Mix Examples-Thermodynamics Questions in English

(C) For an ideal gas at constant pressure,the heat supplied is $\Delta Q = n C_p \Delta T$.
The work done is $W = P \Delta V = n R \Delta T$.
The change in internal energy is $\Delta U = n C_v \Delta T$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.
The fraction of heat converted into work is $\frac{W}{\Delta Q} = \frac{\Delta Q - \Delta U}{\Delta Q} = 1 - \frac{\Delta U}{\Delta Q}$.
Since $\Delta U = n C_v \Delta T$ and $\Delta Q = n C_p \Delta T$,we have $\frac{\Delta U}{\Delta Q} = \frac{C_v}{C_p} = \frac{1}{\gamma}$.
Thus,the fraction is $1 - \frac{1}{\gamma} = 1 - \frac{1}{5/3} = 1 - \frac{3}{5} = \frac{2}{5}$.

(D) In an isothermal process,the temperature remains constant,so $\Delta T = 0$. However,heat exchange $Q$ is not necessarily zero; it is equal to the work done $(Q = W)$.
In an adiabatic process,the heat exchange $Q$ is zero.
Therefore,the statement 'In an isothermal process,$Q = 0$' is incorrect.

(A) For the isothermal process,the relation is $P_1 V_1 = P_2 V_2$. Given $P_1 = P$,$V_1 = V$,and $V_2 = 2V$.
Substituting these values: $P \times V = P_2 \times 2V$,which gives $P_2 = \frac{P}{2}$.
For the adiabatic process,the relation is $P_2 V_2^\gamma = P_3 V_3^\gamma$. Given $V_2 = 2V$,$V_3 = 16V$,and $\gamma = \frac{5}{3}$.
Substituting these values: $P_3 = P_2 \left(\frac{V_2}{V_3}\right)^\gamma = \frac{P}{2} \left(\frac{2V}{16V}\right)^{5/3}$.
$P_3 = \frac{P}{2} \left(\frac{1}{8}\right)^{5/3} = \frac{P}{2} \times \left(\left(\frac{1}{2}\right)^3\right)^{5/3} = \frac{P}{2} \times \left(\frac{1}{2}\right)^5 = \frac{P}{2} \times \frac{1}{32} = \frac{P}{64}$.

(D) At constant pressure, the work done is given by $W = P \Delta V$.
For an ideal gas, $PV = nRT$, so $P \Delta V = nR \Delta T$.
Thus, $W = nR \Delta T$.
Heat supplied at constant volume is given by $Q = \Delta U = nC_v \Delta T$.
For a monoatomic gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$.
Substituting this into the heat equation, we get $Q = n \left( \frac{3}{2}R \right) \Delta T = \frac{3}{2} (nR \Delta T)$.
Since $W = nR \Delta T$, we substitute $W$ into the equation for $Q$:
$Q = \frac{3}{2} W$.

(B) For isothermal expansion,the process follows Boyle's Law: $P_1 V_1 = P_2 V_2$.
Given $P_1 = P$,$V_1 = V$,and $V_2 = 2V$.
Therefore,$P_i = P \times (V / 2V) = P / 2$.
For adiabatic expansion,the process follows the relation $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.
Given $P_1 = P$,$V_1 = V$,and $V_2 = 2V$.
Therefore,$P_a = P \times (V / 2V)^{\gamma} = P / 2^{\gamma}$.
Now,calculating the ratio $\frac{P_i}{P_a}$:
$\frac{P_i}{P_a} = \frac{P/2}{P/2^{\gamma}} = \frac{2^{\gamma}}{2} = 2^{\gamma-1}$.

(C) Given: $\gamma = \frac{C_p}{C_v} = \frac{3}{2}$.
Case $I$: Isothermal process $(T = \text{constant})$
For an isothermal process,$P_1 V_1 = P_2 V_2$.
Given $P_1 = P$,$V_1 = V$,and $V_2 = 9V$.
$P \times V = P_2 \times 9V$
$P_2 = \frac{P}{9}$.
Case $II$: Adiabatic process $(PV^\gamma = \text{constant})$
For an adiabatic process,$P_2 V_2^\gamma = P_3 V_3^\gamma$.
Given $P_2 = \frac{P}{9}$,$V_2 = 9V$,and $V_3 = V$.
$\frac{P}{9} \times (9V)^{3/2} = P_3 \times (V)^{3/2}$
$P_3 = \frac{P}{9} \times \left(\frac{9V}{V}\right)^{3/2}$
$P_3 = \frac{P}{9} \times (9)^{3/2}$
$P_3 = \frac{P}{9} \times (3^2)^{3/2} = \frac{P}{9} \times 3^3 = \frac{P}{9} \times 27 = 3P$.
Therefore,the final pressure is $3P$.

(B) For a process at constant pressure,the heat supplied is given by $\Delta Q = n C_p \Delta T$.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
The work done is given by $\Delta W = P \Delta V = n R \Delta T$.
Thus,the ratio $\Delta Q : \Delta U : \Delta W = C_p : C_v : R$.
For a rigid diatomic gas,the degrees of freedom $f = 5$.
Therefore,$C_v = \frac{f}{2} R = \frac{5}{2} R$ and $C_p = C_v + R = \frac{7}{2} R$.
Substituting these values,we get $\Delta Q : \Delta U : \Delta W = \frac{7}{2} R : \frac{5}{2} R : R$.
Multiplying by $\frac{2}{R}$,we obtain the ratio $7 : 5 : 2$.

(D) Given: $\gamma = \frac{5}{3}$.
Step $1$: Isothermal expansion from volume $V$ to $2V$.
For an isothermal process,$P_1 V_1 = P_2 V_2$.
$P \times V = P_2 \times 2V$.
Therefore,$P_2 = \frac{P}{2}$.
Step $2$: Adiabatic expansion from volume $2V$ to $16V$.
For an adiabatic process,$P_2 V_2^{\gamma} = P_3 V_3^{\gamma}$.
$P_3 = P_2 \left( \frac{V_2}{V_3} \right)^{\gamma}$.
$P_3 = \frac{P}{2} \left( \frac{2V}{16V} \right)^{5/3} = \frac{P}{2} \left( \frac{1}{8} \right)^{5/3}$.
$P_3 = \frac{P}{2} \left( (2^{-3})^{5/3} \right) = \frac{P}{2} \times 2^{-5} = \frac{P}{2 \times 32} = \frac{P}{64}$.

(A) Since the gas flow is suddenly stopped in an insulated tube,the process is adiabatic.
The kinetic energy of the gas is given by $K.E. = \frac{1}{2}m(2V)^2 = 2mV^2$.
In an adiabatic process,the work done to compress the gas is equal to the change in internal energy,which is given by $\Delta U = \frac{nR\Delta T}{\gamma-1}$.
Since the number of moles $n = \frac{m}{M}$,we have $\Delta U = \frac{mR\Delta T}{M(\gamma-1)}$.
Equating the kinetic energy to the change in internal energy:
$2mV^2 = \frac{mR\Delta T}{M(\gamma-1)}$.
Solving for $\Delta T$:
$\Delta T = \frac{2MV^2(\gamma-1)}{R}$.

(A) Given: Heat supplied at constant pressure,$Q_p = 80 \ J$. Work done by the gas,$W = 20 \ J$.
According to the First Law of Thermodynamics,$Q_p = \Delta U + W$,where $\Delta U$ is the change in internal energy.
$\Delta U = Q_p - W = 80 \ J - 20 \ J = 60 \ J$.
We know that $Q_p = n C_p \Delta T$ and $\Delta U = n C_v \Delta T$.
Therefore,the ratio $\frac{Q_p}{\Delta U} = \frac{n C_p \Delta T}{n C_v \Delta T} = \frac{C_p}{C_v} = \gamma$.
Substituting the values,$\gamma = \frac{80}{60} = \frac{4}{3}$.
Thus,the ratio of the specific heat capacities is $4/3$.

(B) The change in internal energy $\Delta U$ of an ideal gas is given by the formula $\Delta U = n C_v \Delta T$.
We know that $C_v = \frac{R}{\gamma-1}$.
Substituting this,we get $\Delta U = n \left( \frac{R}{\gamma-1} \right) \Delta T = \frac{n R \Delta T}{\gamma-1}$.
From the ideal gas equation at constant pressure,$PV = nRT$,so $n R \Delta T = P \Delta V$.
Here,the change in volume is $\Delta V = 3V - V = 2V$.
Therefore,$n R \Delta T = P(2V) = 2PV$.
Substituting this into the internal energy equation: $\Delta U = \frac{2PV}{\gamma-1}$.

(C) Given: Heat supplied $Q = 100 \ J$,Change in internal energy $\Delta U = 60 \ J$.
According to the First Law of Thermodynamics,$Q = \Delta U + W$,so the work done $W = Q - \Delta U = 100 - 60 = 40 \ J$.
We know that $\Delta U = n C_v \Delta T$ and $Q = n C_p \Delta T$.
Therefore,$\frac{Q}{\Delta U} = \frac{C_p}{C_v} = \gamma$.
Substituting the values,$\gamma = \frac{100}{60} = \frac{5}{3} \approx 1.67$.
Since the adiabatic index $\gamma = 1.67$ corresponds to a monoatomic gas,the gas is monoatomic.

(C) The internal energy $U$ of an ideal gas depends only on its temperature $T$.
For a monoatomic gas,the degrees of freedom $f = 3$.
The change in internal energy is given by the formula $\Delta U = \frac{n f R \Delta T}{2}$.
$(i)$ At constant volume: $\Delta U_1 = \frac{n(3)R(T_2 - T_1)}{2} = \frac{3}{2} nR(T_2 - T_1)$.
(ii) At constant pressure: $\Delta U_2 = \frac{n(3)R(T_2 - T_1)}{2} = \frac{3}{2} nR(T_2 - T_1)$.
Since the change in temperature $\Delta T = T_2 - T_1$ is the same in both cases,the change in internal energy is the same in both cases.

(A) For a monatomic gas,the degrees of freedom $f = 3$.
$C_p = \frac{5}{2}R$ and $C_v = \frac{3}{2}R$.
Total heat supplied at constant pressure is $Q = n C_p \Delta T = \frac{5}{2} nR \Delta T$.
Heat used to increase internal energy is $\Delta U = n C_v \Delta T = \frac{3}{2} nR \Delta T$.
Percentage of heat used to increase internal energy $= \frac{\Delta U}{Q} \times 100 = \frac{\frac{3}{2} nR \Delta T}{\frac{5}{2} nR \Delta T} \times 100 = 60\%$.
Heat used to do external work is $W = P \Delta V = nR \Delta T$.
Percentage of heat used to do external work $= \frac{W}{Q} \times 100 = \frac{nR \Delta T}{\frac{5}{2} nR \Delta T} \times 100 = 40\%$.
Thus,the percentages are $40\%$ and $60\%$ respectively.

(A) For an isobaric process,the heat supplied is $\Delta Q = C_p \Delta T$ and the change in internal energy is $\Delta U = C_V \Delta T$.
Thus,$\frac{\Delta Q}{\Delta U} = \frac{C_p \Delta T}{C_V \Delta T} = \frac{C_p}{C_V} = \gamma$. So,$A \rightarrow 4$.
For an isobaric process,$\Delta Q = \Delta U + \Delta W$,so $\Delta W = \Delta Q - \Delta U = (C_p - C_V) \Delta T$.
Thus,$\frac{\Delta Q}{\Delta W} = \frac{C_p \Delta T}{(C_p - C_V) \Delta T} = \frac{C_p}{C_p - C_V} = \frac{C_p/C_V}{(C_p/C_V) - 1} = \frac{\gamma}{\gamma - 1}$. So,$B \rightarrow 3$.
For a refrigerator,the coefficient of performance $\beta = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2} = \frac{T_2}{T_1 - T_2}$. So,$C \rightarrow 2$.
For a heat pump,the coefficient of performance $\alpha = \frac{Q_1}{W} = \frac{Q_1}{Q_1 - Q_2} = \frac{T_1}{T_1 - T_2}$. So,$D \rightarrow 1$.
Therefore,the correct match is $A \rightarrow 4, B \rightarrow 3, C \rightarrow 2, D \rightarrow 1$.

$(A)$ The gas is compressed from $V_1 = 2 \,m^3$ to $V_2 = 1 \,m^3$ at a constant pressure $P = 100 \,N m^{-2}$.
Work done on the gas is $W = -P \Delta V = -100 \times (1 - 2) = 100 \,J$.
Note: Work done $BY$ the gas is $P \Delta V = -100 \,J$,so work done $ON$ the gas is $+100 \,J$.
Heat supplied to the system is $Q = 150 \,J$.
According to the first law of thermodynamics,$\Delta U = Q + W_{on}$.
$\Delta U = 150 \,J + 100 \,J = 250 \,J$.
Therefore,the internal energy increases by $250 \,J$.

(D) Given the relation $V = k T^{2/3}$.
From the ideal gas law,$PV = RT$ (for $1 \ mole$),so $P = \frac{RT}{V}$.
The work done is given by $W = \int P \ dV = \int \frac{RT}{V} \ dV$.
From $V = k T^{2/3}$,we differentiate with respect to $T$:
$dV = k \cdot \frac{2}{3} T^{-1/3} \ dT$.
Substituting $V$ and $dV$ into the work integral:
$W = \int \frac{RT}{k T^{2/3}} \cdot (k \cdot \frac{2}{3} T^{-1/3} \ dT) = \int R \cdot \frac{2}{3} \cdot \frac{T \cdot T^{-1/3}}{T^{2/3}} \ dT = \int \frac{2}{3} R \ dT$.
Given the change in temperature $\Delta T = 60 \ K$,the work done is:
$W = \frac{2}{3} R \int_{T_1}^{T_2} dT = \frac{2}{3} R \Delta T = \frac{2}{3} R (60) = 40 R$.

(D) Given the relation $V = K T^2$.
Using the ideal gas equation $PV = RT$ for one mole of gas,we have $P = \frac{RT}{V}$.
Substituting $V = K T^2$ into the expression for $P$,we get $P = \frac{RT}{K T^2} = \frac{R}{KT}$.
The work done $W$ is given by $W = \int P dV$.
Since $V = K T^2$,differentiating with respect to $T$ gives $dV = 2KT dT$.
Substituting $P$ and $dV$ into the work integral:
$W = \int_{T_1}^{T_2} \left( \frac{R}{KT} \right) (2KT dT) = \int_{T_1}^{T_2} 2R dT$.
Given the change in temperature $\Delta T = T_2 - T_1 = 60 \text{ K}$.
Therefore,$W = 2R (T_2 - T_1) = 2R(60) = 120R$.

(C) The efficiency of a Carnot heat engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$.
The coefficient of performance of a Carnot refrigerator is given by $\beta = \frac{T_2}{T_1 - T_2}$.
We need to find the ratio of the efficiency of the heat engine to the coefficient of performance of the refrigerator:
Ratio $= \frac{\eta}{\beta} = \frac{\frac{T_1 - T_2}{T_1}}{\frac{T_2}{T_1 - T_2}}$.
Simplifying the expression:
Ratio $= \frac{T_1 - T_2}{T_1} \times \frac{T_1 - T_2}{T_2} = \frac{(T_1 - T_2)^2}{T_1 T_2}$.
Thus,the correct option is $C$.

(B) For a heat engine,efficiency is defined as $\eta = \frac{W}{Q_H} = 1 - \frac{Q_L}{Q_H}$.
For a refrigerator,the coefficient of performance is defined as $\alpha = \frac{Q_L}{W} = \frac{Q_L}{Q_H - Q_L}$.
Taking the reciprocal,$\frac{1}{\alpha} = \frac{Q_H - Q_L}{Q_L} = \frac{Q_H}{Q_L} - 1$.
Thus,$\frac{Q_H}{Q_L} = 1 + \frac{1}{\alpha} = \frac{\alpha + 1}{\alpha}$.
Therefore,$\frac{Q_L}{Q_H} = \frac{\alpha}{1 + \alpha}$.
Substituting this into the efficiency formula: $\eta = 1 - \frac{Q_L}{Q_H} = 1 - \frac{\alpha}{1 + \alpha} = \frac{1 + \alpha - \alpha}{1 + \alpha} = \frac{1}{1 + \alpha}$.

(A) In the given $P-V$ diagram:
$1$. Process $I$ is a vertical line,meaning volume $V$ is constant. This is an isochoric process. Thus,$I-c$.
$2$. Process $IV$ is a horizontal line,meaning pressure $P$ is constant. This is an isobaric process. Thus,$IV-b$.
$3$. Processes $II$ and $III$ are curves representing changes in both $P$ and $V$. Since $II$ connects the $300 \ K$ and $500 \ K$ isotherms and $III$ connects the $500 \ K$ and $700 \ K$ isotherms,they represent adiabatic processes (as they cross isotherms). Specifically,$II$ is adiabatic and $III$ is isothermal (or vice versa depending on the specific slope analysis). However,looking at the options,$I-c$ and $IV-b$ are fixed. Comparing with options,$I-c, II-a, III-d, IV-b$ fits the standard interpretation where $II$ is adiabatic and $III$ is isothermal. Therefore,the correct match is $I-c, II-a, III-d, IV-b$.

(C) Initial state: $P_1 = P$,$V_1 = V$.
Step $1$: Isothermal expansion to $V_2 = 27 V$.
For an isothermal process,$P_1 V_1 = P_2 V_2$.
$P_2 = \frac{P_1 V_1}{V_2} = \frac{P \times V}{27 V} = \frac{P}{27}$.
Step $2$: Adiabatic compression to $V_3 = V$.
For a monatomic gas,the adiabatic index $\gamma = \frac{5}{3}$.
For an adiabatic process,$P_2 V_2^\gamma = P_3 V_3^\gamma$.
$P_3 = P_2 \left( \frac{V_2}{V_3} \right)^\gamma = \frac{P}{27} \left( \frac{27 V}{V} \right)^{5/3}$.
$P_3 = \frac{P}{27} \times (27)^{5/3} = \frac{P}{27} \times (3^3)^{5/3} = \frac{P}{27} \times 3^5$.
$P_3 = \frac{P}{27} \times 243 = 9 P$.
The final pressure of the gas is $9 P$.

(C) The change in internal energy $\Delta U$ is a state function,meaning it depends only on the initial and final states.
For an isothermal process,the temperature remains constant,so the change in internal energy $\Delta U_{\text{iso}} = 0$.
Given the total change in internal energy $\Delta U_{\text{total}} = -30 J$,and since the path consists of two isothermal processes and one adiabatic process,we have $\Delta U_{\text{total}} = \Delta U_{\text{iso1}} + \Delta U_{\text{adiabatic}} + \Delta U_{\text{iso2}}$.
Substituting the values: $-30 J = 0 + \Delta U_{\text{adiabatic}} + 0$,which gives $\Delta U_{\text{adiabatic}} = -30 J$.
For an adiabatic process,the first law of thermodynamics states $\Delta Q = \Delta U + W$. Since $\Delta Q = 0$,we have $W_{\text{adiabatic}} = -\Delta U_{\text{adiabatic}}$.
Therefore,$W_{\text{adiabatic}} = -(-30 J) = 30 J$.

(B) The slope of the isothermal process $(S_I)$ is given by $\frac{dp}{dV} = -\frac{p}{V}$.
The slope of the adiabatic process $(S_A)$ is given by $\frac{dp}{dV} = -\gamma \frac{p}{V}$.
Therefore,the relationship between the slopes is $S_A = \gamma \times S_I$.
This implies $\frac{S_I}{S_A} = \frac{1}{\gamma}$.
Given the heat capacity ratio $\gamma = \frac{3}{2}$,we substitute this value into the expression:
$\frac{S_I}{S_A} = \frac{1}{3/2} = \frac{2}{3}$.

(A) For an ideal monoatomic gas,the ratio of specific heats is $\gamma = \frac{5}{3}$.
For the adiabatic process $A \rightarrow B$:
$p_A V_A^\gamma = p_B V_B^\gamma$
Substituting the values from the graph ($p_A = 32 \times 10^5 \text{ Pa}$,$p_B = 1 \times 10^5 \text{ Pa}$,$V_B = 1 \text{ m}^3$):
$32 \times 10^5 \times V_A^{5/3} = 1 \times 10^5 \times (1)^{5/3}$
$V_A^{5/3} = \frac{1}{32} = (2^{-5}) \Rightarrow V_A = (2^{-5})^{3/5} = 2^{-3} = \frac{1}{8} \text{ m}^3$.
For the adiabatic process $C \rightarrow D$:
$p_C V_C^\gamma = p_D V_D^\gamma$
Substituting the values from the graph ($p_C = 1 \times 10^5 \text{ Pa}$,$V_C = 8 \text{ m}^3$,$p_D = 32 \times 10^5 \text{ Pa}$):
$1 \times 10^5 \times (8)^{5/3} = 32 \times 10^5 \times V_D^{5/3}$
$(2^3)^{5/3} = 32 \times V_D^{5/3} \Rightarrow 2^5 = 32 \times V_D^{5/3} \Rightarrow 32 = 32 \times V_D^{5/3} \Rightarrow V_D = 1 \text{ m}^3$.
The total work done in the cycle $A \rightarrow B \rightarrow C \rightarrow D \rightarrow A$ is:
$W = W_{AB} + W_{BC} + W_{CD} + W_{DA}$
$W_{AB} = \frac{p_A V_A - p_B V_B}{\gamma - 1} = \frac{(32 \times 10^5 \times 1/8) - (1 \times 10^5 \times 1)}{5/3 - 1} = \frac{(4 - 1) \times 10^5}{2/3} = 4.5 \times 10^5 \text{ J}$.
$W_{BC} = p_B(V_C - V_B) = 1 \times 10^5 \times (8 - 1) = 7 \times 10^5 \text{ J}$.
$W_{CD} = \frac{p_C V_C - p_D V_D}{\gamma - 1} = \frac{(1 \times 10^5 \times 8) - (32 \times 10^5 \times 1)}{5/3 - 1} = \frac{(8 - 32) \times 10^5}{2/3} = -36 \times 10^5 \text{ J}$.
$W_{DA} = p_D(V_A - V_D) = 32 \times 10^5 \times (1/8 - 1) = 32 \times 10^5 \times (-7/8) = -28 \times 10^5 \text{ J}$.
Total work $W = (4.5 + 7 - 36 - 28) \times 10^5 \text{ J} = -52.5 \times 10^5 \text{ J}$.

(C) The internal energy of a monoatomic ideal gas is $U = \frac{3}{2} n R T$,so $U \propto T$.
Density $\rho = \frac{m}{V}$,so $\rho \propto \frac{1}{V}$.
For process $AB$,the line passes through origin in $U$-$\rho$ plot,so $U = k \rho$. Substituting $U \propto T$ and $\rho \propto 1/V$,we get $T \propto 1/V$,or $TV = \text{constant}$.
Using $PV = nRT$,we have $(PV/nR)V = \text{constant}$,so $PV^2 = \text{constant}$. This is a polytropic process with index $m = 2$.
The molar heat capacity is $C = C_V + \frac{R}{1-m} = \frac{3R}{2} + \frac{R}{1-2} = \frac{3R}{2} - R = \frac{R}{2}$. Thus,option $A$ is correct.
In process $BC$,$\rho$ is constant,so it is an isochoric process. Since $U$ decreases,temperature decreases,so heat is rejected. Thus,option $B$ is correct.
For an isochoric process,the molar heat capacity is $C_V = \frac{3}{2} R$. Option $C$ states it is $\frac{2R}{3}$,which is incorrect.
In process $CA$,$U$ is constant,so $T$ is constant (isothermal). $W_{CA} = nRT \ln(V_A/V_C) = nRT \ln(\rho_C/\rho_A) = (\frac{2}{3} U_0) \ln(4\rho_0/\rho_0) = \frac{2}{3} U_0 \ln 4$. Option $D$ is correct.

(A) Initial state: $V_A = V_B = V$,$p_A = p_B = p$.
For container $A$ (isothermal process): $p_A V_A = p_A^{\prime} V_A^{\prime}$.
Given $V_A^{\prime} = V/2$,so $p V = p_A^{\prime} (V/2) \Rightarrow p_A^{\prime} = 2p$.
For container $B$ (adiabatic process): $p_B V_B^{\gamma} = p_B^{\prime} (V_B^{\prime})^{\gamma}$.
Given $V_B^{\prime} = V/2$,so $p V^{\gamma} = p_B^{\prime} (V/2)^{\gamma} \Rightarrow p_B^{\prime} = p \cdot 2^{\gamma}$.
The ratio of final pressures is $\frac{p_B^{\prime}}{p_A^{\prime}} = \frac{p \cdot 2^{\gamma}}{2p} = 2^{\gamma-1}$.

(C) For an isothermal process,the equation is $pV = K$.
Differentiating both sides with respect to $V$,we get $p + V \frac{dp}{dV} = 0$,which implies $\left(\frac{dp}{dV}\right)_{\text{iso}} = -\frac{p}{V}$.
For an adiabatic process,the equation is $pV^{\gamma} = K'$.
Differentiating both sides with respect to $V$,we get $\frac{dp}{dV} V^{\gamma} + p \gamma V^{\gamma-1} = 0$.
Rearranging this gives $\left(\frac{dp}{dV}\right)_{\text{adia}} = -\gamma \frac{p}{V}$.
Taking the ratio of the slope of the isothermal curve to the slope of the adiabatic curve:
$\frac{(\frac{dp}{dV})_{\text{iso}}}{(\frac{dp}{dV})_{\text{adia}}} = \frac{-p/V}{-\gamma p/V} = \frac{1}{\gamma}$.

(C) The given graph is a $P-T$ diagram. The work done by an ideal gas in a cycle is given by the area enclosed by the cycle in a $P-V$ diagram. However, for a $P-T$ diagram, we can use the relation $PV = nRT$.
For an ideal gas, the work done in a process is $W = \int P dV$.
From the ideal gas equation, $V = \frac{nRT}{P}$, so $dV = \frac{nR}{P} dT - \frac{nRT}{P^2} dP$.
In the given $P-T$ diagram, the processes $AB$ and $CD$ are isochoric (since they lie on lines passing through the origin, $P \propto T \implies V = \text{constant}$), and processes $BC$ and $DA$ are isobaric (since $P$ is constant).
Work done in isochoric processes ($AB$ and $CD$) is $0$.
Work done in isobaric processes is $W = P \Delta V = nR \Delta T$.
For process $BC$ (isobaric at $P_B$): $W_{BC} = nR(T_C - T_B) = 3R(2400 - 800) = 3R(1600) = 4800R$.
For process $DA$ (isobaric at $P_A$): $W_{DA} = nR(T_A - T_D) = 3R(400 - 1200) = 3R(-800) = -2400R$.
Total work done $W_{net} = W_{AB} + W_{BC} + W_{CD} + W_{DA} = 0 + 4800R + 0 - 2400R = 2400R$.

(C) The correct statements are $1$ and $4$.
$1$. For an isothermal process, $pV = \text{constant}$. Differentiating with respect to $V$, we get $p + V \frac{dp}{dV} = 0$, which gives the slope $\frac{dp}{dV} = -\frac{p}{V}$. This is correct.
$2$. For an adiabatic process, $pV^{\gamma} = \text{constant}$. Differentiating with respect to $V$, we get $\frac{dp}{dV} V^{\gamma} + p \gamma V^{\gamma-1} = 0$, which gives the slope $\frac{dp}{dV} = -\gamma \frac{p}{V}$. Thus, statement $2$ is incorrect.
$3$. In an isochoric process, volume $V$ is constant, so $dV = 0$. The graph is a vertical line, and the slope $\frac{dp}{dV}$ is undefined (or $\infty$). Thus, statement $3$ is incorrect.
$4$. In an isobaric process, pressure $p$ is constant, so $dp = 0$. The graph is a horizontal line parallel to the volume axis, and the slope $\frac{dp}{dV} = 0$. This is correct.

(A) Heat is absorbed by the system when $\Delta Q > 0$. In a cyclic process,heat is absorbed during the segments where the internal energy increases and work is done by the gas. Looking at the cycle $ABCDA$:
$1$. Path $A \rightarrow B$ (isochoric): $W = 0$,$\Delta U = n C_V \Delta T = \frac{3}{2} V_0 (3p_0 - p_0) = 3 p_0 V_0$. Since $\Delta U > 0$,heat is absorbed: $\Delta Q_{AB} = 3 p_0 V_0$.
$2$. Path $B \rightarrow C$ (isobaric): $W = p \Delta V = 3p_0 (2V_0 - V_0) = 3 p_0 V_0$. $\Delta U = n C_V \Delta T = \frac{3}{2} (p_C V_C - p_B V_B) = \frac{3}{2} (6 p_0 V_0 - 3 p_0 V_0) = 4.5 p_0 V_0$. Heat is absorbed: $\Delta Q_{BC} = \Delta U + W = 4.5 p_0 V_0 + 3 p_0 V_0 = 7.5 p_0 V_0$.
$3$. Path $C \rightarrow D$ and $D \rightarrow A$: The gas releases heat as internal energy decreases and work is done on the gas.
Total heat absorbed = $\Delta Q_{AB} + \Delta Q_{BC} = 3 p_0 V_0 + 7.5 p_0 V_0 = 10.5 p_0 V_0$.

(D) The given equation is $V = \frac{aT^3}{P}$.
For a constant pressure $P$,the initial volume is $V_1 = \frac{aT^3}{P}$.
When the temperature is doubled $(T_2 = 2T)$,the new volume becomes $V_2 = \frac{a(2T)^3}{P} = \frac{8aT^3}{P}$.
The work done by the gas at constant pressure is given by $W = P \Delta V = P(V_2 - V_1)$.
Substituting the values,we get $W = P \left( \frac{8aT^3}{P} - \frac{aT^3}{P} \right) = P \left( \frac{7aT^3}{P} \right) = 7aT^3$.

(A) The work done $W$ in different thermodynamic processes is given by:
$1$. Isothermal process: $W = \int_{V_1}^{V_2} P dV = \mu RT \ln(\frac{V_2}{V_1})$. Thus,$i-c$.
$2$. Isobaric process: Pressure $P$ is constant,so $W = P \int_{V_1}^{V_2} dV = P(V_2 - V_1)$. Thus,$ii-d$.
$3$. Isochoric process: Volume $V$ is constant,so $dV = 0$,which implies $W = 0$. Thus,$iii-a$.
$4$. Adiabatic process: $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{1}{\gamma - 1}[P_2 V_2 - P_1 V_1]$ (with a sign convention adjustment). Thus,$iv-b$.
Therefore,the correct matching is $i-c, ii-d, iii-a, iv-b$.

(A) Initial state: $n = 2 \ mol$,$T_1 = 27^{\circ} C = 300 \ K$,$P_1 = P$,$V_1 = V$.
Step $1$ (Isobaric expansion): Volume doubles,so $V_2 = 2V$. Since $P$ is constant,$\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies T_2 = T_1 \left(\frac{V_2}{V_1}\right) = 300 \times 2 = 600 \ K$.
Step $2$ (Adiabatic process): The gas expands/compresses until $T_3 = T_1 = 300 \ K$.
Work done in an adiabatic process is $W = \frac{nR(T_2 - T_3)}{\gamma - 1}$.
Given $\gamma = \frac{5}{3}$,so $\gamma - 1 = \frac{2}{3}$.
$W = \frac{2 \times 8.3 \times (600 - 300)}{2/3} = \frac{2 \times 8.3 \times 300 \times 3}{2} = 8.3 \times 900 = 7470 \ J$.

(D) The given $p-V$ diagram shows a clockwise cyclic process $A \rightarrow B \rightarrow C \rightarrow D \rightarrow A$.
$1$. $A \rightarrow B$: Isobaric expansion (pressure $p$ is constant,volume $V$ increases,so temperature $T$ increases).
$2$. $B \rightarrow C$: Isothermal compression ($p \propto 1/V$,so $T$ is constant,volume $V$ decreases).
$3$. $C \rightarrow D$: Isochoric compression (volume $V$ is constant,pressure $p$ decreases,so temperature $T$ decreases).
$4$. $D \rightarrow A$: Isothermal expansion ($p \propto 1/V$,so $T$ is constant,volume $V$ increases).
Comparing these processes with the given options,the $p-T$ diagram in option $D$ correctly represents this cycle: $A \rightarrow B$ (isobaric,$T$ increases),$B \rightarrow C$ (isothermal,$p$ decreases),$C \rightarrow D$ (isochoric,$T$ decreases),$D \rightarrow A$ (isothermal,$p$ increases). Thus,the correct representation is option $D$.

$(B)$. Zeroth law of thermodynamics defines thermal equilibrium between systems in contact $(A-III)$.
$B$. First law of thermodynamics is based on the law of conservation of energy $(B-IV)$.
$C$. In the free expansion of a gas, no work is done by the gas against an external pressure, so work done is zero $(C-II)$.
$D$. Second law of thermodynamics provides the criteria for the direction of heat flow $(D-I)$.
Therefore, the correct matching is $A-III, B-IV, C-II, D-I$.

(B) The correct matches are as follows:
$A$. Steam engine operates on the $\text{Laws}$ $\text{of}$ $\text{thermodynamics}$ $(II)$.
$B$. Electron microscope utilizes the $\text{Wave}$ $\text{nature}$ $\text{of}$ $\text{electrons}$ $(III)$.
$C$. Non-reflecting coatings are based on the $\text{Interference}$ $\text{of}$ $\text{light}$ $(IV)$.
$D$. Tokamak uses $\text{Magnetic}$ $\text{confinement}$ $\text{of}$ $\text{plasma}$ $(I)$.
Therefore, the correct sequence is $A-II, B-III, C-IV, D-I$.

(B) The heat supplied $Q$ is equal to the change in internal energy $\Delta U$ since the process is at constant temperature and the cylinder is insulating (no work done by expansion against external pressure,assuming rigid walls):
$Q = U_f - U_i$.
Initially,the cylinder contains $4 \text{ moles}$ of diatomic gas $(f = 5)$.
$U_i = n \left( \frac{f}{2} RT \right) = 4 \left( \frac{5}{2} RT \right) = 10 RT$.
After dissociation,$2 \text{ moles}$ of diatomic gas break into $4 \text{ moles}$ of monoatomic gas $(f = 3)$. The remaining $2 \text{ moles}$ stay diatomic.
Final state: $4 \text{ moles}$ monoatomic $(f = 3)$ and $2 \text{ moles}$ diatomic $(f = 5)$.
$U_f = n_{mono} \left( \frac{3}{2} RT \right) + n_{dia} \left( \frac{5}{2} RT \right) = 4 \left( \frac{3}{2} RT \right) + 2 \left( \frac{5}{2} RT \right) = 6 RT + 5 RT = 11 RT$.
Thus,$Q = U_f - U_i = 11 RT - 10 RT = RT$.

(D) For a diatomic gas,the degrees of freedom $f = 5$.
At constant pressure,the heat absorbed is given by $dQ = n C_p dT$.
The work done is $dW = P dV = n R dT$.
The change in internal energy is $dU = n C_v dT$.
We know that $C_v = \frac{f}{2} R = \frac{5}{2} R$ and $C_p = C_v + R = \frac{7}{2} R$.
Thus,the ratio $dW : dU : dQ$ is $nR dT : \frac{5}{2} nR dT : \frac{7}{2} nR dT$.
Dividing by $nR dT$,we get $1 : \frac{5}{2} : \frac{7}{2}$.
Multiplying by $2$,we get $2 : 5 : 7$.

(D) At constant pressure,$W = P \Delta V = nR \Delta T$,which implies $W \propto \Delta T$.
Given $W_1 = W$ and $W_2 = 2W$,we have $\frac{(\Delta T)_2}{(\Delta T)_1} = \frac{W_2}{W_1} = 2$,so $(\Delta T)_2 = 2(\Delta T)_1$.
For a monatomic gas,the degrees of freedom $f_1 = 3$. The heat supplied is $Q_1 = \Delta U_1 + W_1 = \frac{f_1}{2} nR(\Delta T)_1 + nR(\Delta T)_1 = (\frac{3}{2} + 1) nR(\Delta T)_1 = \frac{5}{2} nR(\Delta T)_1$.
For a diatomic gas,the degrees of freedom $f_2 = 5$. The heat supplied is $Q_2 = \Delta U_2 + W_2 = \frac{f_2}{2} nR(\Delta T)_2 + 2W = \frac{5}{2} nR(2(\Delta T)_1) + 2nR(\Delta T)_1 = (5 + 2) nR(\Delta T)_1 = 7nR(\Delta T)_1$.
Therefore,the ratio is $\frac{Q_1}{Q_2} = \frac{\frac{5}{2} nR(\Delta T)_1}{7nR(\Delta T)_1} = \frac{5}{14}$.
Thus,$Q_1 : Q_2 = 5 : 14$.

(B) Given: $n = 2 \text{ moles}$,$\gamma = 4/3$,$T_1 = 327^{\circ} C = 600 \text{ K}$.
Step $1$: Adiabatic expansion.
$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given $V_2 = 8 V_1$,so $T_2 = T_1 (V_1/V_2)^{\gamma-1} = 600 \times (1/8)^{(4/3 - 1)} = 600 \times (1/8)^{1/3} = 600 \times (1/2) = 300 \text{ K}$.
Work done in adiabatic process: $W_1 = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{2R(600 - 300)}{4/3 - 1} = \frac{2R(300)}{1/3} = 1800 R$.
Step $2$: Isochoric process.
In an isochoric process,volume is constant,so work done $W_2 = 0$.
Total work done $W = W_1 + W_2 = 1800 R + 0 = 1800 R$.

(D) The process consists of three steps:
Step $I$: Expansion from $V_0$ to $3V_0$ where $p = p_0(V/V_0)$.
Work done $W_I = \int_{V_0}^{3V_0} p dV = \int_{V_0}^{3V_0} \frac{p_0}{V_0} V dV = \frac{p_0}{V_0} \left[ \frac{V^2}{2} \right]_{V_0}^{3V_0} = \frac{p_0}{2V_0} (9V_0^2 - V_0^2) = 4p_0 V_0$.
Step $II$: Isochoric process (volume is constant at $3V_0$),so work done $W_{II} = 0$.
Step $III$: Isobaric compression from $3V_0$ to $V_0$ at constant pressure $p_0$.
Work done $W_{III} = \int_{3V_0}^{V_0} p_0 dV = p_0 (V_0 - 3V_0) = -2p_0 V_0$.
Total work done $W_{\text{total}} = W_I + W_{II} + W_{III} = 4p_0 V_0 + 0 - 2p_0 V_0 = 2p_0 V_0$.

(C) The change in internal energy $(\Delta U)$ of an ideal gas is given by the formula $\Delta U = n C_v \Delta T$.
Given:
Number of moles $(n) = 2000 = 2 \times 10^3 \text{ mol}$.
Initial temperature $(T_i) = 34^{\circ} C$,Final temperature $(T_f) = 24^{\circ} C$.
Temperature change $(\Delta T) = 24 - 34 = -10 \text{ K}$.
For air,$\gamma = 1.4$. The molar heat capacity at constant volume is $C_v = \frac{R}{\gamma - 1} = \frac{R}{1.4 - 1} = \frac{R}{0.4}$.
Substituting the values:
$\Delta U = n \left( \frac{R}{0.4} \right) \Delta T$
$\Delta U = (2 \times 10^3) \times \left( \frac{8.314}{0.4} \right) \times (-10)$
$\Delta U = - \frac{2 \times 8.314 \times 10^4}{0.4}$
$\Delta U = - \frac{16.628 \times 10^4}{0.4} = -41.57 \times 10^4 \text{ J} \approx -4.2 \times 10^5 \text{ J}$.

(B) For a cyclic process, the change in internal energy over the complete cycle is zero, i.e., $\sum \Delta U = 0$.
Using the first law of thermodynamics, $\Delta U = Q - W$.
Thus, $\sum (Q_i - W_i) = 0$.
$(Q_1 - W_1) + (Q_2 - W_2) + (Q_3 - W_3) + (Q_4 - W_4) = 0$
$(6000 - 2500) + (-5500 + 1000) + (-3000 + 1200) + (3500 - x) = 0$
$3500 - 4500 - 1800 + 3500 - x = 0$
$700 - x = 0 \implies x = 700 \ J$.
Net work done $W_{net} = W_1 + W_2 + W_3 + W_4 = 2500 - 1000 - 1200 + 700 = 1000 \ J$.
Total heat absorbed $Q_{in} = Q_1 + Q_4 = 6000 + 3500 = 9500 \ J$.
Efficiency $\eta = \frac{W_{net}}{Q_{in}} \times 100 = \frac{1000}{9500} \times 100 = 10.5 \%$.

(D) For a monoatomic gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$ and at constant pressure is $C_p = \frac{5}{2}R$.
In cylinder $B$, the piston is fixed, so the process is isochoric (constant volume). The heat supplied is $\Delta Q = n C_v \Delta T_B = n (\frac{3}{2}R) \Delta T_B$.
In cylinder $A$, the piston is free to move, so the process is isobaric (constant pressure). The heat supplied is $\Delta Q = n C_p \Delta T_A = n (\frac{5}{2}R) \Delta T_A$.
Since the heat supplied is the same in both cases, we have $n (\frac{3}{2}R) \Delta T_B = n (\frac{5}{2}R) \Delta T_A$.
Given $\Delta T_A = 42 \,K$, we substitute the values: $\frac{3}{2} \Delta T_B = \frac{5}{2} \times 42$.
$3 \Delta T_B = 5 \times 42 = 210$.
$\Delta T_B = \frac{210}{3} = 70 \,K$.

(A) The total work done in the cycle $ABCA$ is $W_{ABCA} = W_{AB} + W_{BC} + W_{CA}$.
$1$. Process $AB$ is an isothermal process since $T_A = T_B = T_1$. The work done is $W_{AB} = nRT_1 \ln(V_B/V_A) = 3RT_1 \ln(3V_1/V_1) = 3RT_1 \ln(3)$.
$2$. Process $BC$ is an isobaric process since $P_B = P_C = P_2$. The work done is $W_{BC} = P_2(V_C - V_B) = P_2(V_1 - 3V_1) = -2P_2V_1$. From the ideal gas law at point $B$,$P_2(3V_1) = nRT_1 = 3RT_1$,so $P_2V_1 = RT_1$. Thus,$W_{BC} = -2RT_1$.
$3$. Process $CA$ is an isochoric process since $V_C = V_A = V_1$. The work done is $W_{CA} = 0$.
$4$. Total work $W_{ABCA} = 3RT_1 \ln(3) - 2RT_1 + 0 = RT_1[3 \ln(3) - 2]$.

(D) For a diatomic gas at constant pressure:
$1$. The work done is $dW = P dV = nR dT$.
$2$. The change in internal energy is $dU = n C_v dT$. For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{5}{2} R$,so $dU = n \left( \frac{5}{2} R \right) dT$.
$3$. The heat absorbed is $dQ = n C_p dT$. For a diatomic gas,the molar heat capacity at constant pressure is $C_p = \frac{7}{2} R$,so $dQ = n \left( \frac{7}{2} R \right) dT$.
$4$. The ratio $dW : dU : dQ$ is:
$dW : dU : dQ = nR dT : n \left( \frac{5}{2} R \right) dT : n \left( \frac{7}{2} R \right) dT$
Dividing by $nR dT$,we get:
$1 : \frac{5}{2} : \frac{7}{2}$
Multiplying by $2$ to simplify:
$2 : 5 : 7$.