Reversible and Irreversible process , Free Expansion Questions in English

(A) In a free expansion process,the work done by the gas is $W = 0$ and the heat exchanged is $Q = 0$. According to the first law of thermodynamics,$\Delta U = Q - W = 0$. For a Van der Waals gas,the internal energy $U$ depends on both temperature $T$ and volume $V$. As the gas expands,the intermolecular forces of attraction perform negative work internally,which leads to a decrease in the kinetic energy of the molecules. Consequently,the final temperature of a Van der Waals gas after free expansion is less than the initial temperature.

(A) This is a case of free expansion of an ideal gas into a vacuum.
In free expansion,the gas does no work because it expands against zero external pressure,so $\Delta W = 0$.
Since the system is thermally isolated from its surroundings,there is no heat exchange,so $\Delta Q = 0$.
According to the first law of thermodynamics,$\Delta U = \Delta Q - \Delta W$.
Substituting the values,$\Delta U = 0 - 0 = 0$.
For an ideal gas,the internal energy $U$ is a function of temperature only $(U \propto T)$.
Since $\Delta U = 0$,the change in temperature $\Delta T$ must also be $0$.
Therefore,the final temperature remains the same as the initial temperature,which is $300 \ K$.

(A) During free expansion of a perfect gas,no work is done $(W = 0)$ because the external pressure is zero.
Also,no heat is exchanged with the surroundings $(Q = 0)$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
Since $Q = 0$ and $W = 0$,the change in internal energy $\Delta U = 0$.
For a perfect gas,internal energy depends only on temperature $(U \propto T)$.
Since internal energy remains constant,the temperature of the gas remains constant.

(A) Free expansion is an adiabatic process where a gas expands into a vacuum.
Since the gas expands into a vacuum,there is no external pressure to work against,so the work done by the gas is $W = 0$.
Because the process is adiabatic,there is no heat exchange with the surroundings,so $Q = 0$.
According to the first law of thermodynamics,$\Delta E_{\text{int}} = Q - W$.
Substituting the values,$\Delta E_{\text{int}} = 0 - 0 = 0$.
Therefore,the correct option is $A$.

(C) This is a case of free expansion of an ideal gas into a vacuum.
For an ideal gas,the internal energy $U$ depends only on temperature $T$.
In free expansion,the gas does no work against an external pressure because the external pressure is zero $(W = 0)$.
Since the container is thermally insulated,there is no heat exchange with the surroundings $(Q = 0)$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
Substituting the values,$\Delta U = 0 - 0 = 0$.
Since the internal energy remains constant $(\Delta U = 0)$,the temperature of the ideal gas remains the same.

(C) When a gas cylinder explodes suddenly,the process is extremely rapid,meaning there is no time for heat exchange with the surroundings,which characterizes an adiabatic process.
Since the process occurs spontaneously and rapidly against external atmospheric pressure,it is irreversible.
According to the first law of thermodynamics,$dQ = dU + dW$. For an adiabatic process,$dQ = 0$,so $dU = -dW$.
As the gas expands rapidly,it performs work on the surroundings $(dW > 0)$,which leads to a decrease in internal energy $(dU < 0)$.
Since the internal energy of an ideal gas is directly proportional to its temperature,a decrease in internal energy results in a fall in temperature.

(B) The process in which a gas or liquid at a higher pressure $P_1$ flows into a region of lower pressure $P_2$ through a porous plug or a narrow opening,without any significant change in kinetic energy,is known as the Joule-Thomson expansion.
This expansion occurs spontaneously and involves a pressure drop without performing external work,which leads to an increase in the entropy of the system and its surroundings.
Therefore,the Joule-Thomson expansion is inherently an irreversible process.

(D) reversible process is defined as a process that can be reversed without leaving any trace on the surroundings.
For a process to be reversible,it must satisfy two main conditions:
$1$. The process must be quasistatic (infinitely slow),ensuring the system remains in thermal and mechanical equilibrium with its surroundings at every stage.
$2$. There should be no dissipative effects such as friction,viscosity,or electrical resistance,meaning the loss of energy must be zero.
Therefore,option $(d)$ is the most comprehensive answer as it covers both the absence of energy loss and the requirement for the process to be quasistatic.
Mathematically,for a reversible cycle,$\oint \frac{dQ}{T} = 0$.

(D) process is reversible if it occurs infinitely slowly such that the system remains in thermodynamic equilibrium with its surroundings at every stage.
$(a)$ Heat transfer by radiation involves a temperature gradient and is irreversible.
$(b)$ Electrical heating (Joule heating) is irreversible due to the dissipation of energy as heat.
$(c)$ Heat transfer by conduction requires a temperature difference and is irreversible.
$(d)$ An isothermal compression of an ideal gas,if performed infinitely slowly,allows the system to remain in equilibrium,making it a reversible process.

(A) The Joule effect,which describes the production of heat when an electric current flows through a conductor $(H = I^2Rt)$,is an irreversible process because the heat produced is always positive regardless of the direction of the current.
In contrast,the Peltier effect,Seebeck effect,and Thomson effect are all reversible thermoelectric phenomena where the direction of heat exchange changes with the direction of the electric current.

(D) This process is known as free expansion.
In free expansion,the gas expands into a vacuum,so the external pressure $P_{ext} = 0$.
Therefore,the work done $W = \int P_{ext} dV = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.
Since the process is adiabatic,$\Delta Q = 0$.
Substituting these values,we get $0 = \Delta U + 0$,which implies $\Delta U = 0$.
For an ideal gas,internal energy $U$ is a function of temperature only $(U \propto T)$.
Since $\Delta U = 0$,the change in temperature $\Delta T = 0$,meaning the temperature remains constant.
Thus,both the heat supplied is zero and the temperature remains constant.

(A) This process is known as free expansion of an ideal gas into a vacuum.
Since the walls are insulating,the heat exchange with the surroundings is zero $(Q = 0)$.
Since the gas expands into a vacuum,the work done by the gas is zero $(W = 0)$.
According to the first law of thermodynamics,$\Delta U = Q - W = 0 - 0 = 0$.
For an ideal gas,internal energy $U$ depends only on temperature. Since $\Delta U = 0$,the temperature remains constant,so $T_2 = T$.
Using the ideal gas law for the initial and final states: $P_1 V_1 = P_2 V_2$.
Initially,the gas occupies volume $V_1 = V$ at pressure $P_1 = P$.
After opening the valve,the gas occupies the total volume $V_2 = 2V$.
Substituting these values: $P \cdot V = P_2 \cdot (2V)$.
Therefore,$P_2 = P/2$.

(A) The process described is free expansion of an ideal gas into a vacuum.
Work done by a gas is given by the formula:
$W = \int P_{ext} dV$
Since the gas expands into a vacuum,the external pressure $P_{ext} = 0$.
Therefore,the work done is:
$W = \int 0 \cdot dV = 0$
Since the container is fully insulated,there is no heat exchange with the surroundings,so $Q = 0$.
According to the first law of thermodynamics,$\Delta U = Q - W = 0 - 0 = 0$.
For an ideal gas,internal energy $U$ depends only on temperature $(U \propto T)$. Since $\Delta U = 0$,the temperature remains constant $(\Delta T = 0)$.
Thus,no work is done by the gas,and its internal energy and temperature remain unchanged.

(A) Free expansion is an adiabatic process where a gas expands into a vacuum.
Since the gas expands against zero external pressure $(P_{ext} = 0)$,the work done is $W = P_{ext} \Delta V = 0$.
Because the process is adiabatic and occurs in an isolated system,no heat is exchanged with the surroundings,so $Q = 0$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
Substituting the values,$\Delta U = 0 - 0 = 0$.
Therefore,for free expansion,$Q = 0, W = 0$ and $\Delta U = 0$.

(A) Most phenomena in nature are irreversible. $A$ process becomes irreversible if some energy is dissipated,typically converted into heat energy due to friction,viscosity,or other resistive forces. This is known as a dissipative effect. Since these dissipative effects cannot be completely eliminated in any real-world process,all natural thermodynamic processes are irreversible.

(A) In any real-world process,some energy is inevitably converted into heat due to friction,viscosity,or other resistive forces,which is dissipative in nature.
Because of this energy loss,the system cannot be returned to its original state without leaving a change in the surroundings.
Therefore,all natural processes are irreversible,making perfectly reversible systems impossible to find in reality.
Thus,the $Assertion$ is correct,and the $Reason$ is the correct explanation for it.

(A) Free expansion of an ideal gas is an adiabatic process $(Q = 0)$ where no work is done $(W = 0)$. Since $\Delta U = Q - W$,the internal energy remains constant,meaning the temperature of the ideal gas does not change. However,the gas occupies a larger volume,which increases the number of available microstates. The change in entropy for an ideal gas is given by $\Delta S = nR \ln(V_f/V_i)$. Since $V_f > V_i$,$\Delta S > 0$,so entropy increases. Furthermore,all natural (irreversible) processes are accompanied by an increase in the total entropy of the universe. Thus,both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

(A) $0.5 \text{ atm}$: The volume available to the gas is doubled as soon as the stopcock between cylinders $A$ and $B$ is opened. Since volume is inversely proportional to pressure at constant temperature,the pressure will decrease to one-half of the original value. Since the initial pressure of the gas is $1 \text{ atm}$,the final pressure in each cylinder will be $0.5 \text{ atm}$.
$(b)$ Zero: The internal energy of an ideal gas depends only on its temperature. Since this is a case of free expansion in a thermally insulated system,no work is done $(W = 0)$ and no heat is exchanged $(Q = 0)$. According to the first law of thermodynamics,$\Delta U = Q - W = 0$. Thus,the internal energy does not change.
$(c)$ Zero: Since the internal energy of an ideal gas is a function of temperature only and $\Delta U = 0$,the temperature of the gas remains constant.
$(d)$ No: The given process is a case of free expansion. It is rapid and uncontrolled. The intermediate states are non-equilibrium states and do not satisfy the ideal gas equation $PV = nRT$; therefore,they do not lie on the $P-V-T$ surface.

(N/A) Suppose the external pressure is suddenly reduced (by lifting the weight on the movable piston in the container). The piston will accelerate outward. During this process,the gas passes through states that are not equilibrium states. These non-equilibrium states do not have well-defined pressure and temperature. If a finite temperature difference exists between the gas and its surroundings,there will be a rapid exchange of heat during which the gas will pass through non-equilibrium states,and after some time,the gas will settle into an equilibrium state.
An ideal process in which at every stage the system is in an equilibrium state is known as a quasi-static process. Such a process is infinitely slow.
$A$ quasi-static process is obviously a hypothetical construct. In practice,processes that are sufficiently slow and do not involve accelerated motion of the piston or large temperature gradients are approximations to an ideal quasi-static process.
The system changes its variables $(P, T, V)$ so slowly that it remains in thermal and mechanical equilibrium with its surroundings throughout.
In a quasi-static process,at every stage,the difference in the pressure of the system and the external pressure is infinitesimally small.
To take a gas from the state $(P, T)$ to another state $(P', T')$ via a quasi-static process,we change the external pressure by a very small amount,allow the system to equalize its pressure with that of the surroundings,and continue the process infinitely slowly until the system achieves the pressure $P'$.
Similarly,to change the temperature,we introduce an infinitesimal temperature difference between the system and the surrounding reservoirs. By choosing reservoirs of progressively different temperatures from $T$ to $T'$,the system achieves the temperature $T'$.
In a quasi-static process,the temperature of the surrounding reservoir and the external pressure differ only infinitesimally from the temperature and pressure of the system.

(N/A) process is said to be reversible if it can be retraced in the opposite direction such that both the system and the surroundings return to their original states without leaving any net change in the universe. Reversible processes are ideal and cannot be fully realized in practice.
An irreversible process is one that cannot be reversed to restore the system and surroundings to their initial states. Most natural processes are irreversible. Examples include:
$(1)$ Heat transfer from a hot body to a cold body.
$(2)$ Diffusion of gases.
$(3)$ Combustion of fuel.
$(4)$ Rusting of iron.
$(5)$ Friction and viscosity effects.
Irreversibility arises primarily due to:
$(i)$ The system passing through non-equilibrium states during the process.
$(ii)$ The presence of dissipative effects like friction and viscosity,which cannot be completely eliminated.

(N/A) reversible process is a process that can be reversed by means of infinitesimal changes in a property of the system without leaving any trace of change on the surroundings. It is an ideal process that occurs in equilibrium at every stage.
An irreversible process is a process that cannot be reversed to its initial state without leaving a permanent change in the surroundings. Most natural processes,such as friction,diffusion,and heat transfer across a finite temperature difference,are irreversible.

(B) quasi-static process is an idealized thermodynamic process that occurs infinitely slowly such that the system remains in thermal and mechanical equilibrium with its surroundings at every instant.
In this process,the state variables (like pressure,volume,and temperature) are well-defined at every point along the path.
Since the change is infinitesimal,the system is always infinitesimally close to an equilibrium state.

(B) In nature,all spontaneous processes are irreversible. $A$ reversible process is an idealization that cannot be perfectly achieved in the real world because it requires the system to be in equilibrium at every stage,which implies an infinitely slow process. Therefore,only irreversible processes are acceptable and occur naturally.

(B) process is called quasi-static if it occurs infinitely slowly such that the system remains in thermodynamic equilibrium with its surroundings at every stage.
Energy dissipation (due to friction,viscosity,or turbulence) occurs in irreversible processes.
$A$ process that is both quasi-static and frictionless is defined as a reversible process.
In a reversible process,there is no loss of energy due to dissipative forces,and the system can be returned to its initial state without leaving any change in the surroundings.
Therefore,the correct answer is a reversible process.

(B) In thermodynamics,an $Irreversible$ $process$ is one that cannot be reversed to restore the system and its surroundings to their initial states without leaving any change in the universe.
Such processes are always associated with dissipative effects like friction,viscosity,or electrical resistance,which convert mechanical or useful energy into internal energy (heat).
Because some energy is dissipated as heat that cannot be converted back into work,the efficiency of any real-world engine operating through such processes is always less than that of an ideal Carnot engine.

(B) No, the rusting of iron is not a reversible process.
Rusting is a chemical change where iron reacts with oxygen and moisture to form hydrated iron $(III)$ oxide $(Fe_2O_3 \cdot nH_2O)$.
Since the chemical composition of the substance changes permanently and cannot be easily reversed to obtain the original iron, it is classified as an irreversible process.

(B) The bursting of a balloon is a sudden and spontaneous process,which makes it irreversible.
Since the process occurs very rapidly,there is no sufficient time for heat exchange with the surroundings $(dQ = 0)$.
Therefore,the expansion of helium during the bursting of the balloon is an irreversible adiabatic process.

(C) The process described is known as free expansion,where a gas expands into a vacuum.
Since the system is thermally insulated,there is no heat exchange with the surroundings,meaning $Q = 0$.
Because the gas expands into a vacuum,it does no work against any external pressure,meaning $W = 0$.
According to the first law of thermodynamics,$\Delta U = Q - W$. Since $Q = 0$ and $W = 0$,the change in internal energy $\Delta U = 0$.
For an ideal gas,internal energy depends only on temperature. Therefore,if $\Delta U = 0$,the temperature remains constant.
However,the defining characteristic of a process where $Q = 0$ is that it is an adiabatic process.

(A) The expansion occurs in a vacuum,so the work done by the expanding gas is $\Delta W = 0$.
The container is perfectly insulated,so the heat exchanged with the surroundings is $\Delta Q = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Substituting the values,we get $0 = \Delta U + 0$,which implies $\Delta U = 0$.
Therefore,$U_i = U_f$,meaning the initial and final internal energies of the gas are equal.
Since the gas is not ideal,the expansion results in an increase in the intermolecular potential energy. Because the total internal energy remains constant,this increase in potential energy leads to a decrease in the kinetic energy of the molecules,which consequently causes the temperature of the gas to decrease.

(C) The correct answer is $(c)$.
$A$ process is considered reversible if it can be reversed without leaving any trace on the surroundings. For a process to be reversible,it must satisfy two main conditions:
$1$. It must be quasi-static: The process must occur infinitely slowly so that the system remains in thermodynamic equilibrium at every stage.
$2$. It must be non-dissipative: There should be no dissipative forces such as friction,viscosity,or electrical resistance,which would cause an irreversible loss of energy as heat.

(C) For an adiabatic process,the relation between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
When a gas is shrunk to half its initial volume,$V_2 = V_1 / 2$.
Substituting this,$T_2 = T_1 (V_1 / V_2)^{\gamma-1} = T_1 (2)^{\gamma-1}$.
Since $\gamma > 1$ for all gases,$T_2 > T_1$,meaning the temperature increases,not decreases. Thus,Assertion $(A)$ is false.
Free expansion of an ideal gas occurs against zero external pressure $(P_{ext} = 0)$,so no work is done $(W = 0)$. Since the container is insulated,$Q = 0$. By the first law of thermodynamics,$\Delta U = Q - W = 0$,implying the temperature remains constant. This is an irreversible adiabatic process. Thus,Reason $(R)$ is true.

(C) Free expansion is an adiabatic process where work done $W = 0$ and internal energy change $\Delta U = 0$.
Because the process is rapid and uncontrolled, the system does not pass through a series of equilibrium states.
Since a $p-V$ diagram requires the system to be in thermodynamic equilibrium at every point to define pressure and volume, free expansion cannot be plotted as a continuous path on a $p-V$ diagram.
Therefore, the statement that free expansion can be plotted on a $p-V$ diagram is incorrect.

(A) i. Free expansions are adiabatic processes where there is no exchange of heat between the system and its environment,so $Q=0$.
ii. In a free expansion,the gas expands into a vacuum,meaning there is no external pressure to push against,so no work is done by the system,$W=0$.
iii. According to the first law of thermodynamics,$\Delta U = Q - W$. Since $Q=0$ and $W=0$,the change in internal energy $\Delta U = 0$.
iv. An example is the sudden rupture of a balloon or a tire,where gas rushes out into the surroundings without performing work on a piston or boundary.
v. $A$ free expansion is an uncontrolled,irreversible,and instantaneous change where the system is not in thermodynamic equilibrium during the process,and it cannot be represented as a continuous path on a $p-V$ diagram.

(B) reversible process is one that can be reversed without leaving any trace on the surroundings.
Melting of ice,isothermal expansion,and adiabatic expansion (if performed quasi-statically) can be considered reversible processes.
Conduction of heat is an irreversible process because it occurs due to a finite temperature difference between two bodies.
Heat always flows spontaneously from a higher temperature to a lower temperature,and this process cannot be reversed without external work,thus increasing the entropy of the universe.

(C) In free expansion,an ideal gas is allowed to expand into a vacuum.
Since the expansion occurs against zero external pressure $(P_{ext} = 0)$,the work done by the gas is $W = P_{ext} \Delta V = 0$.
Additionally,free expansion is typically considered adiabatic $(Q = 0)$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
Substituting the values,$\Delta U = 0 - 0 = 0$.
Therefore,there is no change in the internal energy of an ideal gas during free expansion.

(A) This process is known as free expansion of an ideal gas.
Since the system is thermally isolated,there is no heat exchange with the surroundings,so $Q = 0$.
Since the gas expands into a vacuum,there is no external pressure against which the gas does work,so the work done $W = 0$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
Since $Q = 0$ and $W = 0$,the change in internal energy $\Delta U = 0$.
For an ideal gas,the internal energy depends only on its temperature $(U \propto T)$.
Since $\Delta U = 0$,the temperature of the ideal gas remains constant.
Therefore,the final temperature attained at equilibrium will be $T$.