Isothermal Process Questions in English

(C) For an isothermal process, the equation of state for an ideal gas is given by $PV = \text{constant}$.
Differentiating both sides with respect to volume $V$, we get $P + V(dP/dV) = 0$.
This implies $P = -V(dP/dV)$.
By definition, the bulk modulus (or elasticity) $K$ is given by $K = -V(dP/dV)$.
Comparing the two expressions, we find that the isothermal elasticity $K_i$ is equal to the pressure $P$ of the gas.

(C) For an ideal gas,the internal energy $U$ is a function of temperature $T$ only,given by $U = nC_vT$.
Since the process is isothermal (temperature is kept constant),the change in temperature $\Delta T = 0$.
Therefore,the change in internal energy $\Delta U = nC_v \Delta T = 0$.
This implies that the internal energy of the gas remains constant throughout the process.

(C) An isothermal process is defined as a thermodynamic process in which the temperature of the system remains constant throughout the process $(T = \text{constant})$.
For an ideal gas, the internal energy $U$ is a function of temperature only $(U = f(T))$.
Since the temperature remains constant in an isothermal process, the internal energy of the ideal gas also remains constant $(\Delta U = 0)$.
Therefore, option $(C)$ is the correct statement.

(A) Two isothermal curves represent the relationship between pressure and volume at a constant temperature. If two isothermal curves were to intersect at a single point,it would imply that at that specific point (defined by a unique pressure and volume),the system has two different temperatures simultaneously. Since a system in equilibrium can only have one unique temperature at a given state,it is physically impossible for two distinct isothermal curves to intersect. Therefore,they can never cut each other.

(C) In an isothermal process,the temperature $T$ of the system remains constant throughout the expansion.
Since the internal energy $U$ of an ideal gas is a function of temperature only $(U = f(T))$,a constant temperature implies that there is no change in the internal energy of the gas.
Therefore,$\Delta U = 0$.

(D) The work done in an isothermal reversible expansion is given by the formula: $W = \mu RT \ln\left(\frac{V_2}{V_1}\right)$.
Here,$\mu$ is the number of moles,$R$ is the universal gas constant,$T$ is the absolute temperature,$V_1$ is the initial volume,and $V_2$ is the final volume.
Given: Mass of oxygen $(m)$ = $96\, g$,Molar mass of $O_2$ $(M)$ = $32\, g/mol$.
Number of moles $\mu = \frac{m}{M} = \frac{96}{32} = 3\, mol$.
Temperature $T = 27^{\circ}C = 27 + 273 = 300\, K$.
Initial volume $V_1 = 70\, L$,Final volume $V_2 = 140\, L$.
Converting natural logarithm to base $10$: $\ln(x) = 2.303 \log_{10}(x)$.
Substituting the values: $W = 3 \times R \times 300 \times 2.3 \log_{10}\left(\frac{140}{70}\right)$.
$W = 900 \times 2.3 \times R \times \log_{10}(2)$.
Thus,the work done is $2.3 \times 900\, R\, \log_{10} 2$.

(B) For an isothermal process,the temperature $T$ remains constant.
According to the ideal gas equation,$PV = nRT$.
Since $n$,$R$,and $T$ are constant,$PV = \text{constant}$.
Differentiating both sides with respect to $V$,we get:
$P \Delta V + V \Delta P = 0$
Rearranging the terms,we get $V \Delta P = -P \Delta V$.
Dividing both sides by $PV$,we obtain:
$\frac{\Delta P}{P} = -\frac{\Delta V}{V}$.

(D) For an isothermal process,the work done $W$ is given by the formula: $W = -\mu RT \ln\left(\frac{V_2}{V_1}\right)$.
Given: $\mu = 1 \text{ mole}$,$T = 273 \text{ K}$,$V_1 = 22.4 \text{ litres}$,$V_2 = 11.2 \text{ litres}$,and $R = 8.31 \text{ J/mol K}$.
Substituting the values:
$W = -1 \times 8.31 \times 273 \times \ln\left(\frac{11.2}{22.4}\right)$.
$W = -8.31 \times 273 \times \ln(0.5)$.
Since $\ln(0.5) = -\ln(2) \approx -0.693$,
$W = -8.31 \times 273 \times (-0.693) \approx 1572.5 \text{ J}$.
Note: The work done on the gas is positive,but the work done by the gas is negative. Given the options,the value $-1572.5 \text{ J}$ represents the work done by the gas during compression.

(C) For an isothermal process,the temperature $T$ remains constant.
From the ideal gas equation for $1$ mole of gas,$PV = RT$,which implies $P = \frac{RT}{V}$.
The work done $W$ during an isothermal expansion from volume $V_1$ to $V_2$ is given by the integral:
$W = \int_{V_1}^{V_2} P \, dV$
Substituting the value of $P$:
$W = \int_{V_1}^{V_2} \frac{RT}{V} \, dV$
Since $R$ and $T$ are constants,we can take them out of the integral:
$W = RT \int_{V_1}^{V_2} \frac{1}{V} \, dV$
$W = RT [\ln V]_{V_1}^{V_2}$
$W = RT (\ln V_2 - \ln V_1)$
$W = RT \log_{e} \left( \frac{V_2}{V_1} \right)$.

(B) For an isothermal process,the temperature $T$ remains constant.
By definition,the isothermal compressibility $\beta_T$ is given by $\beta_T = -\frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T$.
For an ideal gas,$PV = nRT$,so $P = \frac{nRT}{V}$.
Since $T$ is constant,the pressure $P$ is inversely proportional to the volume $V$,and the compressibility is defined as $\beta_T = \frac{1}{P}$.
Therefore,the pressure $P$ is determined by the reciprocal of the isothermal compressibility,i.e.,$P = \frac{1}{\beta_T}$.

(A) An isothermal process is a thermodynamic process in which the temperature of the system remains constant, i.e., $T = \text{constant}$.
According to the ideal gas equation, $PV = nRT$.
Since $n$, $R$, and $T$ are constant for an isothermal process, the product $PV$ must be constant.
Therefore, $PV = \text{constant}$, which is the mathematical statement of Boyle's law.
Thus, an ideal gas obeys Boyle's law during an isothermal change.

(C) In an isothermal process,the temperature of the system remains constant $(dT = 0)$.
Since internal energy of an ideal gas depends only on temperature $(U = f(T))$,the internal energy also remains constant $(dU = 0)$.
However,for an isothermal process to occur,heat must be exchanged between the system and the surroundings to compensate for the work done by or on the system $(dQ = dW)$.
Therefore,the statement that there is 'No exchange of energy' is incorrect.

(A) In an isothermal process,the temperature of the system remains constant,which means the change in temperature $\Delta T = 0$.
By definition,the specific heat capacity $C$ is given by the formula $C = \frac{Q}{m \Delta T}$,where $Q$ is the heat supplied,$m$ is the mass,and $\Delta T$ is the change in temperature.
Since $\Delta T = 0$ in an isothermal process,the denominator becomes zero.
Therefore,$C = \frac{Q}{m \times 0} = \infty$.
Thus,the specific heat of a gas in an isothermal process is infinite.

(A) An isothermal process occurs at a constant temperature.
To maintain a constant temperature,the system must be in thermal equilibrium with its surroundings.
This requires the container to have highly conducting walls so that heat can be exchanged freely between the system and the surroundings.
Among the given options,Copper is a metal and a good conductor of heat,whereas Glass,Wood,and Cloth are insulators.
Therefore,a container made of Copper is suitable for an isothermal process.

(C) For an isothermal process,the temperature $T$ remains constant. Since the internal energy $U$ of an ideal gas depends only on temperature,the change in internal energy $dU = 0$.
The work done by the gas is given by $dW = P dV$.
When the volume is halved,the final volume $V_2 = V_1 / 2$.
Therefore,the change in volume $dV = V_2 - V_1 = V_1 / 2 - V_1 = -V_1 / 2$.
Since $dV$ is negative,the work done by the gas $dW = P dV$ is negative. This indicates that work is done on the gas.

(B) thermodynamic process is defined by the state variables $P$,$V$,and $T$.
An $Isochoric$ process is one where volume $V$ remains constant.
An $Isobaric$ process is one where pressure $P$ remains constant.
An $Isothermal$ process is defined as a process in which the temperature $T$ of the system remains constant throughout the change,even if pressure $P$ and volume $V$ vary.
Therefore,the correct answer is $B$.

(B) In an isothermal process,the temperature $T$ of the system remains constant. According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. For an ideal gas,the internal energy $U$ depends only on temperature. Since the process is isothermal,$\Delta T = 0$,which implies $\Delta U = 0$. When a gas is compressed,work is done on the gas,so $\Delta W < 0$. Substituting these into the first law,$\Delta Q = 0 + \Delta W$. Since $\Delta W < 0$,$\Delta Q$ must also be negative,meaning heat is released by the gas to the surroundings to maintain a constant temperature.

(A) For an isothermal process,the change in internal energy $\Delta U = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Since $\Delta U = 0$,we have $\Delta Q = \Delta W$.
In this case,work is done on the gas,so $\Delta W = -1.5 \times 10^4 \ J$ (by convention,work done on the system is negative).
Therefore,$\Delta Q = -1.5 \times 10^4 \ J$.
To convert Joules to calories,we use the conversion factor $1 \ \text{cal} \approx 4.18 \ J$.
$\Delta Q = \frac{-1.5 \times 10^4}{4.18} \ \text{cal} \approx -3.588 \times 10^3 \ \text{cal} \approx -3.6 \times 10^3 \ \text{cal}$.
The negative sign indicates that heat flowed out of the gas.

(A) In an isothermal process,the temperature of the system remains constant.
Since the internal energy of an ideal gas depends only on its temperature,the change in internal energy $\Delta U$ is $0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Substituting $\Delta U = 0$,we get $\Delta Q = \Delta W$.
Therefore,the heat supplied to the gas is entirely used in doing external work.

(B) For an isothermal process,the work done $W$ by an ideal gas is given by the formula: $W = \mu RT \ln\left(\frac{V_2}{V_1}\right)$.
Given values are: $\mu = 1 \, mol$,$R = 8.31 \, J/mol \cdot K$,$T = 300 \, K$,$V_1 = 10 \, L$,and $V_2 = 20 \, L$.
Substituting these values into the formula:
$W = 1 \times 8.31 \times 300 \times \ln\left(\frac{20}{10}\right)$
$W = 2493 \times \ln(2)$
Using $\ln(2) \approx 0.6931$:
$W = 2493 \times 0.6931 \approx 1728 \, J$.

(B) Since the process occurs very slowly and the system remains in thermal equilibrium with the surroundings,it is an isothermal process.
For an isothermal process,the work done $W$ is given by the formula:
$W = \mu RT \ln\left(\frac{V_2}{V_1}\right)$
Given:
Number of moles $\mu = 0.2 \, mol$
Temperature $T = 27^{\circ}C = 27 + 273 = 300 \, K$
Universal gas constant $R = 8.3 \, J/(mol \cdot K)$
Final volume $V_2 = 2V_1$,so $\frac{V_2}{V_1} = 2$
Substituting the values:
$W = 0.2 \times 8.3 \times 300 \times \ln(2)$
$W = 0.2 \times 8.3 \times 300 \times 0.693$
$W = 498 \times 0.693 \approx 345 \, J$
Thus,the work done is approximately $345 \, J$.

(A) For an isothermal process,the relationship between pressure and volume is given by Boyle's Law: $P_1 V_1 = P_2 V_2$.
Given:
Initial volume $V_1 = 1 \, L = 1000 \, cm^3$.
Initial pressure $P_1 = 72 \, cm$ of $Hg$.
Final volume $V_2 = 900 \, cm^3$.
Using the formula: $P_2 = \frac{P_1 V_1}{V_2} = \frac{72 \times 1000}{900} = 80 \, cm$ of $Hg$.
The increase in pressure (stress) is $\Delta P = P_2 - P_1 = 80 - 72 = 8 \, cm$ of $Hg$.

(D) For an ideal gas,the internal energy $U$ is a function of temperature $T$ only,i.e.,$U = f(T)$.
During an isothermal process,the temperature $T$ remains constant.
Therefore,the change in internal energy $\Delta U = 0$,which means the internal energy does not change.
According to the First Law of Thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Since $\Delta U = 0$,we have $\Delta Q = \Delta W$.
This implies that the heat absorbed by the gas is equal to the work done by the gas.
Thus,both statements $(b)$ and $(c)$ are correct.

(A) An isothermal process is a thermodynamic process in which the temperature of the system remains constant throughout the process.
For the temperature to remain constant,the system must exchange heat with its surroundings very slowly to maintain thermal equilibrium.
If the process were fast,the system would not have sufficient time to exchange heat,and the temperature would change,making it adiabatic rather than isothermal.
Therefore,isothermal processes are inherently slow processes.

(C) The process is isothermal,therefore $T = \text{constant}$.
Since $PV = \mu RT$,we have $P = \frac{\mu RT}{V}$.
For chamber $A$,the change in pressure is $\Delta P = P_i - P_f = \frac{\mu_A RT}{V} - \frac{\mu_A RT}{2V} = \frac{\mu_A RT}{2V} \dots (i)$.
For chamber $B$,the change in pressure is $1.5 \Delta P = P_i - P_f = \frac{\mu_B RT}{V} - \frac{\mu_B RT}{2V} = \frac{\mu_B RT}{2V} \dots (ii)$.
Dividing equation $(i)$ by $(ii)$,we get $\frac{\Delta P}{1.5 \Delta P} = \frac{\mu_A}{\mu_B} \implies \frac{1}{1.5} = \frac{\mu_A}{\mu_B} \implies \frac{\mu_A}{\mu_B} = \frac{2}{3}$.
Since $\mu = \frac{m}{M}$,where $M$ is the molar mass,we have $\frac{m_A/M}{m_B/M} = \frac{2}{3} \implies \frac{m_A}{m_B} = \frac{2}{3}$.
Therefore,$3m_A = 2m_B$.

(A) According to the given Van der Waals' equation:
$(V - n\beta) \left( P + \frac{\alpha n^2}{V^2} \right) = nRT$
Rearranging for $P$:
$P = \frac{nRT}{V - n\beta} - \frac{\alpha n^2}{V^2}$
Work done $W$ is given by the integral:
$W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \left( \frac{nRT}{V - n\beta} - \frac{\alpha n^2}{V^2} \right) dV$
$W = nRT \int_{V_1}^{V_2} \frac{dV}{V - n\beta} - \alpha n^2 \int_{V_1}^{V_2} V^{-2} dV$
Integrating the terms:
$W = nRT [\log_e(V - n\beta)]_{V_1}^{V_2} - \alpha n^2 \left[ -\frac{1}{V} \right]_{V_1}^{V_2}$
$W = nRT \log_e \left( \frac{V_2 - n\beta}{V_1 - n\beta} \right) + \alpha n^2 \left( \frac{1}{V_2} - \frac{1}{V_1} \right)$
$W = nRT \log_e \left( \frac{V_2 - n\beta}{V_1 - n\beta} \right) + \alpha n^2 \left( \frac{V_1 - V_2}{V_1 V_2} \right)$

(B) In an isothermal process, the temperature $T$ remains constant.
According to the ideal gas equation, $PV = nRT$.
Since $n$, $R$, and $T$ are constants, we have $PV = \text{constant}$, which implies $P \propto \frac{1}{V}$.
This relationship represents a rectangular hyperbola in a $PV$ diagram.
Among the given options, the graph in option $B$ shows this hyperbolic relationship.

(B) In thermodynamics, an isothermal process is a type of thermodynamic process in which the temperature $T$ of the system remains constant $(dT = 0)$.
During this process, the heat transfer into or out of the system typically occurs at such a slow rate that thermal equilibrium is maintained.
According to the ideal gas law $PV = nRT$, if $T$ is constant, then the product $PV$ must also be constant $(PV = \text{constant})$.
Therefore, while $T$ remains constant, both pressure $P$ and volume $V$ change to satisfy the equation.
Thus, the correct option is $B$.

(B) For an isothermal process,the work done by an ideal gas is given by the formula:
$W = \mu RT \ln\left(\frac{V_2}{V_1}\right)$
Given:
Number of moles $\mu = 1 \; mol$
Gas constant $R = 8.31 \; J/mol \cdot K$
Temperature $T = 300 \; K$
Initial volume $V_1 = 10 \; L$
Final volume $V_2 = 20 \; L$
Substituting the values:
$W = 1 \times 8.31 \times 300 \times \ln\left(\frac{20}{10}\right)$
$W = 2493 \times \ln(2)$
Using $\ln(2) \approx 0.693$:
$W \approx 2493 \times 0.693 \approx 1727.65 \; J$
Rounding to the nearest integer,we get $W \approx 1728 \; J$.

(B) For an isothermal process,the work done $W$ is given by the formula: $W = nRT \ln\left(\frac{V_f}{V_i}\right)$.
Given: $n = 5 \, \text{moles}$,$T = 500 \, K$,$V_f = 2V_i$,and $R \approx 8.314 \, J/(mol \cdot K)$.
Substituting the values: $W = 5 \times 8.314 \times 500 \times \ln(2)$.
Since $\ln(2) \approx 0.693$,we have $W = 5 \times 8.314 \times 500 \times 0.693$.
$W = 2500 \times 8.314 \times 0.693 \approx 14412 \, J$.
Rounding to the nearest provided option,the work done is approximately $14400 \, J$.

(D) According to the first law of thermodynamics,$\Delta U = Q + W$,where $\Delta U$ is the change in internal energy,$Q$ is the heat added to the system,and $W$ is the work done on the system.
For an isothermal process involving an ideal gas,the internal energy depends only on temperature. Since the temperature remains constant,$\Delta U = 0$.
Therefore,the equation becomes $0 = Q + W$,which implies $Q = -W$.
Given that the gas does work against its surroundings,the work done on the gas is $W = -150 \, J$.
Substituting this value,we get $Q = -(-150 \, J) = +150 \, J$.
$A$ positive value for $Q$ indicates that $150 \, J$ of heat has been added to the gas.

(A) The internal energy $U$ of an ideal gas is given by the formula $U = \frac{f}{2} \mu RT = \frac{f}{2} \left( \frac{N}{N_A} \right) RT$,which implies $U \propto NT$.
In an isothermal process,the temperature $T$ remains constant. Therefore,the internal energy $U$ becomes directly proportional to the number of molecules $N$ $(U \propto N)$.
Consequently,the internal energy of an ideal gas increases during an isothermal process if the number of molecules $N$ is increased.

(B) For an ideal gas undergoing an isothermal process,the temperature $T$ remains constant. According to the ideal gas law,$PV = nRT$. Since $n$,$R$,and $T$ are constants,the product $PV$ must be constant. Therefore,$P = \text{constant} / V$,which implies $P \propto 1/V$.

(B) In the first case,the pressure of the gas $P_1$ is given by $P_1 = P_0 + \frac{Mg}{A}$.
In the second case,when the tube is inverted,the pressure of the gas $P_2$ is given by $P_2 = P_0 - \frac{Mg}{A}$.
Since the process is isothermal,$P_1 V_1 = P_2 V_2$. Since $V = A \times L$,we have $P_1 L_1 = P_2 L_2$.
Therefore,$\frac{L_2}{L_1} = \frac{P_1}{P_2} = \frac{P_0 + \frac{Mg}{A}}{P_0 - \frac{Mg}{A}} = \frac{P_0 A + Mg}{P_0 A - Mg}$.
Substituting the values: $P_0 A = 10^5 \times 20 \times 10^{-6} = 2 \, N$ and $Mg = 0.002 \times 10 = 0.02 \, N$.
$\frac{L_2}{L_1} = \frac{2 + 0.02}{2 - 0.02} = \frac{2.02}{1.98} = \frac{202}{198} = \frac{101}{99}$.

(D) Since the walls are heat-conducting and the process is slow,the temperature remains constant at $T_0$. This is an isothermal process.
$1$. For an isothermal process,$\Delta U = 0$ (Internal energy change is zero).
$2$. Work done by the gas $W = \int P \, dV = \int_{V_0}^{\eta V_0} \frac{RT_0}{V} \, dV = RT_0 \ln \eta$.
$3$. The piston is light,so the pressure of the gas is always equal to the atmospheric pressure $P_{atm}$. Since $PV = RT_0$,$P = \frac{RT_0}{V}$. Initially $P_i = \frac{RT_0}{V_0}$ and finally $P_f = \frac{RT_0}{\eta V_0} = \frac{P_i}{\eta}$. Thus,the final pressure is $\frac{1}{\eta}$ times the initial pressure.
$4$. Work done against the atmosphere $W_{atm} = P_{atm} \Delta V = P_i (\eta V_0 - V_0) = P_i V_0 (\eta - 1) = RT_0(\eta - 1)$.
Comparing these with the options,option $D$ states the final pressure is $\frac{1}{(\eta - 1)}$ times the initial pressure,which is incorrect.

(A) The process $DA$ is an isochoric process because the temperature $T$ is constant at $300 \text{ K}$ while the pressure changes from $1 \times 10^5 \text{ Pa}$ to $2 \times 10^5 \text{ Pa}$.
However,looking at the $P-T$ diagram,the line $DA$ is vertical,meaning it is an isothermal process at $T = 300 \text{ K}$.
For an isothermal process,the work done by the gas is given by $W_{\text{by}} = nRT \ln\left(\frac{V_f}{V_i}\right)$.
Using the ideal gas law $PV = nRT$,we have $V = \frac{nRT}{P}$,so $\frac{V_f}{V_i} = \frac{P_i}{P_f}$.
Thus,$W_{\text{by}} = nRT \ln\left(\frac{P_D}{P_A}\right) = 2.303 nRT \log_{10}\left(\frac{P_D}{P_A}\right)$.
Given $n = 2$,$T = 300 \text{ K}$,$P_D = 1 \times 10^5 \text{ Pa}$,and $P_A = 2 \times 10^5 \text{ Pa}$.
$W_{\text{by}} = 2.303 \times 2 \times R \times 300 \times \log_{10}\left(\frac{1 \times 10^5}{2 \times 10^5}\right) = 2.303 \times 600 \times R \times \log_{10}(0.5)$.
$W_{\text{by}} = 1381.8 \times R \times (-0.301) \approx -416 R$ (using $\log_{10}(0.5) \approx -0.301$).
Using the standard approximation $\log_{10}(0.5) \approx -0.3$,$W_{\text{by}} = 2.303 \times 600 \times R \times (-0.3) = -414.54 R \approx -414 R$.
The work done $ON$ the gas is $W_{\text{on}} = -W_{\text{by}} = -(-414 R) = +414 R$.

(B) For an isothermal process,the temperature remains constant,so the change in internal energy $dU = 0$.
According to the first law of thermodynamics,$dQ = dU + dW$.
Given that the gas does $600\,J$ of work,$dW = 600\,J$.
Substituting these values,$dQ = 0 + 600\,J = 600\,J$.
Since $dQ$ is positive,the gas absorbs $600\,J$ of heat from the surroundings.

(B) The molar specific heat capacity $C$ is defined as $C = \frac{\Delta Q}{n \Delta T}$,where $\Delta Q$ is the heat supplied,$n$ is the number of moles,and $\Delta T$ is the change in temperature.
In an isothermal process,the temperature of the system remains constant,which means $\Delta T = 0$.
Substituting $\Delta T = 0$ into the formula,we get $C = \frac{\Delta Q}{n \times 0} = \infty$.
Therefore,the specific heat of an isothermal process is infinite.

(D) For an isothermal process,the work done on the gas is given by the formula $W = -nRT \ln\left(\frac{V_f}{V_i}\right)$.
Since the process is isothermal,$P_i V_i = P_f V_f$,which implies $\frac{V_f}{V_i} = \frac{P_i}{P_f}$.
Given that the pressure is doubled,$P_f = 2P_i$,so $\frac{P_i}{P_f} = \frac{1}{2}$.
Thus,$\frac{V_f}{V_i} = \frac{1}{2}$.
The work done on the gas is $W = -nRT \ln\left(\frac{1}{2}\right) = nRT \ln(2)$.
Given $n = 1$ mole and $T = 27 + 273 = 300\,K$.
Substituting the values,$W = 1 \times R \times 300 \times \ln(2) = 300R \ln(2)$.

(D) From the given $P-V$ diagram,the process follows the relation $P = \frac{\text{constant}}{V}$,which implies $PV = \text{constant}$.
According to the ideal gas equation $PV = nRT$,if $PV$ is constant,then $T$ must also be constant $(T = \text{constant})$.
This represents an isothermal process.
In a $T-P$ diagram,an isothermal process is represented by a horizontal line (where $T$ remains constant as $P$ changes).
Looking at the $P-V$ graph,the pressure $P$ decreases as we move from point $1$ to point $2$ (since $V$ increases).
Therefore,in the $T-P$ diagram,the process should be a horizontal line starting from $1$ and moving towards $2$ as $P$ decreases.
Among the given options,the graph where $T$ is constant and the transition is from $1$ to $2$ as $P$ decreases is represented by option $(d)$.

(D) For an isothermal process,the temperature $T$ remains constant.
For $n$ moles of an ideal gas,the heat exchanged $\Delta Q$ during an isothermal expansion from $V_1$ to $V_2$ is equal to the work done $W$,because the change in internal energy $\Delta U = 0$.
$W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV = nRT \ln \frac{V_2}{V_1}$.
Given $n = 1$ mole,the heat exchanged is $\Delta Q = RT \ln \frac{V_2}{V_1}$.
The change in entropy $\Delta S$ is defined as $\Delta S = \frac{\Delta Q}{T}$.
Substituting the value of $\Delta Q$,we get $\Delta S = \frac{RT \ln \frac{V_2}{V_1}}{T} = R \ln \frac{V_2}{V_1}$.

(D) For an isothermal process,the work done $BY$ the gas is given by $W_{by} = \int_{V_i}^{V_f} P \, dV$. Since $PV = nRT$,we have $P = nRT/V$.
Integrating this,$W_{by} = nRT \int_{V_i}^{V_f} \frac{1}{V} dV = nRT \ln(V_f/V_i)$.
By Boyle's Law,$P_i V_i = P_f V_f$,so $V_f/V_i = P_i/P_f$.
Thus,$W_{by} = nRT \ln(P_i/P_f) = -nRT \ln(P_f/P_i)$.
The work done $ON$ the gas is the negative of the work done by the gas: $W_{on} = -W_{by} = -nRT \ln(V_f/V_i) = nRT \ln(P_f/P_i)$.

(D) The first law of thermodynamics is given by $\Delta Q = \Delta U + P\Delta V$.
For an ideal gas,the internal energy $U$ is a function of temperature $T$ only $(U = f(T))$.
In an isothermal process,the temperature remains constant,so $\Delta T = 0$.
Since $\Delta U = nC_v\Delta T$,it follows that $\Delta U = 0$ for an isothermal process.
Substituting this into the first law: $\Delta Q = 0 + P\Delta V$,which means $\Delta Q = P\Delta V$.
Therefore,the heat supplied is converted into work done by the system,not internal energy.
Thus,the Assertion is incorrect,and the Reason is a correct statement of the first law of thermodynamics.

(D) The $Assertion$ is incorrect. Two isothermal curves for different temperatures cannot intersect each other. If they were to intersect at a point,it would imply that at that specific $(P, V)$ state,the system has two different temperatures simultaneously,which is impossible.
The $Reason$ is correct. An isothermal process is a slow process that allows the system to remain in thermal equilibrium with its surroundings. The slope of an isothermal curve on a $P-V$ diagram is given by $-\frac{dP}{dV} = \frac{P}{V}$. Since $P$ and $V$ are positive,the slope is finite and relatively small compared to adiabatic curves (which have a slope of $\gamma \frac{P}{V}$).

(D) An isothermal process is a process that occurs at a constant temperature. For an ideal gas,the equation of state is $PV = nRT$. Since $T$ is constant,$P = (nRT)/V$,which implies $P \propto 1/V$.
Two different isothermal curves correspond to two different temperatures ($T_1$ and $T_2$). If they were to intersect,it would mean that at the point of intersection,the system would have two different temperatures simultaneously,which is physically impossible.
Furthermore,isothermal processes are inherently slow (quasi-static) to maintain thermal equilibrium,contrary to the claim in the $Reason$ that they occur rapidly.
Therefore,both the $Assertion$ and the $Reason$ are incorrect.

(N/A) In an isothermal process of an ideal gas,the temperature remains constant.
Since the internal energy $U$ of an ideal gas depends only on its temperature,for an isothermal process,the change in internal energy $\Delta U = 0$.
According to the first law of thermodynamics:
$\Delta Q = \Delta U + \Delta W$
Substituting $\Delta U = 0$ into the equation:
$\Delta Q = \Delta W$
This means that in an isothermal process,the entire amount of heat supplied to the system is used to perform work by the system.

(N/A) The first law of thermodynamics is given by the equation: $\Delta Q = \Delta U + \Delta W$,where $\Delta Q$ is the heat supplied to the system,$\Delta U$ is the change in internal energy,and $\Delta W$ is the work done by the system.
In an isothermal process,the temperature of the system remains constant $(T = \text{constant})$.
Since the internal energy of an ideal gas depends only on its temperature,for an isothermal process,the change in internal energy is zero $(\Delta U = 0)$.
Substituting this into the first law of thermodynamics,we get: $\Delta Q = \Delta W$.
This means that in an isothermal expansion,the entire amount of heat supplied to the system is used to perform work by the system.

(N/A) Yes,it is possible for the temperature of a system to remain constant while it is being heated.
According to the first law of thermodynamics,$dQ = dU + dW$.
If the heat supplied $(dQ)$ is entirely used to perform work $(dW)$ against the surroundings,then the change in internal energy $(dU)$ will be zero.
Since internal energy is a function of temperature $(U \propto T)$,if $dU = 0$,then the change in temperature $(dT)$ is also zero.
This occurs during an isothermal process,such as the melting of ice at $0^{\circ}C$ or the boiling of water at $100^{\circ}C$ under constant pressure.