Second Law of Thermodynamics and Entropy Questions in English

(B) In thermodynamics,a state function is a property whose value depends only on the current state of the system and not on the path taken to reach that state.
Internal energy $(U)$ and entropy $(S)$ are both state functions because they are defined by the macroscopic variables of the system (such as pressure,volume,and temperature) at a given equilibrium state.
Therefore,the statement that internal energy and entropy are state functions is correct.

(D) The change in entropy $(dS)$ for a reversible process is defined as $dS = \frac{dQ_{rev}}{T}$.
For an adiabatic reversible process,the heat exchange $dQ_{rev} = 0$,which implies $dS = 0$.
Among the given options,the conversion of work into heat isochorically (if done reversibly) or any process that is internally reversible and adiabatic results in no change in entropy.
However,in the context of standard thermodynamic problems,a reversible process is defined as an isentropic process $(dS = 0)$.
Option $(d)$ represents a process where work is converted to heat; if this is performed reversibly,the entropy of the system remains constant.

(D) Entropy is defined as a measure of the degree of disorder or randomness of a system.
In thermodynamics,it represents the thermal energy per unit temperature that is unavailable for doing useful work.
Since work is derived from ordered molecular motion,higher entropy corresponds to higher molecular disorder within the system.

(A) The statement provided is known as the Clausius statement of the Second Law of Thermodynamics.
It asserts that it is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body.
Therefore,the correct option is $A$.

(B) The process involves two parts: the melting of ice and the heat loss from the water.
$1$. Entropy change of ice $({\Delta S}_{ice})$: The ice melts at $T = 0^{\circ}C = 273 \, K$. The heat absorbed is $Q = mL = 100 \, g \times 80 \, cal/g = 8000 \, cal$.
${\Delta S}_{ice} = \frac{Q}{T} = \frac{8000}{273} \approx 29.30 \, cal/K$.
$2$. Entropy change of water $({\Delta S}_{water})$: The water loses the same amount of heat at $T = 50^{\circ}C = 323 \, K$.
${\Delta S}_{water} = -\frac{Q}{T} = -\frac{8000}{323} \approx -24.77 \, cal/K$.
$3$. Total entropy change $({\Delta S}_{total})$: ${\Delta S}_{total} = {\Delta S}_{ice} + {\Delta S}_{water} = 29.30 - 24.77 = +4.53 \, cal/K \approx +4.5 \, cal/K$.

(B) The heat $Q$ required to melt $1\, kg$ of ice at $0^\circ C$ $(273\, K)$ to water at $0^\circ C$ is given by $Q = m \cdot L$.
Given $m = 1\, kg = 1000\, g$ and $L = 80\, cal/g$.
$Q = 1000\, g \times 80\, cal/g = 80,000\, cal = 8 \times 10^4\, cal$.
The change in entropy $\Delta S$ for an isothermal process is given by $\Delta S = \frac{Q}{T}$.
Here,$T = 0^\circ C = 273\, K$.
$\Delta S = \frac{80,000\, cal}{273\, K} \approx 293\, cal/K$.

(A) Entropy is a state function,which means its change depends only on the initial and final states of the system,not on the path taken to reach the final state.
The formula for the change in entropy $(\Delta S)$ for a body of constant heat capacity $C$ heated from temperature $T_i$ to $T_f$ is given by:
$\Delta S = \int_{T_i}^{T_f} \frac{dQ}{T} = \int_{T_i}^{T_f} \frac{C dT}{T} = C \ln\left(\frac{T_f}{T_i}\right)$.
Given $C = 1 \ J/^oC$,$T_i = 100^oC = 373 \ K$,and $T_f = 200^oC = 473 \ K$. However,in thermodynamic problems of this type,temperatures are often treated in Celsius if the ratio is defined by the problem context or if the change is independent of the scale. Using the absolute temperature ratio:
$\Delta S = 1 \cdot \ln\left(\frac{200+273}{100+273}\right) = \ln\left(\frac{473}{373}\right)$.
Since the initial and final temperatures are the same in both cases $(i)$ and $(ii)$,the entropy change of the body is identical in both scenarios.
Note: If the problem implies the ratio of temperatures as $200/100 = 2$,then $\Delta S = \ln(2)$. Thus,the entropy change in both cases is $\ln(2)$.

(D) Entropy $(S)$ is a state function, meaning the change in entropy $(\Delta S)$ between two states depends only on the initial and final states, not the path taken. Thus, option $(A)$ is true.
Entropy is defined as a quantitative measure of the degree of disorder or randomness in a system. Thus, option $(B)$ is true.
A Carnot engine operates in a reversible cycle. Since entropy is a state function, the change in entropy over a complete cycle $(\Delta S_{cycle})$ must be zero. Thus, option $(C)$ is true.
According to the Second Law of Thermodynamics, for any process occurring in an isolated system, the total entropy change must be greater than or equal to zero $(\Delta S \ge 0)$. Therefore, the entropy of an isolated system can never decrease; it can only increase (for irreversible processes) or remain constant (for reversible processes). Thus, option $(D)$ is false.

(C) The change in entropy is given by the formula $dS = \frac{dQ}{T}$.
In the process of freezing water into ice,heat is released by the system to the surroundings,meaning $dQ$ is negative.
Since the temperature $T$ is positive,the change in entropy $dS$ is negative.
Therefore,the entropy of the water decreases during the formation of ice cubes.

(B) According to the $Second$ $Law$ of $Thermodynamics$,for any spontaneous process in an isolated system,the entropy of the system must increase $(dS > 0)$.
An isolated system is defined as a system that cannot exchange energy (heat or work) or matter with its surroundings.
Since an adiabatic process is defined as a process where there is no exchange of heat $(dQ = 0)$ between the system and the surroundings,all processes occurring within an isolated system are indeed adiabatic.
However,the fact that the processes are adiabatic does not directly explain why the entropy must increase; the increase in entropy is a consequence of the $Second$ $Law$ of $Thermodynamics$ regarding spontaneous processes.
Therefore,both statements are correct,but the $Reason$ is not the correct explanation for the $Assertion$.

(B) The entropy change of a system is given by $dS = \frac{dQ}{T}$. When a body cools,it loses heat,so $dQ$ is negative. Since $T$ is positive,$dS$ is negative,meaning the entropy of the milk decreases. Thus,the $Assertion$ is correct.
The second law of thermodynamics states that the total entropy of the universe (system + surroundings) must increase for any spontaneous process. The cooling of a hot object in a room involves an increase in the entropy of the surroundings that is greater than the decrease in the entropy of the system,so it does not violate the second law. Thus,the $Reason$ is also correct.
However,the $Reason$ explains why the process is possible,not why the entropy of the milk itself decreases. Therefore,the $Reason$ is not the correct explanation of the $Assertion$.

(N/A) The reason for both phenomena is the $Second$ $Law$ of $Thermodynamics$,which states that heat naturally flows from a body at a higher temperature to a body at a lower temperature until thermal equilibrium is reached.
$1$. In the case of the cup of ice water,the surrounding air (at a higher temperature) transfers heat to the ice water (at a lower temperature),causing the ice water to heat up.
$2$. In the case of the cup of hot tea,the tea (at a higher temperature) transfers heat to the surrounding air (at a lower temperature),causing the tea to cool down.
In both scenarios,the system and the surroundings are moving toward thermal equilibrium with the ambient temperature.

(N/A) heat engine operates by taking heat $Q_H$ from a high-temperature reservoir,performing work $W$,and rejecting the remaining heat $Q_L$ to a low-temperature sink. According to the Second Law of Thermodynamics,specifically the Kelvin-Planck statement,it is impossible for any device operating in a cycle to receive heat from a single reservoir and convert it entirely into work. For an engine to be $100\%$ efficient,it would require $Q_L = 0$,which implies that all heat absorbed is converted into work. This would violate the Second Law of Thermodynamics,as some energy must always be dissipated to the surroundings to maintain the entropy of the universe. Therefore,the efficiency $\eta = 1 - (Q_L / Q_H)$ is always less than $1$.

(N/A) The efficiency of a heat engine is defined as $\eta = 1 - \frac{Q_2}{Q_1}$,where $Q_1$ is the heat absorbed from the source and $Q_2$ is the heat rejected to the sink.
According to the Second Law of Thermodynamics (Kelvin-Planck statement),it is impossible to construct a device that operates in a cycle and produces no effect other than the extraction of heat from a single reservoir and the performance of an equivalent amount of work.
For the efficiency to be $100 \%$,$Q_2$ must be $0$,meaning all heat absorbed from the source is converted into work.
This would require the sink temperature to be absolute zero $(0 \ K)$,which is unattainable according to the Third Law of Thermodynamics.
Therefore,some amount of heat must always be rejected to the sink,making $100 \%$ efficiency impossible.

(N/A) An example of this is given below.
Nobody has ever seen a book lying on a table jumping to a height by itself. Such a phenomenon would be consistent with the first law of thermodynamics (principle of conservation of energy) because the table could cool spontaneously,converting some of its internal energy into an equal amount of mechanical energy for the book,which would then hop to a height with potential energy equal to the mechanical energy it acquired. However,this never happens in reality.
This implies that there is another principle that disallows many phenomena consistent with the first law of thermodynamics,which is known as the second law of thermodynamics.
The second law of thermodynamics provides a fundamental limitation to the efficiency of a heat engine and the coefficient of performance of a refrigerator.
$1$. For a heat engine,the efficiency $\eta = 1 - \frac{Q_2}{Q_1}$ can never be unity (i.e.,$100 \%$),because the heat released $Q_2$ to the cold reservoir (sink) can never be zero. This means the absorbed heat $Q_1$ cannot be converted completely into work.
$2$. For a refrigerator,the second law states that the coefficient of performance $\alpha = \frac{Q_2}{W}$ can never be infinite. Since external work $W$ can never be zero,$\alpha$ remains finite.
Based on these observations,Kelvin-Planck and Clausius formulated statements for the second law of thermodynamics:
$(i)$ Kelvin-Planck Statement: No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work.
$(ii)$ Clausius Statement: No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.

(A) The second law of thermodynamics,specifically the Kelvin-Planck statement,asserts that it is impossible for any device that operates on a thermodynamic cycle to receive heat from a single thermal reservoir and produce a net amount of work.
This implies that a heat engine must reject some amount of heat to a cold reservoir (sink) to complete the cycle.
Consequently,no heat engine can have an efficiency of $100\%$,as some energy must always be dissipated as waste heat.
Therefore,the second law of thermodynamics imposes the limitation that a heat engine cannot convert all the heat supplied to it into work.

(N/A) The Kelvin-Planck statement of the second law of thermodynamics states that:
It is impossible for any device that operates on a thermodynamic cycle to receive heat from a single thermal reservoir and produce a net amount of work.
In simpler terms,no heat engine can have a thermal efficiency of $100\%$.
This implies that for a heat engine to operate,it must exchange heat with at least two reservoirs at different temperatures.

(N/A) The Clausius statement of the second law of thermodynamics states that it is impossible to construct a device operating in a cycle that produces no effect other than the transfer of heat from a body at a lower temperature to a body at a higher temperature without the input of external work.

(B) No,the whole of heat cannot be converted into work. According to the second law of thermodynamics (specifically the Kelvin-Planck statement),it is impossible for any device that operates on a thermodynamic cycle to receive heat from a single thermal reservoir and produce a net amount of work. Some amount of heat must always be rejected to a cold reservoir (sink) to complete the cycle.

(N/A) The second law of thermodynamics can be stated through two primary perspectives:
$1$. Kelvin-Planck Statement: It is impossible for any device that operates on a thermodynamic cycle to receive heat from a single reservoir and produce a net amount of work. In other words,$100\%$ conversion of heat into work is impossible.
$2$. Clausius Statement: It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body without the application of external work.

(B) The change in entropy for an isobaric process is given by the formula:
$\Delta S = n C_{P} \ln \left( \frac{T_{2}}{T_{1}} \right)$
For a diatomic gas like $N_{2}$, the molar heat capacity at constant pressure is $C_{P} = \frac{7}{2} R$.
Given: $n = 1\, \text{mole}$, $T_{1} = 300\, K$, $T_{2} = 600\, K$, and $R \approx 8.314\, J/(mol \cdot K)$.
Substituting the values:
$\Delta S = 1 \times \frac{7}{2} \times 8.314 \times \ln \left( \frac{600}{300} \right)$
$\Delta S = 3.5 \times 8.314 \times \ln(2)$
Using $\ln(2) \approx 0.693$:
$\Delta S = 3.5 \times 8.314 \times 0.693 \approx 20.16\, J/K$.
Rounding to the nearest integer, the change in entropy is $20\, J/K$.

(D) Entropy is an extensive property of a thermodynamic system.
An extensive property is a property whose value depends on the quantity or size of the matter present in the system.
Since entropy is additive,the total entropy of the system when the two parts are combined by removing the piston is the sum of the entropies of the individual parts.
Therefore,the total entropy $S_{\text{total}} = S_{1} + S_{2}$.

(C) The temperature $T$ of a system is defined by the relation $\frac{1}{T} = \frac{dS}{dU}$,which is the slope of the entropy versus energy graph.
From the given graphs,at the initial states $U_{1, i}$ and $U_{2, i}$,the slope of the graph for system $1$ is steeper than the slope of the graph for system $2$. Therefore,$\left( \frac{dS}{dU} \right)_1 > \left( \frac{dS}{dU} \right)_2$,which implies $\frac{1}{T_1} > \frac{1}{T_2}$,or $T_1 < T_2$.
Since system $2$ is at a higher temperature than system $1$,heat will flow from system $2$ to system $1$ until they reach thermal equilibrium. As a result,the internal energy $U_1$ of system $1$ increases and the internal energy $U_2$ of system $2$ decreases.
According to the second law of thermodynamics,for any spontaneous process in an isolated system (the combined system $1+2$ is isolated),the total entropy must increase until it reaches a maximum at equilibrium. Thus,the total entropy increases.

(C) Entropy mein parivartan ka sutra $\Delta S = \int \frac{dQ}{T}$ hai.
Tambe ke liye,entropy mein parivartan $\Delta S_{Cu} = \int_{T_i}^{T_f} \frac{mc dT}{T} = mc \ln(\frac{T_f}{T_i})$ hai,jo ki negative hai (tamba entropy khota hai).
Talab ke liye,usne $Q = mc(T_i - T_f)$ ushma prapt ki hai. Talab ka taapman $T_{pond} = 30^{\circ} C$ sthir rehta hai,isliye talab dwara prapt entropy $\Delta S_{pond} = \frac{Q}{T_{pond}} = \frac{mc(100 - 30)}{303.15}$ hai.
Kyuki yeh ek anutkramniya (irreversible) prakriya hai,isliye kul entropy mein vriddhi honi chahiye $(\Delta S_{total} > 0)$.
Isliye,$\Delta S_{pond} + \Delta S_{Cu} > 0$,jiska arth hai ki $\Delta S_{pond} > |\Delta S_{Cu}|$.
Atah,talab tambe dwara khoyi gayi entropy se adhik entropy prapt karta hai.

(B) The change in entropy $(dS)$ for a reversible process is given by the formula $dS = \frac{dQ}{T}$,where $dQ$ is the heat exchanged and $T$ is the absolute temperature.
If heat is supplied to the system,$dQ > 0$,so $dS > 0$ (entropy increases).
If heat is taken out from the system,$dQ < 0$,so $dS < 0$ (entropy decreases).
Therefore,the entropy of a system decreases when heat is removed from it at a constant temperature.

(D) The entropy change $(\Delta S)$ of a system depends only on its initial and final states, as entropy is a state function.
The formula for the change in entropy of a body with constant heat capacity $C$ is given by:
$\Delta S = \int_{T_i}^{T_f} \frac{dQ}{T} = \int_{T_i}^{T_f} \frac{C dT}{T} = C \ln\left(\frac{T_f}{T_i}\right)$
Given:
$C = 1\,J/^{\circ}C$
$T_i = 100^{\circ}C = 373.15\,K$
$T_f = 200^{\circ}C = 473.15\,K$
Since the initial and final temperatures are the same in both cases $(i)$ and $(ii)$, the entropy change for the body will be identical in both scenarios.
$\Delta S = 1 \cdot \ln\left(\frac{200+273.15}{100+273.15}\right) \approx \ln(1.27)$
However, in the context of idealized problems where temperatures are often treated as absolute values or ratios of Kelvin, the change is simply $\ln(T_f/T_i)$. Since $T_f/T_i$ is constant in both cases, the entropy change is the same: $\ln 2, \ln 2$ (assuming the ratio simplifies to $2$ in the problem's specific context).
Therefore, the correct option is $(d)$.

(C) The change in entropy of the surroundings is given by $\Delta S_{\text{surr}} = -\frac{q_{\text{sys}}}{T}$.
Since the gas expands against a constant external pressure $P_{\text{ext}}$,the work done by the gas is $W = -P_{\text{ext}}(V_2 - V_1)$.
According to the first law of thermodynamics,$\Delta U = q + W$. For an isothermal process of an ideal gas,$\Delta U = 0$,so $q = -W = P_{\text{ext}}(V_2 - V_1)$.
Given $P_{\text{ext}} = 3.0 \ atm$,$V_1 = 1.0 \ L$,$V_2 = 2.0 \ L$,and $T = 300 \ K$.
$q = 3.0 \ atm \times (2.0 \ L - 1.0 \ L) = 3.0 \ L \ atm$.
Converting to Joules: $q = 3.0 \times 101.3 \ J = 303.9 \ J$.
The heat absorbed by the surroundings is $-q = -303.9 \ J$.
Therefore,$\Delta S_{\text{surr}} = \frac{-303.9 \ J}{300 \ K} = -1.013 \ J \ K^{-1}$.

(D) The heat supplied to the water is given by $dQ = m s dT$,where $m$ is the mass and $s$ is the specific heat capacity.
Given $s = 1 \ J \ g^{-1} \ K^{-1}$.
The change in entropy $dS$ is defined as $dS = \frac{dQ}{T}$.
Substituting $dQ$,we get $dS = \frac{m s dT}{T}$.
To find the total change in entropy $\Delta S$,we integrate from $T_1$ to $T_2$:
$\Delta S = \int_{T_1}^{T_2} \frac{m s dT}{T} = m s \int_{T_1}^{T_2} \frac{dT}{T}$.
$\Delta S = m s \ln \left(\frac{T_2}{T_1}\right)$.
Since $s = 1$,the change in entropy is $\Delta S = m \ln \left(\frac{T_2}{T_1}\right)$.

(D) The statement "Heat cannot flow by itself from a body at a lower temperature to a body at a higher temperature" is known as the Clausius statement of the second law of thermodynamics.
It implies that heat transfer from a colder body to a hotter body requires external work to be performed on the system.

(D) The coefficient of performance $(COP)$ of a refrigerator is defined as the ratio of the heat extracted from the cold reservoir $(Q_L)$ to the work input $(W)$.
$COP = \frac{Q_L}{W} = \frac{Q_L}{Q_H - Q_L}$.
According to the second law of thermodynamics,heat cannot spontaneously flow from a colder body to a hotter body without the expenditure of external work.
This law imposes a theoretical upper limit on the efficiency of heat engines and the $COP$ of refrigerators (Carnot limit).
Therefore,the fundamental limitation to the $COP$ of a refrigerator is governed by the second law of thermodynamics.

(D) The entropy of a system during a phase change (phase transition) is determined by the heat exchange involved.
Entropy change is given by $\Delta S = \frac{Q}{T}$,where $Q$ is the heat exchanged and $T$ is the absolute temperature.
During melting or vaporization,heat is absorbed by the system $(Q > 0)$,so entropy increases.
During freezing or condensation,heat is released by the system $(Q < 0)$,so entropy decreases.
Therefore,during a phase change,the entropy of the system may increase or decrease depending on the direction of the transition.