Mix Examples-Thermodynamics Questions in English

(A) For an adiabatic process,the heat exchange $Q = 0$.
For an isothermal process,the heat rejected is given by $Q_{\text{rejected}} = -W = -nRT_0 \ln(V_f/V_i) = nRT_0 \ln(V_i/V_f)$. Since $V_f = V_0/2$,we have $Q_{\text{rejected}} = nRT_0 \ln(2) \approx 0.693 nRT_0$.
For an isobaric process,the heat rejected is $Q_{\text{rejected}} = -n C_p \Delta T$. Since $V \propto T$,if $V$ becomes $V_0/2$,then $T$ becomes $T_0/2$,so $\Delta T = -T_0/2$. Thus,$Q_{\text{rejected}} = -n (\frac{f}{2} + 1) R (-T_0/2) = (\frac{f}{4} + 0.5) nRT_0$.
For a monoatomic gas $(f=3)$,$Q_{\text{rejected}} = (0.75 + 0.5) nRT_0 = 1.25 nRT_0$.
Comparing the values,$1.25 nRT_0 > 0.693 nRT_0$,therefore,the isobaric process releases the maximum amount of heat.

(B) The molar heat capacity $C$ is defined by the relation $dQ = \mu C dT$,where $\mu$ is the number of moles and $dT$ is the change in temperature.
The internal energy change for an ideal gas is given by $dU = \mu C_v dT$.
Given the process equation: $dQ = 2dU$.
Substituting the expressions for $dQ$ and $dU$:
$\mu C dT = 2(\mu C_v dT)$
Dividing both sides by $\mu dT$:
$C = 2C_v$
For a monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$.
Therefore,$C = 2 \times (\frac{3}{2}R) = 3R$.
The molar heat capacity for the process is $3R$.

(B) The work done $(W)$ in the cycle $ABCDA$ is the area enclosed by the rectangle:
$W = (2P_0 - P_0) \times (2V_0 - V_0) = P_0 V_0$
Heat is absorbed during processes $AB$ and $BC$:
For process $AB$ (isochoric): $Q_{AB} = n C_v \Delta T = n (1.5 R) (T_B - T_A) = 1.5 (P_B V_B - P_A V_A) = 1.5 (2P_0 V_0 - P_0 V_0) = 1.5 P_0 V_0$
For process $BC$ (isobaric): $Q_{BC} = n C_p \Delta T = n (2.5 R) (T_C - T_B) = 2.5 (P_C V_C - P_B V_B) = 2.5 (4P_0 V_0 - 2P_0 V_0) = 5 P_0 V_0$
Total heat absorbed $(Q_{in})$ = $Q_{AB} + Q_{BC} = 1.5 P_0 V_0 + 5 P_0 V_0 = 6.5 P_0 V_0 = \frac{13}{2} P_0 V_0$
Thermal efficiency $(\eta)$ = $\frac{W}{Q_{in}} = \frac{P_0 V_0}{6.5 P_0 V_0} = \frac{1}{6.5} = \frac{2}{13} \approx 0.154$

(A) In the given $V - T$ diagram:
$1$. Process $a \to b$: The line passes through the origin,so $V \propto T$. From the ideal gas law $PV = nRT$,this implies $P$ is constant. Thus,$a \to b$ is an isobaric process where temperature increases,meaning volume also increases.
$2$. Process $b \to c$: This is given as an adiabatic process. Since $V$ decreases and $T$ increases (from the graph),the pressure $P$ must increase significantly.
$3$. Process $c \to d$: The line passes through the origin,so $V \propto T$,which implies $P$ is constant. Thus,$c \to d$ is an isobaric process where temperature decreases,meaning volume also decreases.
$4$. Process $d \to a$: This is given as an adiabatic process. Since $V$ increases and $T$ decreases (from the graph),the pressure $P$ must decrease.
Comparing these characteristics with the given options,the $P - V$ diagram that correctly represents these processes is Option $A$.

(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Step $1$: Calculate the work done during the compression process.
Work done $\Delta W = P \Delta V = 100 \times (1 - 2) = -100\,J$.
Step $2$: Calculate the change in internal energy during the heating process.
Given $\Delta Q = 150\,J$ and since it is a constant volume process,$\Delta W = 0$.
Therefore,$\Delta Q = \Delta U = 150\,J$.
Step $3$: Total change in internal energy.
The total change in internal energy is the sum of the changes in both processes. However,the question asks for the net result of the entire sequence.
Using $\Delta Q_{total} = \Delta U_{total} + \Delta W_{total}$,where $\Delta Q_{total} = 150\,J$ and $\Delta W_{total} = -100\,J$.
$150 = \Delta U + (-100)$.
$\Delta U = 150 + 100 = 250\,J$.
Thus,the internal energy of the gas increases by $250\,J$.

(A) In a $P-V$ diagram:
$1$. An isobaric process is represented by a horizontal line where pressure $P$ is constant, so the slope $\frac{dP}{dV} = 0$.
$2$. An isochoric process is represented by a vertical line where volume $V$ is constant, so the slope $\frac{dP}{dV} = \infty$.
$3$. An isothermal process is represented by a curve where $PV = \text{constant}$, so the slope $\frac{dP}{dV} = -\frac{P}{V}$.
Looking at the given options, option $A$ shows two horizontal lines (isobars), one vertical line (isochore), and one curved line (isothermal). Thus, it correctly represents the cycle $(A-B-C-D-A)$.

(D) For an isothermal process, the temperature remains constant, so $PV = \text{constant}$.
Initial state: $(P, V)$. Final state: $(P_i, 2V)$.
$PV = P_i(2V) \implies P_i = \frac{P}{2} \implies P = 2P_i \quad ...(i)$
For an adiabatic process, $PV^{\gamma} = \text{constant}$.
Initial state: $(P, V)$. Final state: $(P_a, 2V)$.
$PV^{\gamma} = P_a(2V)^{\gamma} \implies P_a = P \left(\frac{V}{2V}\right)^{\gamma} = P(2)^{-\gamma}$
For a monoatomic gas, the adiabatic index $\gamma = \frac{5}{3}$.
Substituting $P = 2P_i$ and $\gamma = \frac{5}{3}$ into the adiabatic equation:
$P_a = (2P_i)(2)^{-5/3} = P_i \cdot 2^1 \cdot 2^{-5/3} = P_i \cdot 2^{1 - 5/3} = P_i \cdot 2^{-2/3}$
Therefore, the ratio $\frac{P_a}{P_i} = 2^{-2/3}$.

(A) Given the function $y = 1 - e^{-x}$.
For $x = 0$,$y = 1 - e^{0} = 1 - 1 = 0$. Thus,the graph starts from the origin $(0, 0)$.
As $x \to \infty$,$e^{-x} \to 0$,so $y \to 1 - 0 = 1$. This means the graph approaches the horizontal asymptote $y = 1$ as $x$ increases.
The derivative $\frac{dy}{dx} = e^{-x}$ is always positive for all $x$,meaning the function is strictly increasing.
Comparing these properties with the given options,the graph that starts at $(0, 0)$ and asymptotically approaches $y = 1$ is represented by Graph $A$.

(A) For process $AB$: The volume $V$ is constant $(V = V_1)$. Therefore, the work done $W_{AB} = \int P dV = 0$.
For process $BC$: The temperature $T$ is constant $(T = T_2)$. The work done for an isothermal process is $W_{BC} = nRT_2 \ln \left( \frac{V_f}{V_i} \right)$. Here, $n = 1$, $V_i = V_1$, and $V_f = V_2$. So, $W_{BC} = R T_2 \ln \left( \frac{V_2}{V_1} \right)$.
For process $CA$: This is a process where the graph is a straight line passing through the origin in the $V-T$ diagram, which implies $V \propto T$, meaning pressure $P$ is constant (isobaric process). The work done is $W_{CA} = nR \Delta T = R(T_1 - T_2)$.

(D) कार्नोट चक्र में चार चरण होते हैं:
$1$. $AB$: समतापीय प्रसार ($T$ स्थिर, $P$ घटता है, $V$ बढ़ता है)।
$2$. $BC$: रुद्धोष्म प्रसार ($Q=0$, $T$ घटता है, $V$ बढ़ता है)।
$3$. $CD$: समतापीय संपीड़न ($T$ स्थिर, $P$ बढ़ता है, $V$ घटता है)।
$4$. $DA$: रुद्धोष्म संपीड़न ($Q=0$, $T$ बढ़ता है, $V$ घटता है)।
$T-V$ आरेख में, समतापीय प्रक्रियाएं ($AB$ और $CD$) क्षैतिज रेखाओं के रूप में दिखाई देती हैं क्योंकि $T$ स्थिर रहता है। रुद्धोष्म प्रक्रियाएं ($BC$ और $DA$) वक्र रेखाओं के रूप में दिखाई देती हैं क्योंकि $T$ और $V$ के बीच का संबंध $TV^{\gamma-1} = \text{constant}$ है। अतः, सही विकल्प $D$ है।

(B) The work done by the system during an isobaric expansion is given by the formula $W = P(V_2 - V_1)$.
Given:
Pressure $P = 1 \text{ atm} \approx 1.013 \times 10^5 \text{ Pa}$. For simplicity in calculation,we use $P = 10^5 \text{ Pa}$.
Initial volume $V_1 = 1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3$.
Final volume $V_2 = 1671 \text{ cm}^3 = 1671 \times 10^{-6} \text{ m}^3$.
Change in volume $\Delta V = V_2 - V_1 = (1671 - 1) \times 10^{-6} \text{ m}^3 = 1670 \times 10^{-6} \text{ m}^3$.
Work done in Joules: $W = 10^5 \times 1670 \times 10^{-6} \text{ J} = 167 \text{ J}$.
To convert work from Joules to calories,divide by the mechanical equivalent of heat $(J \approx 4.2 \text{ J/cal})$:
$W = \frac{167}{4.2} \approx 39.76 \text{ cal}$.
Rounding to the nearest integer,the work done is approximately $40 \text{ cal}$.

(D) In the $PT$ diagram:
$A \rightarrow B$: $P$ is constant and $T$ decreases. Since $PV = nRT$,$V$ must decrease. This is an isobaric compression.
$B \rightarrow C$: The line passes through the origin,so $P \propto T$,which means $V$ is constant. Since $P$ decreases,$T$ also decreases. This is an isochoric cooling.
$C \rightarrow A$: The line is not passing through the origin. From the ideal gas law $PV = nRT$,we have $P = (nR/V)T$. The slope of the $PT$ graph is $nR/V$. Since $P$ and $T$ both increase from $C$ to $A$,and the slope is positive,this represents a process where $V$ changes. Specifically,for an isothermal process,$P \propto 1/V$. The correct representation in the $PV$ diagram is shown in image $818-$s925.

(B) For an isothermal process,the temperature remains constant,so $P_1 V_1 = P_2 V_2$.
Given $P_1 = P$,$V_1 = V$,and $V_2 = 4V$,we have $P \times V = P_2 \times (4V)$.
Thus,$P_2 = \frac{P}{4}$.
For an adiabatic process,the relation is $P_2 V_2^{\gamma} = P_f V_f^{\gamma}$.
Given $\gamma = \frac{5}{3}$,$V_2 = 4V$,and $V_f = 16V$,we have $\frac{P}{4} \times (4V)^{5/3} = P_f \times (16V)^{5/3}$.
$P_f = \frac{P}{4} \times \left(\frac{4V}{16V}\right)^{5/3} = \frac{P}{4} \times \left(\frac{1}{4}\right)^{5/3} = \frac{P}{4} \times \frac{1}{4^{5/3}} = \frac{P}{4^{1 + 5/3}} = \frac{P}{4^{8/3}}$.
Note: If $\gamma = 1.5$ (or $3/2$) as per the original prompt,$P_f = \frac{P}{4} \times (1/4)^{3/2} = \frac{P}{4} \times \frac{1}{8} = \frac{P}{32}$.

(C) The work done in a cycle is given by $W = \oint P dV$. For an ideal gas,$PV = nRT$,so $dV = \frac{nR}{P} dT$ for isobaric processes.
Process $AB$ and $CD$ are isochoric (constant volume) as they lie on lines passing through the origin in a $P-T$ diagram,so $W_{AB} = 0$ and $W_{CD} = 0$.
Process $BC$ is isobaric at pressure $P_2$. Work $W_{BC} = P_2(V_C - V_B) = nR(T_C - T_B) = 6 \times R \times (2200 - 800) = 6R \times 1400 = 8400R$.
Process $DA$ is isobaric at pressure $P_1$. Work $W_{DA} = P_1(V_A - V_D) = nR(T_A - T_D) = 6 \times R \times (600 - 1200) = 6R \times (-600) = -3600R$.
Total work $W = W_{AB} + W_{BC} + W_{CD} + W_{DA} = 0 + 8400R + 0 - 3600R = 4800R$.
Using $R \approx 8.314\, J/(mol\cdot K)$ or $R \approx 25/3\, J/(mol\cdot K)$:
$W = 4800 \times (25/3) = 1600 \times 25 = 40000\, J = 40\, kJ$.

(C) For an ideal gas,the work done in a process is given by $\Delta W = \int P \, dV$. Using the ideal gas equation $PV = nRT$,we have $V = \frac{nRT}{P}$. Thus,$dV = \frac{nR}{P} dT$ (at constant pressure) or $dV = -\frac{nRT}{P^2} dP$ (at constant temperature).
$1$. Path $AB$ (Constant pressure $P = 2P_0$): $\Delta W_{AB} = P \Delta V = nR \Delta T = 1 \cdot R \cdot (2T - T) = RT$.
$2$. Path $BC$ (Constant temperature $T = 2T_0$): $\Delta W_{BC} = nRT \ln \left( \frac{V_f}{V_i} \right) = nRT \ln \left( \frac{P_i}{P_f} \right) = 1 \cdot R \cdot (2T) \ln \left( \frac{2P}{P} \right) = 2RT \ln 2$.
$3$. Path $CD$ (Constant pressure $P = P_0$): $\Delta W_{CD} = P \Delta V = nR \Delta T = 1 \cdot R \cdot (T - 2T) = -RT$.
$4$. Path $DA$ (Constant temperature $T = T_0$): $\Delta W_{DA} = nRT \ln \left( \frac{V_f}{V_i} \right) = nRT \ln \left( \frac{P_i}{P_f} \right) = 1 \cdot R \cdot (T) \ln \left( \frac{P}{2P} \right) = RT \ln \left( \frac{1}{2} \right) = -RT \ln 2$.
Net work done $\Delta W = \Delta W_{AB} + \Delta W_{BC} + \Delta W_{CD} + \Delta W_{DA} = RT + 2RT \ln 2 - RT - RT \ln 2 = RT \ln 2$.

(A) In the given $V-T$ diagram:
$1$. Process $A \rightarrow B$: This is a straight line passing through the origin,so $V \propto T$. According to the ideal gas law $PV = nRT$,this implies $P$ is constant. Thus,$A \rightarrow B$ is an isobaric process.
$2$. Process $B \rightarrow C$: This is a horizontal line,meaning $V$ is constant. Thus,$B \rightarrow C$ is an isochoric process.
$3$. Process $C \rightarrow A$: This is a vertical line,meaning $T$ is constant. Thus,$C \rightarrow A$ is an isothermal process.
Now,let's analyze the $P-V$ diagrams:
- In option $A$,$A \rightarrow B$ is an isobaric process (horizontal line),$B \rightarrow C$ is an isochoric process (vertical line),and $C \rightarrow A$ is an isothermal process (hyperbolic curve). This matches the given cycle.

(A) From the given $P-V$ diagram,we analyze each process:
$1$. Process $A-B$: $V$ is constant (isochoric). From the ideal gas equation $PV = nRT$,since $V$ is constant,$P \propto T$. This represents a straight line passing through the origin in a $P-T$ diagram.
$2$. Process $B-C$: $P$ is constant (isobaric). From $PV = nRT$,since $P$ is constant,$V \propto T$. In a $P-T$ diagram,this is a line with a positive slope passing through the origin.
$3$. Process $C-D$: $V$ is constant (isochoric). Similar to $A-B$,this is a straight line passing through the origin.
$4$. Process $D-A$: $P$ is constant (isobaric). Similar to $B-C$,this is a line passing through the origin.
Comparing these characteristics with the given options,the graph in option $(A)$ correctly represents these processes in the $P-T$ plane.

(D) In a $T-S$ diagram,the heat exchanged $Q$ is given by the area under the process curve,i.e.,$Q = \int T dS$.
For the given cycle:
$1$. Heat absorbed $(Q_1)$ during the expansion process (top diagonal line) is the area under the line from $S_0$ to $2S_0$:
$Q_1 = \text{Area of rectangle} + \text{Area of triangle} = (T_0 \times S_0) + \frac{1}{2} \times (T_0 \times S_0) = \frac{3}{2} T_0 S_0$.
$2$. Heat rejected $(Q_2)$ during the constant temperature process (bottom horizontal line) is the area under the line from $2S_0$ to $S_0$:
$Q_2 = T_0 \times (2S_0 - S_0) = T_0 S_0$.
$3$. The process $Q_3$ is an adiabatic process (vertical line),so $Q_3 = 0$.
$4$. The efficiency $\eta$ of the cycle is given by:
$\eta = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}$.
Substituting the values:
$\eta = 1 - \frac{T_0 S_0}{\frac{3}{2} T_0 S_0} = 1 - \frac{2}{3} = \frac{1}{3}$.

(B) For path $iaf$:
$Q = 50 \, cal$,$W = 20 \, cal$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
$\Delta U = 50 - 20 = 30 \, cal$.
Since internal energy is a state function,$\Delta U$ is the same for any path from $i$ to $f$.
$(i)$ For path $ibf$:
$Q = 36 \, cal$,$\Delta U = 30 \, cal$.
$W = Q - \Delta U = 36 - 30 = 6 \, cal$.
$(ii)$ For path $fi$ (reverse path):
$W = -13 \, cal$,$\Delta U_{fi} = -\Delta U_{if} = -30 \, cal$.
$Q = \Delta U + W = -30 + (-13) = -43 \, cal$.
$(iii)$ Given $E_{int,i} = 10 \, cal$:
$E_{int,f} = E_{int,i} + \Delta U = 10 + 30 = 40 \, cal$.

(A) The change in internal energy $\Delta U$ for an ideal gas is given by the formula $\Delta U = n C_v \Delta T$.
For a monoatomic ideal gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.
Given: $n = 2\, mol$,$\Delta T = 10\, K$ (since a change of $10\,^{\circ}C$ is equivalent to a change of $10\, K$),and $R = 8.31\, J/mol\cdot K$.
Substituting these values into the formula:
$\Delta U = 2 \times \left( \frac{3}{2} \times 8.31 \right) \times 10$
$\Delta U = 3 \times 8.31 \times 10$
$\Delta U = 24.93 \times 10 = 249.3\, J$.
Rounding to the nearest integer,the approximate change in internal energy is $250\, J$.

(A) Step $1$: Adiabatic compression from $V_{1}$ to $V_{2} = V_{1}/16$.
For an adiabatic process,$T_{1}V_{1}^{\gamma-1} = T_{2}V_{2}^{\gamma-1}$.
Given $T_{1} = 300 \; K$,$\gamma = 1.4 = 7/5$,so $\gamma-1 = 0.4 = 2/5$.
$300 \times V_{1}^{2/5} = T_{2} \times (V_{1}/16)^{2/5}$.
$T_{2} = 300 \times (16)^{2/5} = 300 \times (2^{4})^{2/5} = 300 \times 2^{8/5}$.
$T_{2} = 300 \times 3.0314 = 909.42 \; K$.
Step $2$: Isobaric expansion from $V_{2}$ to $2V_{2}$.
For an isobaric process,$V/T = \text{constant}$,so $V_{2}/T_{2} = (2V_{2})/T_{f}$.
$T_{f} = 2 \times T_{2} = 2 \times 909.42 = 1818.84 \; K$.
Rounding to the nearest integer,the final temperature is $1819 \; K$. Given the options,$1818$ is the closest match.

(D) In the given $P-V$ diagram:
$1$. Process $1 \rightarrow 2$ is an adiabatic expansion (pressure decreases, volume increases).
$2$. Process $2 \rightarrow 3$ is an isobaric compression (pressure is constant, volume decreases).
$3$. Process $3 \rightarrow 1$ is an isochoric heating (volume is constant, pressure increases).
Now, let's analyze the $V-T$ diagrams:
- For process $1 \rightarrow 2$ (adiabatic): $TV^{\gamma-1} = \text{constant}$. Since volume increases, temperature must decrease.
- For process $2 \rightarrow 3$ (isobaric): $V \propto T$. Since volume decreases, temperature must decrease.
- For process $3 \rightarrow 1$ (isochoric): $P \propto T$. Since pressure increases, temperature must increase.
Comparing these with the given options, the $V-T$ diagram that shows these trends is represented in option $D$.

(N/A) No. At $1 \; atm$ pressure and $-60^{\circ} C$,$CO_{2}$ is in the vapour phase. As it is compressed isothermally,it moves from the vapour region directly into the solid region,bypassing the liquid phase.
$(b)$ At $4 \; atm$ pressure,which is below the triple point pressure of $5.11 \; atm$,cooling $CO_{2}$ from room temperature causes it to transition directly from the vapour phase to the solid phase (deposition).
$(c)$ At $10 \; atm$ pressure,which is above the triple point pressure,heating solid $CO_{2}$ from $-65^{\circ} C$ causes it to first melt into the liquid phase at the fusion curve and then vaporise into the gas phase at the vaporisation curve as it reaches room temperature.
$(d)$ Since $70^{\circ} C$ is above the critical temperature of $CO_{2}$ $(31.1^{\circ} C)$,it cannot be liquefied by compression alone. It will remain in the vapour phase,but its density will increase,and it will deviate from ideal gas behaviour as pressure increases.

(A) For an adiabatic process,the heat exchange $\Delta Q = 0$. The work done on the system is $W_{adiabatic} = 22.3 \; J$. By the sign convention,work done on the system is negative,so $\Delta W = -22.3 \; J$.
From the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,we get $0 = \Delta U - 22.3 \; J$,which implies $\Delta U = 22.3 \; J$.
Since internal energy is a state function,$\Delta U$ remains the same for any process between states $A$ and $B$.
In the second process,the heat absorbed is $\Delta Q' = 9.35 \; cal = 9.35 \times 4.19 \; J = 39.1765 \; J$.
Using the first law again: $\Delta Q' = \Delta U + \Delta W'$,where $\Delta W'$ is the work done by the system.
$\Delta W' = \Delta Q' - \Delta U = 39.1765 \; J - 22.3 \; J = 16.8765 \; J$.
Rounding to two decimal places,the work done by the system is $16.88 \; J$.

(N/A) $1$. Work to Heat: In winter,when we rub our palms together,the mechanical work done in rubbing is converted into thermal energy (heat),which makes our palms feel warmer.
$2$. Heat to Work: In a steam engine,the thermal energy (heat) released from steam is used to perform mechanical work by pushing pistons,which in turn rotate the wheels of the train.

(N/A) Mechanics deals with the motion of particles or bodies under the action of forces and torques. It focuses on the mechanical state of a system,such as its position,velocity,and kinetic energy.
Thermodynamics,on the other hand,is concerned with the internal macroscopic state of a system,defined by state variables like pressure,volume,and temperature. It does not focus on the motion of the system as a whole.
For example,when a bullet is fired from a gun,its mechanical state (kinetic energy) changes. However,when the bullet hits a piece of wood and stops,its kinetic energy is converted into internal energy (heat),which changes the temperature of both the bullet and the wood.
In summary,temperature is related to the energy of the internal,disordered motion of particles,whereas mechanics is concerned with the ordered motion of the system as a whole.

(A) For a heat engine,the efficiency $\eta$ is defined as the ratio of work done to the heat absorbed from the source: $\eta = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1}$.
For a Carnot engine,this is also expressed in terms of temperatures as $\eta = \frac{T_1 - T_2}{T_1}$. Thus,$(a)$ matches with $(ii)$ and $(iii)$.
For a heat pump,the coefficient of performance $(COP)$ is defined as the ratio of heat extracted from the cold reservoir to the work done: $\text{COP} = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}$. Thus,$(b)$ matches with $(i)$.
Therefore,the correct matching is $(a-ii, iii)$ and $(b-i)$.

(B) The work done by a gas during expansion is given by the area under the $P-V$ curve.
$(1)$ For isothermal expansion $(a)$,the pressure $P$ decreases as volume $V$ increases according to $P = nRT/V$. The curve is a rectangular hyperbola.
$(2)$ For isobaric expansion $(b)$,the pressure $P$ remains constant at $P_i$. The curve is a horizontal line.
$(3)$ Since the gas expands from $V_i$ to $V_f$ starting from the same initial pressure $P_i$,the isobaric process maintains the pressure at $P_i$ throughout the expansion,whereas the isothermal process involves a decrease in pressure.
$(4)$ Graphically,the area under the horizontal line $(P = P_i)$ from $V_i$ to $V_f$ is a rectangle with area $P_i(V_f - V_i)$. The area under the isothermal curve is smaller because the pressure drops below $P_i$ during the process.
$(5)$ Therefore,the work done is greater in the isobaric expansion $(b)$.

(N/A) Heat is supplied during the processes where the internal energy and work done increase. In the isothermal expansion $B \to C$,heat is absorbed $(Q_{BC} = nRT_B \ln(V_C/V_B) > 0)$. In the isochoric heating $A \to B$,heat is absorbed $(Q_{AB} = nC_V(T_B - T_A) > 0)$. Thus,heat is supplied in $A \to B$ and $B \to C$.
$(b)$ The engine gives energy to the surroundings during the cooling processes: isochoric cooling $C \to D$ and isothermal compression $D \to A$.
$(c)$ Work done $W = W_{AB} + W_{BC} + W_{CD} + W_{DA}$. Since $AB$ and $CD$ are isochoric,$W_{AB} = W_{CD} = 0$. $W_{BC} = nRT_B \ln(V_C/V_B) = P_B V_B \ln(2)$. $W_{DA} = nRT_A \ln(V_A/V_D) = P_A V_A \ln(1/2) = -P_A V_A \ln(2)$. Total work $W = (P_B - P_A) V_A \ln(2)$.
$(d)$ Efficiency $\eta = W / Q_{in}$. $Q_{in} = Q_{AB} + Q_{BC} = C_V(T_B - T_A) + P_B V_B \ln(2) = \frac{3}{2}(P_B - P_A)V_A + P_B V_A \ln(2)$. $\eta = \frac{(P_B - P_A) V_A \ln(2)}{\frac{3}{2}(P_B - P_A)V_A + P_B V_A \ln(2)} = \frac{(P_B - P_A) \ln(2)}{\frac{3}{2}(P_B - P_A) + P_B \ln(2)}$.

(N/A) For a process,the heat exchanged is given by $Q = nC\Delta T$.
$(a)$ $A$ to $B$ (Isochoric): $W = 0$,so $Q_{AB} = \Delta U = nC_V(T_B - T_A) = \frac{3}{2}nR(T_B - T_A)$. Since $P_B > P_A$ and $V$ is constant,$T_B > T_A$,so heat is absorbed.
$(b)$ $B$ to $C$ (Isobaric): $Q_{BC} = nC_P(T_C - T_B) = n(\frac{5}{2}R)(T_C - T_B)$. Since $V_C > V_B$,$T_C > T_B$,so heat is absorbed.
$(c)$ $C$ to $D$ (Adiabatic): By definition,$Q_{CD} = 0$.
$(d)$ $D$ to $A$ (Isobaric): $Q_{DA} = nC_P(T_A - T_D) = n(\frac{5}{2}R)(T_A - T_D)$. Since $V_A < V_D$,$T_A < T_D$,so heat is released.

(N/A) The heat capacity $C$ for a process is given by $C = C_V + \frac{R}{1 - x}$,where $x$ is the polytropic index or related to the slope of the process curve.
For an adiabatic process,the slope is given by $(\frac{dP}{dV})_{ad} = -\gamma \frac{P_0}{V_0}$.
For the given process $P = f(V)$,let the slope be $(\frac{dP}{dV})_{proc}$.
We are given that $(\frac{dP}{dV})_{proc} > (\frac{dP}{dV})_{ad}$,which means $(\frac{dP}{dV})_{proc} > -\gamma \frac{P_0}{V_0}$.
The molar heat capacity for a general process is $C = C_V + \frac{P}{n(\frac{dT}{dV})}$.
Using the ideal gas law $PV = nRT$,we have $P + V(\frac{dP}{dV}) = nR(\frac{dT}{dV})$.
Substituting this into the heat capacity formula,we get $C = C_V + \frac{R}{1 - \frac{V}{P}(\frac{dP}{dV})}$.
Since the slope of the process is greater than the adiabatic slope,the denominator $(1 - \frac{V}{P}(\frac{dP}{dV}))$ becomes smaller than the denominator for the adiabatic process (which is $1 - (-\gamma) = 1 + \gamma$).
Specifically,if the slope is greater than the adiabatic slope,the process index $n_{poly}$ is less than $\gamma$.
Since $C = C_V \frac{\gamma - n_{poly}}{1 - n_{poly}}$,and given $n_{poly} < \gamma$,the heat capacity $C$ becomes positive.
Therefore,the gas absorbs heat during the expansion.

(B) The splitting of water,also known as water electrolysis,is the process of breaking down water molecules $(H_2O)$ into hydrogen gas $(H_2)$ and oxygen gas $(O_2)$.
This reaction requires an input of energy (usually in the form of electricity) to break the strong covalent bonds between hydrogen and oxygen atoms.
Since energy is absorbed from the surroundings to drive the reaction,it is classified as an endothermic process.

(N/A) Thermodynamic processes are defined as follows:
$(1)$ Isothermal process: The process during which the temperature of the system remains constant.
$(2)$ Isobaric process: The process during which the pressure of the system remains constant.
$(3)$ Isochoric process: The process during which the volume of the system remains constant.
$(4)$ Adiabatic process: $A$ process in which a system undergoes physical changes such that no heat is exchanged between the system and its surroundings $(\Delta Q = 0)$.
$(5)$ Cyclic process: $A$ process in which a system undergoes a series of changes that return the system back to its initial state.
Some special thermodynamic processes are summarized in the table below:

Type of process	Feature
Isothermal	Temperature constant
Isobaric	Pressure constant
Isochoric	Volume constant
Adiabatic	No heat flow $(\Delta Q = 0)$

(N/A) The $P-V$ diagram represents a Carnot cycle,which consists of two isothermal processes and two adiabatic processes.
$1$. Isothermal expansion: The gas expands at a constant temperature.
$2$. Adiabatic expansion: The gas expands without any heat exchange with the surroundings.
$3$. Isothermal compression: The gas is compressed at a constant temperature.
$4$. Adiabatic compression: The gas is compressed without any heat exchange with the surroundings.
The adiabatic curves are steeper than the isothermal curves because the adiabatic bulk modulus is $\gamma$ times the isothermal bulk modulus,where $\gamma > 1$.

(N/A) $1$. Isothermal Process: $A$ process in which the temperature of the system remains constant $(T = \text{constant})$. For an ideal gas, the internal energy remains constant $(dU = 0)$.
$2$. Adiabatic Process: $A$ process in which there is no exchange of heat between the system and its surroundings $(dQ = 0)$.
$3$. Isobaric Process: $A$ process in which the pressure of the system remains constant $(P = \text{constant})$.
$4$. First Law of Thermodynamics: For an ideal gas, the first law is given by $dQ = dU + dW$, where $dQ$ is the heat supplied to the system, $dU$ is the change in internal energy, and $dW$ is the work done by the system.

(N/A) $1$. Isochoric Process: A thermodynamic process in which the volume of the system remains constant $(dV = 0)$ is called an isochoric process. In this process, no work is done by or on the system $(W = 0)$.
$2$. Cyclic Process: A process is called cyclic if the system returns to its initial state after a series of changes. In a cyclic process, the change in internal energy is zero $(\Delta U = 0)$.
$3$. First Law of Thermodynamics: For an ideal gas, the first law of thermodynamics is given by $\Delta Q = \Delta U + \Delta W$, 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.

(N/A) The work done $W$ by or on a gas during a process is given by the integral $W = \int_{V_i}^{V_f} P \, dV$.
For an ideal gas undergoing an isothermal compression,the equation is $W = nRT \ln\left(\frac{V_f}{V_i}\right)$. Since it is compression,$V_f < V_i$,making $W$ negative (work done on the gas).
For an adiabatic compression,the equation is $W = \frac{nR(T_f - T_i)}{\gamma - 1} = \frac{P_f V_f - P_i V_i}{\gamma - 1}$,where $\gamma$ is the adiabatic index.

(B) In an external combustion engine,the fuel is burned outside the engine cylinder. The heat generated is used to convert water into high-pressure steam,which then acts as the working substance to perform mechanical work. Examples include steam engines and steam turbines.
In an internal combustion engine,the fuel is burned inside the engine cylinder itself. The combustion of the fuel creates high-pressure gases (a mixture of air and fuel),which expand to push the piston,acting as the working substance. Examples include petrol and diesel engines.

(N/A) The coefficient of performance $(COP)$, denoted by $\alpha$, for a refrigerator is defined as the ratio of heat extracted from the cold reservoir $(Q_2)$ to the work done $(W)$ on the system.
$\alpha = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}$
Dividing the numerator and denominator by $Q_1$:
$\alpha = \frac{Q_2 / Q_1}{1 - Q_2 / Q_1} \dots (1)$
For a Carnot engine, the efficiency $\eta$ is defined as:
$\eta = 1 - \frac{Q_2}{Q_1}$
Rearranging this gives:
$\frac{Q_2}{Q_1} = 1 - \eta \dots (2)$
Substituting equation $(2)$ into equation $(1)$:
$\alpha = \frac{1 - \eta}{\eta}$
Thus, the relation is $\alpha = \frac{1}{\eta} - 1$.

(A) For an isothermal process,the temperature change $\Delta T = 0$. Since the specific heat is defined as $C = \frac{\Delta Q}{m \Delta T}$,as $\Delta T \to 0$,the specific heat $C$ becomes infinite.
For an adiabatic process,there is no heat exchange with the surroundings,so $\Delta Q = 0$. Substituting this into the formula $C = \frac{\Delta Q}{m \Delta T}$,we get $C = 0$.

(N/A) $(i)$ It does not specify the direction in which heat flows. It only states that energy is conserved.
$(ii)$ It fails to explain why heat cannot be converted into work automatically and completely in a cyclic process without any external effect.

(A) $1.$ True. For an adiabatic process,the relation between temperature $T$ and volume $V$ is given by $T V^{\gamma - 1} = \text{constant}$.
$2.$ True. The charging of a battery is considered a reversible process if it is done infinitely slowly without any energy dissipation.
$3.$ False. Water falling from a height is an irreversible process because the kinetic energy gained is dissipated as heat due to friction and impact,which cannot be recovered.
$4.$ False. Internal energy,volume,and mass are extensive variables (they depend on the amount of matter),whereas pressure,temperature,and density are intensive variables (they are independent of the amount of matter).

(A) $1.$ True. In a cyclic process,the system returns to its initial state,so the change in state functions like internal energy is zero $(\Delta U = 0)$.
$2.$ False. In an adiabatic process,there is no exchange of heat $(Q = 0)$,but the temperature of the system changes as work is done.
$3.$ False. In an isothermal process,the temperature remains constant. Since internal energy of an ideal gas depends only on temperature,$\Delta U = 0$ for an isothermal process.

(N/A) $1.$ In a cyclic process,the system returns to its initial state. Since internal energy is a state function,the change in internal energy is $0$.
$2.$ The internal energy of a gas increases when work is done on the gas in an adiabatic process (where $Q = 0$,$\Delta U = -W$).
$3.$ For an adiabatic process,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$. Given $V_2 = V_1 / 32$ and $\gamma = 1.4$,we have $T_2 = T_1 (V_1 / V_2)^{\gamma-1} = T_1 (32)^{1.4-1} = T_1 (32)^{0.4} = T_1 (2^5)^{2/5} = T_1 \times 2^2 = 4 T_1$.
$4.$ The triple point of water occurs at a pressure of $4.58 \text{ mm Hg}$ and a temperature of $273.16 \text{ K}$.

(N/A) $1.$ The heat required to raise the temperature of a body by $1\,K$ is defined as its thermal capacity.
$2.$ The relation between temperature differences is $\Delta T_F = \frac{9}{5} \Delta T_C$. Given $\Delta T_C = 10\,^{\circ}C$,then $\Delta T_F = \frac{9}{5} \times 10 = 18\,^{\circ}F$.
$3.$ For an isothermal process,$PV = \text{constant}$. Differentiating both sides,we get $P dV + V dP = 0$. Rearranging gives $V dP = -P dV$,which simplifies to $\frac{dP}{P} = -\frac{dV}{V}$.

(A) For an isothermal process,the temperature $T$ remains constant. The work done is given by $W = \int_{V_1}^{V_2} P dV = \int_{V_1}^{V_2} \frac{\mu RT}{V} dV = \mu RT \ln \left( \frac{V_2}{V_1} \right) = 2.303 \mu RT \log_{10} \left( \frac{V_2}{V_1} \right)$. Thus,$(a)$ matches with $(iii)$.
For an adiabatic process,the work done is given by $W = \frac{\mu R(T_1 - T_2)}{\gamma - 1}$. Thus,$(b)$ matches with $(i)$.
Therefore,the correct matching is $(a-iii, b-i)$.

(A) The first law of thermodynamics is given by $\Delta Q = \Delta U + \Delta W$.
$(a)$ For an adiabatic process,there is no heat exchange with the surroundings,so $\Delta Q = 0$. Substituting this into the first law: $0 = \Delta U + \Delta W$,which implies $\Delta U = -\Delta W$. Thus,$(a)$ matches $(iii)$.
$(b)$ For an isothermal process,the temperature remains constant,which implies the change in internal energy $\Delta U = 0$. Substituting this into the first law: $\Delta Q = 0 + \Delta W$,which implies $\Delta Q = \Delta W$. Thus,$(b)$ matches $(ii)$.
Therefore,the correct matching is $(a-iii), (b-ii)$.

(A) For a heat engine,the efficiency $\eta$ is defined as the ratio of work done to heat input,given by $\eta = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1}$. For a Carnot engine,this simplifies to $\eta = \frac{T_1 - T_2}{T_1}$. Thus,$(a)$ matches with $(ii)$ and $(iii)$.
For a heat pump,the coefficient of performance is defined as $\beta = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}$. Thus,$(b)$ matches with $(i)$.

(B) The situation is shown in the provided $P-V$ graph,where the variation is shown for each process.
Process $1$ represents the isobaric expansion (constant pressure),and process $2$ represents the isothermal expansion (constant temperature).
Since the work done by a gas during expansion is equal to the area under the $P-V$ curve,we compare the areas under the two curves.
From the graph,the area under the curve for process $1$ (isobaric) is clearly larger than the area under the curve for process $2$ (isothermal).
Therefore,the work done by the gas is more in the isobaric process (case $b$).

(N/A) In process $AB$, volume is constant $(dV = 0)$, so work done $dW = 0$. From the first law of thermodynamics, $dQ = dU + dW = dU$. Since pressure increases at constant volume, temperature increases, so $dU > 0$. Thus, heat is supplied in process $AB$.
$(b)$ In process $CD$, volume is constant and pressure decreases, so temperature decreases. Thus, heat is released to the surroundings in process $CD$.
$(c)$ Work done $W = W_{AB} + W_{BC} + W_{CD} + W_{DA}$. Since $W_{AB} = 0$ and $W_{CD} = 0$, $W = W_{BC} + W_{DA}$.
For adiabatic processes, $W = \frac{P_i V_i - P_f V_f}{\gamma - 1}$.
$W_{BC} = \frac{P_B V_B - P_C V_C}{\gamma - 1}$ and $W_{DA} = \frac{P_D V_D - P_A V_A}{\gamma - 1}$.
Given $V_C = V_D = 2V_A = 2V_B$, and adiabatic relations $P_B V_B^\gamma = P_C V_C^\gamma$ and $P_A V_A^\gamma = P_D V_D^\gamma$:
$P_C = P_B(1/2)^{5/3}$ and $P_D = P_A(2)^{5/3}$.
$W = \frac{1}{\gamma - 1} [P_B V_B - P_B(1/2)^{5/3}(2V_B) + P_A(2)^{5/3}(2V_A) - P_A V_A]$
$W = \frac{V_A}{2/3} [P_B(1 - 2^{-2/3}) + P_A(2^{8/3} - 1)] = \frac{3V_A}{2} [P_B(1 - 2^{-2/3}) + P_A(2^{8/3} - 1)]$.
$(d)$ Efficiency $\eta = 1 - \frac{|Q_{out}|}{Q_{in}} = 1 - \frac{C_v(T_C - T_D)}{C_v(T_B - T_A)} = 1 - \frac{P_C V_C - P_D V_D}{P_B V_B - P_A V_A} = 1 - \frac{2(P_C - P_D)}{P_B - P_A} = 1 - \frac{2(P_B 2^{-5/3} - P_A 2^{5/3})}{P_B - P_A}$.