left(\frac{T_2}{T_1}\right). $

Thus, the change in entropy is:

$ \Delta S = m \ln\left(\frac{T_2}{T_1}\right). $

Comparing with the options provided, the correct answer is:

Option D: $m \ln\left(\frac{T_2}{T_1}\right).$

To find the work done by an ideal gas along the path $ABCD$ on the given $P-V$ diagram, we calculate the work done on each segment individually and then sum them up.

Segment $AB$: The work is given by $\text{w}_{AB} = P_0 \times V_0$, contributing $P_0 V_0$.

Segment $BC$: No volume change occurs along this path, resulting in zero work: $\text{w}_{BC} = 0$.

Segment $CD$: The gas compresses, performing negative work: $\text{w}_{CD} = -(2P_0 \times 2V_0) = -4P_0 V_0$.

Adding the work done in each segment:

$ \begin{align*} \text{w}_{ABCD} &= \text{w}_{AB} + \text{w}_{BC} + \text{w}_{CD} \\ &= P_0 V_0 + 0 - 4P_0 V_0 \\ &= -3P_0 V_0 \end{align*} $

Thus, the total work done by the gas along the path $ABCD$ is $-3P_0 V_0$.

Let's analyze each statement step by step:

For (A): “Pressure varies inversely with volume of an ideal gas.”

At constant temperature (isothermal process), Boyle's law applies:

$PV = \text{constant}.$

Thus, pressure and volume have an inverse relationship.

Hence, (A) corresponds to the Isothermal process, which is (III).

For (B): “Heat absorbed goes partly to increase internal energy and partly to do work.”

In a constant pressure (isobaric) process, when heat is supplied, part of it is used to do work (expansion, for example) and the rest increases the internal energy.

Therefore, (B) matches with the Isobaric process, which is (IV).

For (C): “Heat is neither absorbed nor released by a system.”

This implies that the process is adiabatic (no heat exchange).

Thus, (C) matches with the Adiabatic process, which is (I).

For (D): “No work is done on or by a gas.”

Work done by a gas is given by $W = P\Delta V.$

If the volume does not change (constant volume), then no work is done.

So, (D) identifies with the Isochoric process, which is (II).

Putting it all together:

(A) → (III) Isothermal process

(B) → (IV) Isobaric process

(C) → (I) Adiabatic process

(D) → (II) Isochoric process

Thus, the correct answer is:

Option D

A-III, B-IV, C-I, D-II.

Let's determine the mass step by step:

The bullet is heated from 300 K to 600 K, so the increase in temperature is:

$\Delta T = 600\, \text{K} - 300\, \text{K} = 300\, \text{K}.$

The heat required to raise the temperature of the bullet (using the specific heat capacity of lead) is:

$Q_1 = m \times c \times \Delta T = m \times 125\, \text{J/kg K} \times 300\, \text{K} = 37500\, m \, \text{J},$

where $ m $ is the mass in kilograms.

After reaching 600 K, the bullet melts. The heat required for the phase change (melting) is given by:

$Q_2 = m \times L = m \times 2.5 \times 10^4\, \text{J/kg} = 25000\, m \, \text{J}.$

The total heat required for the process is:

$Q = Q_1 + Q_2 = 37500\, m + 25000\, m = 62500\, m \, \text{J}.$

We are given that the total heat is 625 J, so:

$62500\, m = 625.$

Solving for $ m $:

$m = \frac{625}{62500} = 0.01\, \text{kg}.$

Finally, converting the mass from kilograms to grams:

$0.01\, \text{kg} = 0.01 \times 1000\, \text{g} = 10\, \text{g}.$

Thus, the mass of the bullet is 10 grams. This corresponds to Option D.

Check Statement (A):

“Internal energy (U), volume (V), and mass (M) are extensive variables.”

Internal energy is extensive.

Volume is extensive.

Mass is extensive.

Hence, statement (A) is correct.

Check Statement (B):

“Pressure (P), temperature (T), and density (ρ) are intensive variables.”

Pressure is intensive.

Temperature is intensive.

Density is intensive.

Hence, statement (B) is correct.

Check Statement (C):

“Volume (V), temperature (T), and density (ρ) are intensive variables.”

Volume is not intensive; it is an extensive property.

(T) and (ρ) are indeed intensive.

Since volume is miscategorized here, (C) is incorrect.

Check Statement (D):

“Mass (M), temperature (T), and internal energy are extensive variables.”

Mass is extensive.

Temperature is not extensive (it is intensive).

Internal energy is extensive.

Because temperature is miscategorized, (D) is incorrect.

Only (A) and (B) are correct statements.

Therefore, the correct choice is:

Option B: (A) and (B) Only.

Let's analyze the problem step by step:

For a rigid diatomic molecule (without vibrational modes):

Degrees of freedom:

Translation: 3

Rotation: 2

Total: 5

The molar specific heat at constant volume is given by:

$C_v = \frac{5}{2}R$

At constant pressure, it is:

$C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$

Therefore, the ratio is:

$\gamma_1 = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} \approx 1.4$

For the diatomic molecule that also has vibrational modes (assuming the vibrational mode is fully excited):

Additional vibrational mode contributions:

Each vibrational mode contributes 2 degrees of freedom (one kinetic and one potential).

Therefore, the degrees of freedom become:

$3\ (\text{translation}) + 2\ (\text{rotation}) + 2\ (\text{vibration}) = 7$

The molar specific heat at constant volume now is:

$C_v = \frac{7}{2}R$

And at constant pressure:

$C_p = C_v + R = \frac{7}{2}R + R = \frac{9}{2}R$

Thus, the ratio is:

$\gamma_2 = \frac{C_p}{C_v} = \frac{\frac{9}{2}R}{\frac{7}{2}R} = \frac{9}{7} \approx 1.2857$

Comparison:

For the rigid molecule, $\gamma_1 \approx 1.4$.

For the molecule with vibrational modes, $\gamma_2 \approx 1.2857$.

Clearly, $\gamma_2 < \gamma_1$.

Therefore, the correct option is:

Option A: $\gamma_2 < \gamma_1$

To determine the energy radiated by two spherical bodies, both made of the same material but with different radii and temperatures, we can use the Stefan-Boltzmann law. This law states that the energy radiated by a black body per unit time is proportional to the product of the surface area $ A $ and the fourth power of its temperature $ T $.

The formula for power $ P $ is given by:

$ P = \sigma e A T^4 $

Where:

$ \sigma $ is the Stefan-Boltzmann constant

$ e $ is the emissivity of the material

$ A $ is the surface area of the sphere

$ T $ is the temperature in Kelvin

Expressing this relationship as a proportionality:

$ P \propto A T^4 $

For a sphere, the surface area $ A $ is:

$ A = 4\pi R^2 $

Hence, the power radiated is proportional to:

$ P \propto R^2 T^4 $

For two bodies with radii $ R_1 $ and $ R_2 $, and temperatures $ T_1 $ and $ T_2 $, the ratio of their powers is:

$ \frac{P_1}{P_2} = \left( \frac{R_1}{R_2} \right)^2 \left( \frac{T_1}{T_2} \right)^4 $

Given:

$ R_1 = 0.2 \, \text{m} $, $ T_1 = 800 \, \text{K} $

$ R_2 = 0.8 \, \text{m} $, $ T_2 = 400 \, \text{K} $

Substituting into the formula:

$ \frac{E}{E'} = \left( \frac{0.2}{0.8} \right)^2 \left( \frac{800}{400} \right)^4 $

Calculating each part:

$\left( \frac{0.2}{0.8} \right)^2 = \left( \frac{1}{4} \right)^2 = \frac{1}{16}$

$\left( \frac{800}{400} \right)^4 = 2^4 = 16$

Combining these calculations:

$ \frac{E}{E'} = \frac{1}{16} \times 16 = 1 $

This implies that the energy radiated by the bigger body $ E' $ equals the energy radiated by the smaller body $ E $. Thus, $ E' = E $.

$\Delta {Q_1} = m{s_i}\Delta {T_2},\,\Delta {Q_2} = m{L_i}$

$\Delta {Q_3} = m{S_w}\Delta {T_3},\,\Delta {Q_4} = m{L_s},\,\Delta {Q_5} = m{S_s}\Delta {T_5}$

$\Delta {Q_{net}} = m({S_i}\Delta {T_1} + {L_i} + {S_w}\Delta {T_3} + {L_s} + {S_s}\Delta {T_s})$

$ = {10^{ - 3}}(2100 \times 10 + 335000 + 4180 \times 100 + 2250000 + 1920 \times 10)$

$ = {10^{ - 3}}(3043200) = 3043.2\,J$

So, $\Delta {Q_{net}} \approx 3043\,J$

Carnot Engine:

Hot reservoir temperature ($T_H$) = 1000 K

Efficiency ($\eta$) = 0.4

Heat extracted from the hot reservoir ($Q_H$) = 150 J per cycle

Heat Pump:

Coefficient of Performance (COP) = 10

Work input ($W_{pump}$) = Work output from the engine ($W_{engine}$)

Hot reservoir temperature ($T_{H,hp}$) = 300 K

Let's analyze each statement:

Statement A: Work extracted from the Carnot engine in one cycle is 60 J.

The efficiency of an engine is defined as the ratio of the work output to the heat input from the hot reservoir.

$ \eta = \frac{W_{engine}}{Q_H} $

We are given $\eta = 0.4$ and $Q_H = 150$ J. We can find the work done by the engine:

$ W_{engine} = \eta \times Q_H = 0.4 \times 150 \text{ J} = 60 \text{ J} $

This statement is correct.

Statement B: Temperature of the cold reservoir of the Carnot engine is 600 K.

For a Carnot engine, the efficiency can also be expressed in terms of the temperatures of the hot and cold reservoirs:

$ \eta = 1 - \frac{T_C}{T_H} $

where $T_C$ is the temperature of the cold reservoir and $T_H$ is the temperature of the hot reservoir (both in Kelvin).

We know $\eta = 0.4$ and $T_H = 1000$ K. We can solve for $T_C$:

$ 0.4 = 1 - \frac{T_C}{1000} $

$ \frac{T_C}{1000} = 1 - 0.4 = 0.6 $

$ T_C = 0.6 \times 1000 \text{ K} = 600 \text{ K} $

This statement is also correct.

Statement C: Temperature of the cold reservoir of the heat pump is 270 K.

The coefficient of performance (COP) for a heat pump is defined as the ratio of the heat delivered to the hot reservoir ($Q_{H,hp}$) to the work input ($W_{pump}$). For an ideal (Carnot) heat pump operating between temperatures $T_{C,hp}$ and $T_{H,hp}$, the COP is also given by:

$ \text{COP} = \frac{T_{H,hp}}{T_{H,hp} - T_{C,hp}} $

We are given the COP of the heat pump is 10 and its hot reservoir temperature $T_{H,hp} = 300$ K. Let's assume this heat pump is operating ideally with this COP. We can find $T_{C,hp}$:

$ 10 = \frac{300 \text{ K}}{300 \text{ K} - T_{C,hp}} $

Let's solve for $T_{C,hp}$:

$ 10 \times (300 - T_{C,hp}) = 300 $

$ 3000 - 10 T_{C,hp} = 300 $

$ 10 T_{C,hp} = 3000 - 300 = 2700 $

$ T_{C,hp} = \frac{2700}{10} \text{ K} = 270 \text{ K} $

This statement is correct, assuming the heat pump is ideal.

Statement D: Heat supplied to the hot reservoir of the heat pump in one cycle is 540 J.

The COP of a heat pump is also defined as the ratio of the heat delivered to the hot reservoir ($Q_{H,hp}$) to the work input ($W_{pump}$).

$ \text{COP} = \frac{Q_{H,hp}}{W_{pump}} $

We know the COP is 10 and the work input to the heat pump is the work output from the engine, which we found to be $W_{pump} = W_{engine} = 60$ J.

So, the heat supplied to the hot reservoir of the heat pump is:

$ Q_{H,hp} = \text{COP} \times W_{pump} = 10 \times 60 \text{ J} = 600 \text{ J} $

This statement says the heat supplied is 540 J, which does not match our calculation of 600 J.

This statement is incorrect.

Based on our analysis, statements A, B, and C are correct.

Fractional change in time period

$ \begin{aligned} \frac{\Delta T}{T} & =\frac{1}{2} \alpha \Delta T \\ & =\frac{1}{2} \times 12 \times 10^{-6}\left(/{ }^{\circ} \mathrm{C}\right) \times 15^{\circ} \mathrm{C} \\ & =90 \times 10^{-6} \end{aligned} $

∴ The change in time per day

$ \Delta t_{\text {day }}=\left(90 \times 10^{-6}\right) \times 86400 \approx 7.8 \mathrm{~s} $

Since, the temperature increases, the length of the pendulum will increase due to thermal expansion. An increase in the length of the pendulum increases its time period, meaning the clock will tick slower and thus lose time.

The change in length due to change of temperature,

$ \begin{aligned} \Delta L & =L \alpha \Delta T \\ \Rightarrow \quad \Delta T & =\frac{\Delta L}{L \alpha}=\frac{0.5 \times 10^{-3}}{1 \times 10^{-5}} \\ & =0.5 \times 10^2=50^{\circ} \mathrm{C} \end{aligned} $

This means the temperature can vary by $\pm 50^{\circ} \mathrm{C}$ from the calibration temperature.

$ \begin{aligned} \therefore T_{\text {lower }} & =T_0-\Delta T \\ & =25^{\circ} \mathrm{C}-50^{\circ} \mathrm{C}==-25^{\circ} \mathrm{C} \\ T_{\text {upper }} & =T_0+\Delta T=25^{\circ} \mathrm{C}+50^{\circ} \mathrm{C}=75^{\circ} \mathrm{C} \end{aligned} $

For an adiabatic process,

$ \begin{aligned} & T V^{\gamma-1}=\text { constant } \\ & \begin{aligned} T_1\left[V_1\right]^{\gamma-1} & =T_2\left[V_2\right]^{\gamma-1} \\ T_1[V]^{\gamma-1} & =T_2[32 V]^{\gamma-1} \\ T_1\left[\frac{1}{32}\right]^{\gamma-1} & =T_2 \\ T_1\left[\frac{1}{32}\right]^{\frac{7}{5}-1} & =T_2 \\ {[\because \gamma} & \left.=\frac{7}{5} \text { for diatomic gas }\right] \\ T_1\left[\frac{1}{32}\right]^{\frac{2}{5}} & =T_2 \\ \frac{1}{4} & =\frac{T_2}{T_1} \end{aligned} \end{aligned} $

Efficiency of Carnot engine is given by,

$ \begin{aligned} &\eta=1-\frac{T_2}{T_1}=1-\frac{1}{4}=\frac{3}{4}=0.75\\ &\text { Percentage efficiency }=75 \% \end{aligned} $

$ \begin{aligned} &\text { Average translational kinetic energy }\\ &\begin{aligned} & E_{\text {avg }}=\frac{3}{2} K T \\ \Rightarrow & E_{\text {avg }} \propto T \\ \Rightarrow & \frac{\left(E_{\text {avg }}\right)_{\mathrm{H}_2}}{\left(E_{\text {avg }}\right)_{\mathrm{O}_2}}=\frac{T_{\mathrm{H}_2}}{T_{\mathrm{O}_2}}=\frac{T}{T}=1 \text { or } 1: 1 \end{aligned} \end{aligned} $

On combining of drop, surface area decreases, hence, heat get released,

$ \begin{aligned} & \qquad \begin{aligned} v_f & =v_i=v \\ \frac{4}{3} \pi R^3 & =n \times \frac{4}{3} \pi r^3=v \end{aligned} \\ & \Rightarrow \quad R=n^{1 / 3} r \\ & \text { Heat released, } H=S\left[A_i-A_f\right] \\ & \\ & =S\left[4 \pi r^2 \times n-4 \pi R^2\right] \\ & = \\ & =3 S\left[\frac{4 \pi r^3 \times n}{3 r}-\frac{4 \pi R^3}{3 R}\right] \\ & =3 S\left[\frac{V}{r}-\frac{V}{R}\right]=3 S V\left[\frac{1}{r}-\frac{n^{-1 / 3}}{r}\right] \\ & H=\frac{35 V}{r}\left[1-n^{-1 / 3}\right] \end{aligned} $