Mix Examples-Thermodynamics Questions in English

(N/A) For process $A$ to $B$,the volume is constant,hence the work done $dW = 0$. From the first law of thermodynamics,$dQ = dU + dW = dU + 0 = dU = nC_V dT = nC_V(T_B - T_A) = (1)(\frac{3}{2}R)(T_B - T_A) = \frac{3}{2}(RT_B - RT_A)$. Using the ideal gas equation $PV = RT$,we get $dQ = \frac{3}{2}(P_B V_B - P_A V_A)$.
$(b)$ For process $B$ to $C$,the pressure is constant. The work done is $dW = P_B(V_C - V_B)$. From the first law of thermodynamics,$dQ = dU + dW = nC_V(T_C - T_B) + P_B(V_C - V_B) = \frac{3}{2}(P_C V_C - P_B V_B) + P_B(V_C - V_B)$. Since $P_B = P_C$,this simplifies to $dQ = \frac{3}{2}P_B(V_C - V_B) + P_B(V_C - V_B) = \frac{5}{2}P_B(V_C - V_B)$.
$(c)$ For process $C$ to $D$,it is an adiabatic process,so the heat exchanged $dQ = 0$.
$(d)$ For process $D$ to $A$,the pressure is constant at $P_A$. The gas is compressed from volume $V_D$ to $V_A$. Similar to process $(b)$,the heat exchanged is $dQ = \frac{5}{2}P_A(V_A - V_D)$.

(N/A) For an ideal gas,the heat absorbed is given by $dQ = dU + dW = nC_V dT + P dV$.
Since $PV = nRT$,we have $T = \frac{PV}{nR} = \frac{f(V)V}{nR}$.
Differentiating with respect to $V$,we get $dT = \frac{1}{nR} [f(V) + V f'(V)] dV$.
Substituting this into the heat equation: $dQ = n C_V \left( \frac{f(V) + V f'(V)}{nR} \right) dV + f(V) dV$.
Using $C_V = \frac{R}{\gamma - 1}$,we get $dQ = \left( \frac{f(V) + V f'(V)}{\gamma - 1} + f(V) \right) dV$.
For the gas to absorb heat,$dQ > 0$ for $dV > 0$.
This requires $\frac{f(V) + V f'(V)}{\gamma - 1} + f(V) > 0$.
Simplifying,$f(V) + V f'(V) + (\gamma - 1)f(V) > 0$,which leads to $\gamma f(V) + V f'(V) > 0$.
Rearranging,$f'(V) > -\gamma \frac{f(V)}{V}$.
The slope of the adiabatic curve is given by $\frac{dP}{dV} = -\gamma \frac{P}{V}$.
Thus,the condition for heat absorption is that the slope of the process curve $f'(V)$ must be greater than the slope of the adiabatic curve $-\gamma \frac{P_0}{V_0}$.

(C) The efficiency $\eta$ of a heat engine is defined as the ratio of the net work done to the total heat supplied.
$\eta = \frac{W}{Q_{\text{in}}} = \frac{Q_{\text{net}}}{Q_{\text{in}}}$
Here,the net heat exchanged in one cycle is $Q_{\text{net}} = 1915 - 40 + 125 - Q = 2000 - Q$.
The heat supplied $(Q_{\text{in}})$ is the sum of all positive heat exchanges: $Q_{\text{in}} = 1915 + 125 = 2040\, J$.
Given $\eta = 50.0\% = 0.5$,we have:
$0.5 = \frac{2000 - Q}{2040}$
$1020 = 2000 - Q$
$Q = 2000 - 1020 = 980\, J$.

(A) $(I)$ Adiabatic process: $\Delta Q = 0$. No exchange of heat takes place with the surroundings.
$(II)$ Isothermal process: Temperature remains constant $(\Delta T = 0)$. Since internal energy $\Delta U = \frac{f}{2}nR\Delta T$,it follows that $\Delta U = 0$.
$(III)$ Isochoric process: Volume remains constant $(\Delta V = 0)$. Work done $W = \int P \cdot dV = 0$.
$(IV)$ Isobaric process: Pressure remains constant. Here,$W = P\Delta V \neq 0$,$\Delta U = \frac{f}{2}nR\Delta T \neq 0$,and $\Delta Q = nC_p\Delta T \neq 0$.

(D) For the process $A \rightarrow B$ (isochoric): $V_A = V_B$. Using $PV = nRT$,we have $\frac{P_A}{T_A} = \frac{P_B}{T_B}$. Given $\frac{P_B}{P_A} = \frac{1}{5}$,so $T_B = \frac{T_A}{5} = \frac{400}{5} = 80 \, K$.
Heat supplied in $A \rightarrow B$: $Q_1 = n C_V (T_B - T_A) = 1 \times \frac{3}{2} R (80 - 400) = \frac{3}{2} \times 8.31 \times (-320) = -3988.8 \, J$.
For the process $B \rightarrow C$ (isobaric): $P_B = P_C$. Using $PV = nRT$,we have $\frac{V_B}{T_B} = \frac{V_C}{T_C}$. Since $T_C = 400 \, K$ and $T_B = 80 \, K$,$V_C = V_B \frac{400}{80} = 5 V_B$.
Heat supplied in $B \rightarrow C$: $Q_2 = n C_P (T_C - T_B) = 1 \times \frac{5}{2} R (400 - 80) = \frac{5}{2} \times 8.31 \times 320 = 6648 \, J$.
Total heat supplied: $Q = Q_1 + Q_2 = -3988.8 + 6648 = 2659.2 \, J$.

(A) For cylinder $A$ (isobaric process),the heat supplied is $Q_A = n C_p \Delta T_A$.
For cylinder $B$ (isochoric process),the heat supplied is $Q_B = n C_v \Delta T_B$.
Given that $Q_A = Q_B$ and the amount of gas $n$ is the same,we have $n C_p \Delta T_A = n C_v \Delta T_B$.
Thus,$\Delta T_B = \frac{C_p}{C_v} \Delta T_A$.
For a diatomic gas,the adiabatic index $\gamma = \frac{C_p}{C_v} = \frac{7}{5} = 1.4$.
Substituting the values,$\Delta T_B = 1.4 \times 30\, K = 42\, K$.

(A) In thermodynamics,heat and work are classified as path functions.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Here,$\Delta U$ (internal energy) is a state function,meaning it depends only on the initial and final states of the system.
However,the work done $(\Delta W)$ depends on the specific process or path taken to change the state of the system.
Since $\Delta U$ is a state function and $\Delta W$ is a path function,the heat exchanged $(\Delta Q)$ must also depend on the path taken.
Therefore,both heat and work are path functions.

(D) The total work done in a cyclic process is the sum of work done in each individual process.
$1$. For process $A \rightarrow B$ (Isothermal expansion):
$W_{AB} = nRT \ln \left(\frac{V_{2}}{V_{1}}\right) = nRT \ln \left(\frac{2V_{1}}{V_{1}}\right) = nRT \ln 2$.
$2$. For process $B \rightarrow C$ (Isobaric compression):
Since the process is isothermal from $A$ to $B$,$P_{1}V_{1} = P_{2}V_{2}$. Given $V_{2} = 2V_{1}$,we have $P_{1}V_{1} = P_{2}(2V_{1})$,which means $P_{2} = \frac{P_{1}}{2}$.
$W_{BC} = P_{2}(V_{1} - V_{2}) = P_{2}(V_{1} - 2V_{1}) = -P_{2}V_{1}$.
Using the ideal gas law at point $B$,$P_{2}V_{2} = nRT$,so $P_{2}(2V_{1}) = nRT$,which gives $P_{2}V_{1} = \frac{nRT}{2}$.
Thus,$W_{BC} = -\frac{nRT}{2}$.
$3$. For process $C \rightarrow A$ (Isochoric change):
Since the volume is constant $(V_{1})$,the work done $W_{CA} = 0$.
Total work done $W_{net} = W_{AB} + W_{BC} + W_{CA} = nRT \ln 2 - \frac{nRT}{2} + 0 = nRT \left(\ln 2 - \frac{1}{2}\right)$.

(B) Isothermal process: In this process, the temperature of the system remains constant $(\Delta T = 0)$. Thus, $(a) \rightarrow (ii)$.
$(b)$ Isochoric process: In this process, the volume of the system remains constant $(\Delta V = 0)$. Thus, $(b) \rightarrow (iii)$.
$(c)$ Adiabatic process: In this process, there is no exchange of heat between the system and the surroundings $(\Delta Q = 0)$. Thus, the heat content remains constant. So, $(c) \rightarrow (iv)$.
$(d)$ Isobaric process: In this process, the pressure of the system remains constant $(\Delta P = 0)$. Thus, $(d) \rightarrow (i)$.
Therefore, the correct matching is $(a) \rightarrow (ii), (b) \rightarrow (iii), (c) \rightarrow (iv), (d) \rightarrow (i)$.

(C) Given the process equation: $P = kV^{3}$.
Using the ideal gas law: $PV = nRT$,we can write $P = \frac{nRT}{V}$.
Equating the two expressions for $P$: $\frac{nRT}{V} = kV^{3}$,which implies $kV^{4} = nRT$.
Differentiating both sides with respect to temperature: $4kV^{3} dV = nR dT$.
Since $P = kV^{3}$,we can substitute this into the equation: $4P dV = nR dT$,which gives $P dV = \frac{nR dT}{4}$.
The work done $W$ is given by the integral: $W = \int P dV = \int_{T_i}^{T_f} \frac{nR}{4} dT$.
$W = \frac{nR}{4} (T_f - T_i) = \frac{nR}{4} (300 - 100) = \frac{nR}{4} (200) = 50nR$.
Thus,the work done is $50nR$.

(A) For the isothermal process $A-B$:
$W_{AB} = nRT \ln \left(\frac{V_B}{V_A}\right) = RT \ln \left(\frac{2V_1}{V_1}\right) = RT \ln 2$.
For the isochoric process $B-C$:
Since the volume is constant,$W_{BC} = 0$.
For the adiabatic process $C-A$:
The work done in an adiabatic process is $W = \frac{P_i V_i - P_f V_f}{\gamma - 1}$.
Here,$P_C = \frac{P_1}{4}$,$V_C = 2V_1$,$P_A = P_1$,$V_A = V_1$.
$W_{CA} = \frac{P_C V_C - P_A V_A}{\gamma - 1} = \frac{(\frac{P_1}{4})(2V_1) - P_1 V_1}{\gamma - 1} = \frac{\frac{P_1 V_1}{2} - P_1 V_1}{\gamma - 1} = \frac{-\frac{P_1 V_1}{2}}{\gamma - 1} = -\frac{RT}{2(\gamma - 1)}$.
The net work done is $W_{net} = W_{AB} + W_{BC} + W_{CA} = RT \ln 2 + 0 - \frac{RT}{2(\gamma - 1)} = RT \left(\ln 2 - \frac{1}{2(\gamma - 1)}\right)$.

(B) The work done by a gas is given by $W = \int P\,dV$.
From the ideal gas law,$P = \frac{nRT}{V}$.
Substituting this into the work formula: $W = \int \frac{nRT}{V}\,dV$.
Given the relation $V = KT^{2/3}$,we differentiate with respect to $T$: $dV = K \cdot \frac{2}{3} T^{-1/3}\,dT$.
Substituting $V$ and $dV$ into the integral: $W = \int_{T_1}^{T_2} \frac{nRT}{KT^{2/3}} \cdot \left( \frac{2}{3} K T^{-1/3} \right) dT$.
Simplifying the expression: $W = \int_{T_1}^{T_2} \frac{2}{3} nR \cdot \frac{T \cdot T^{-1/3}}{T^{2/3}} \,dT = \int_{T_1}^{T_2} \frac{2}{3} nR \,dT$.
Integrating,we get $W = \frac{2}{3} nR (T_2 - T_1)$.
Given $\Delta T = T_2 - T_1 = 90\,K$ and assuming $n = 1$ mole: $W = \frac{2}{3} \cdot 1 \cdot R \cdot 90 = 60R$.
Thus,$x = 60$.

(B) From the first law of thermodynamics,$Q = \Delta U + W.$
Given that $W = Q/5,$ we have $Q = \Delta U + Q/5.$
Therefore,$\Delta U = Q - Q/5 = 4Q/5.$
For a rigid diatomic gas,the internal energy change is $\Delta U = n C_v \Delta T,$ where $C_v = \frac{5}{2}R.$
Also,$Q = n C \Delta T,$ where $C$ is the molar heat capacity.
Substituting these into the equation $\Delta U = \frac{4}{5}Q,$ we get $n C_v \Delta T = \frac{4}{5} (n C \Delta T).$
$C_v = \frac{4}{5} C \implies C = \frac{5}{4} C_v.$
Substituting $C_v = \frac{5}{2}R,$ we get $C = \frac{5}{4} \times \frac{5}{2} R = \frac{25}{8} R.$
Comparing this with $\frac{xR}{8},$ we find $x = 25.$

(A) In a $P-V$ diagram,the slope of an adiabatic process is given by $\frac{dP}{dV} = -\gamma \frac{P}{V}$,while the slope of an isothermal process is given by $\frac{dP}{dV} = -\frac{P}{V}$.
Since the adiabatic index $\gamma > 1$ for all gases,the magnitude of the slope of the adiabatic curve is greater than that of the isothermal curve.
Therefore,the adiabatic curve is steeper than the isothermal curve.
Looking at the graph:
$1$ represents an isochoric process (constant volume).
$4$ represents an isobaric process (constant pressure).
Between $2$ and $3$,curve $2$ is steeper than curve $3$.
Thus,curve $2$ represents the adiabatic process and curve $3$ represents the isothermal process.

(C) According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.
Given,$W = \frac{Q}{4}$.
Therefore,$\Delta U = \Delta Q - W = Q - \frac{Q}{4} = \frac{3Q}{4}$.
For a monoatomic gas,the change in internal energy is $\Delta U = n C_v \Delta T = n (\frac{3}{2}R) \Delta T$.
So,$n (\frac{3}{2}R) \Delta T = \frac{3Q}{4} \Rightarrow n \Delta T = \frac{Q}{2R}$.
The molar heat capacity $C$ is defined as $C = \frac{Q}{n \Delta T}$.
Substituting $n \Delta T$,we get $C = \frac{Q}{Q / (2R)} = 2R$.
Thus,the value of $x$ is $2$.

(B) The ratio of specific heats is given by $\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f} = 1.4$,where $f$ is the degrees of freedom.
$1 + \frac{2}{f} = 1.4 \Rightarrow \frac{2}{f} = 0.4 \Rightarrow f = 5$.
By the law of conservation of energy,the kinetic energy of the gas is converted into internal energy.
$\Delta U = K.E._{initial}$
$\frac{f}{2} nR \Delta T = \frac{1}{2} (nm) v^2$,where $n$ is the number of moles and $m$ is the mass of one molecule.
Since $M = m \times N_A$ and $R = k_B \times N_A$,we have $m = \frac{M}{N_A}$.
$\frac{5}{2} nR \Delta T = \frac{1}{2} n M v^2$
$\Delta T = \frac{M v^2}{5R}$.

(B) For a cyclic process,the change in internal energy $\Delta U = 0$. Therefore,the total heat absorbed $\Delta Q$ equals the total work done $\Delta W$ by the gas.
$\Delta Q_{\text{cycle}} = \Delta Q_{AB} + \Delta Q_{BC} + \Delta Q_{CA} = 40 + 0 - 60 = -20 \ J$.
Since $\Delta Q_{\text{cycle}} = \Delta W_{\text{cycle}}$,we have $\Delta W_{\text{cycle}} = -20 \ J$.
The total work done is $\Delta W_{\text{cycle}} = W_{AB} + W_{BC} + W_{CA}$.
From the figure,the process $AB$ is isochoric (vertical line),so $W_{AB} = 0$.
Given $W_{BC} = -50 \ J$ (work done on the gas).
Thus,$-20 = 0 + (-50) + W_{CA}$.
$W_{CA} = -20 + 50 = 30 \ J$.

(D) The work done by a gas during an expansion process is given by the area under the $P-V$ curve.
From the given $P-V$ diagram,for the same change in volume from $V_{i}$ to $V_{f}$,the area under the isobaric process curve $(3)$ is the largest,followed by the isothermal process curve $(1)$,and the adiabatic process curve $(2)$ has the smallest area under it.
Therefore,the work done follows the order: $W_{2} < W_{1} < W_{3}$.

(B) Initial state: $P_{1} = 2 \times 10^{7} \; Pa$,$V_{1} = V$,$T_{1} = 300 \; K$.
Step $1$: Isothermal expansion.
For an isothermal process,$P_{1}V_{1} = P_{2}V_{2}$.
Given $V_{2} = 2V_{1} = 2V$.
So,$P_{2} = P_{1} \times (V_{1} / V_{2}) = (2 \times 10^{7}) \times (V / 2V) = 1 \times 10^{7} \; Pa$.
Step $2$: Adiabatic expansion.
For an adiabatic process,$P_{2}V_{2}^{\gamma} = P_{3}V_{3}^{\gamma}$.
Given $V_{3} = 2V_{2} = 4V$ and $\gamma = 1.5$.
So,$P_{3} = P_{2} \times (V_{2} / V_{3})^{\gamma} = (1 \times 10^{7}) \times (V_{2} / 2V_{2})^{1.5} = (1 \times 10^{7}) \times (1/2)^{1.5}$.
$P_{3} = (1 \times 10^{7}) / 2^{1.5} = (1 \times 10^{7}) / 2.8284 = 3.536 \times 10^{6} \; Pa$.

(A) $A.$ According to Newton's law of cooling,the rate of loss of heat $\frac{dQ}{dt} \propto \Delta T$. If $\Delta T$ is doubled,the rate of loss of heat becomes twice. Thus,$A$ is correct.
$B.$ The power radiated by a body is $P = \sigma e A T^4$. The ratio of radiation emitted is $\frac{H_P}{H_Q} = \left(\frac{T_P}{T_Q}\right)^4$. Converting temperatures to Kelvin: $T_P = 10 + 273 = 283 K$ and $T_Q = 20 + 273 = 293 K$. Thus,$\frac{H_P}{H_Q} = \left(\frac{283}{293}\right)^4 \approx (0.9658)^4 \approx 0.87$. However,using the approximation $\frac{293}{283} \approx 1.035$,then $(1.035)^4 \approx 1.15$. Thus,$1:1.15$ is correct. $B$ is correct.
$C.$ Efficiency $\eta = 1 - \frac{T_L}{T_H} = 1 - \frac{100}{400} = 1 - 0.25 = 0.75$ or $75\%$. Thus,$C$ is correct.
$D.$ Since $\frac{dQ}{dt} \propto \Delta T$,if $\Delta T$ is quadrupled,the rate of loss of heat becomes four times,not twice. Thus,$D$ is incorrect.
Therefore,statements $A, B,$ and $C$ are correct.

(B) For a heat engine operating between a variable temperature source $T$ and a fixed temperature sink $T_0 = 200 \,K$,the maximum work is obtained when the engine operates as a Carnot engine at every instant.
The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_0}{T}$.
The infinitesimal work done $dW$ for an infinitesimal heat $dQ_{in} = -C dT$ extracted from the hot body is $dW = \eta dQ_{in} = (1 - \frac{T_0}{T})(-C dT)$.
Integrating from the initial temperature $T_i = 600 \,K$ to the final temperature $T_f = 400 \,K$:
$W = \int_{600}^{400} -(1 - \frac{200}{T}) C dT = C \int_{400}^{600} (1 - \frac{200}{T}) dT$.
Given $C = 1 \,J/K$:
$W = [T - 200 \ln T]_{400}^{600} = (600 - 400) - 200(\ln 600 - \ln 400)$.
$W = 200 - 200 \ln(600/400) = 200 - 200 \ln(3/2) = 200(1 - \ln(3/2)) \,J$.

(B) Let the cycle be $A \rightarrow B \rightarrow C \rightarrow A$.
$A \rightarrow B$ is isobaric,$B \rightarrow C$ is adiabatic,and $C \rightarrow A$ is isothermal.
Let $T_A = T_{min}$ and $T_B = T_{max}$. The ratio is $x = T_B / T_A$.
For isobaric process $A \rightarrow B$: $Q_{AB} = n C_p (T_B - T_A) = n \left( \frac{\gamma R}{\gamma - 1} \right) (T_B - T_A)$.
For isothermal process $C \rightarrow A$: $Q_{CA} = n R T_A \ln(V_A / V_C)$.
Since $B \rightarrow C$ is adiabatic,$T_B V_B^{\gamma-1} = T_C V_C^{\gamma-1}$. Since $T_C = T_A$,$T_B V_B^{\gamma-1} = T_A V_C^{\gamma-1}$.
Thus,$(V_C / V_B)^{\gamma-1} = T_B / T_A = x$,so $V_C / V_B = x^{1/(\gamma-1)}$.
Also,for isobaric process $A \rightarrow B$,$V_B / V_A = T_B / T_A = x$.
So,$V_C / V_A = (V_C / V_B) \cdot (V_B / V_A) = x^{1/(\gamma-1)} \cdot x = x^{\gamma/(\gamma-1)}$.
Efficiency $\eta = 1 - |Q_{CA}| / Q_{AB} = 1 - [n R T_A \ln(V_C / V_A)] / [n R \frac{\gamma}{\gamma-1} (T_B - T_A)]$.
Substituting values: $\eta = 1 - [T_A \cdot \frac{\gamma}{\gamma-1} \ln x] / [\frac{\gamma}{\gamma-1} T_A (x - 1)] = 1 - \frac{\ln x}{x - 1}$.
Given $\eta = 0.5$,so $0.5 = 1 - \frac{\ln x}{x - 1} \Rightarrow \frac{\ln x}{x - 1} = 0.5 \Rightarrow \ln x = 0.5(x - 1) \Rightarrow \ln x = \ln(e^{0.5(x-1)})$.
Thus $x = e^{0.5(x-1)} \Rightarrow x^2 = e^{x-1}$.

(A) The expansion curves are shown in the figure. The slope of an adiabatic curve is given by $\frac{dp}{dV} = -\gamma \left(\frac{p}{V}\right)$.
Since $\gamma_{\text{diatomic}} = \frac{7}{5} = 1.4$ and $\gamma_{\text{monoatomic}} = \frac{5}{3} \approx 1.67$,we have $\gamma_{\text{diatomic}} < \gamma_{\text{monoatomic}}$.
This implies that the adiabatic expansion curve for the monoatomic gas is steeper than that for the diatomic gas. Since both gases start from the same initial state and reach the same final state,the area under the $p-V$ graph represents the work done. Because the diatomic gas curve lies above the monoatomic gas curve during the expansion process,the area under the diatomic gas curve is greater than the area under the monoatomic gas curve.
Therefore,the work done by the diatomic gas is more than that by the monoatomic gas.

(B) For an ideal monoatomic gas,the adiabatic index is $\gamma = 5/3$.
Let the states be $A, B, C, D$ corresponding to the volumes $V_A = 8.0 \, m^3, V_B = 1.0 \, m^3, V_C = 10.0 \, m^3, V_D = 80.0 \, m^3$.
Step $1$ $(A \to B)$: Adiabatic compression.
$T_A V_A^{\gamma-1} = T_B V_B^{\gamma-1} \implies T_2 (8)^{5/3-1} = T_1 (1)^{5/3-1} \implies T_2 (8)^{2/3} = T_1 (1)^{2/3} \implies T_2 (4) = T_1 \implies T_1/T_2 = 4$.
Step $3$ $(C \to D)$: Adiabatic expansion.
$T_C V_C^{\gamma-1} = T_D V_D^{\gamma-1} \implies T_1 (10)^{5/3-1} = T_2 (80)^{5/3-1} \implies T_1 (10)^{2/3} = T_2 (80)^{2/3} \implies T_1/T_2 = (80/10)^{2/3} = 8^{2/3} = 4$.
Both steps confirm that $T_1/T_2 = 4$.

(A) The process equation is $pV^{5/3} \cdot e^{-pV/E_0} = C_1$.
Taking the natural logarithm on both sides: $\ln(p) + \frac{5}{3} \ln(V) = \ln(C_1) + \frac{pV}{E_0}$.
Differentiating with respect to $V$: $\frac{1}{p} \frac{dp}{dV} + \frac{5}{3V} = \frac{1}{E_0} \left( p + V \frac{dp}{dV} \right)$.
Rearranging terms to find $\frac{dp}{dV}$: $\frac{dp}{dV} \left( \frac{1}{p} - \frac{V}{E_0} \right) = \frac{p}{E_0} - \frac{5}{3V}$.
Using $pV = RT$,we have $p = \frac{RT}{V}$. Substituting this,the isothermal compressibility $k = -\frac{1}{V} \frac{dV}{dp}$ is derived.
As $T \to \infty$,the terms involving $\frac{1}{RT}$ approach zero.
Thus,$k = -\frac{1}{V} \frac{dV}{dp} = \frac{1/p - V/E_0}{p/E_0 - 5/3V} \cdot \frac{1}{V} \approx \text{constant}$ as $T \to \infty$.

(A) In the given $T-S$ cycle:
$b-c$ and $a-d$ are isothermal processes. Since internal energy $U$ of an ideal gas depends only on temperature $(U = nC_vT)$, $U$ remains constant during these processes. Thus, on a $U-V$ diagram, these appear as horizontal lines.
$a-b$ and $c-d$ are isentropic (adiabatic) processes $(S = \text{constant})$. For an adiabatic process, $TV^{\gamma-1} = \text{constant}$. Since $U \propto T$, we have $UV^{\gamma-1} = \text{constant}$, which implies $U \propto V^{1-\gamma}$. This represents a curve on the $U-V$ diagram.
Comparing the processes: In $b-c$, $T$ is high, so $U$ is high. In $a-d$, $T$ is low, so $U$ is low. Thus, the cycle on the $U-V$ diagram consists of two horizontal lines at different $U$ levels, connected by two curves. This matches the representation in option $A$.

(C) The first law of thermodynamics states that $dQ = dU + dW$. For an ideal gas,$dU = nC_v dT$. Along the straight line path $XY$,the system moves through different isotherms.
At the start point $X$ and end point $Y$,the temperature is the same as they lie on the same adiabat (if we consider the adiabat as a boundary). However,as the straight line path $XY$ lies above the adiabat,the temperature of the gas first increases as it moves away from the adiabat towards higher temperature isotherms,reaching a maximum at some intermediate point $Z$,and then decreases as it approaches point $Y$.
Since $dQ = nC_v dT + p dV$,and $dV > 0$ throughout the process,the heat exchange depends on the sign of $dT$. From $X$ to $Z$,the temperature increases $(dT > 0)$,so heat is absorbed. From $Z$ to $Y$,the temperature decreases $(dT < 0)$,and the work done by the gas eventually outweighs the internal energy change,leading to heat release. Thus,heat is absorbed from $X$ to $Z$ and released from $Z$ to $Y$.

(D) The total work done in a complete thermodynamic cycle is the sum of the work done in each individual process:
$W_{\text{total}} = W_{12} + W_{23} + W_{31}$
Given that the total work $W_{\text{total}} = 10 \, J$.
In an isochoric process $(2 \rightarrow 3)$,the volume remains constant,so the work done $W_{23} = 0 \, J$.
The work done in the adiabatic process $(3 \rightarrow 1)$ is given as $W_{31} = -20 \, J$.
Substituting these values into the total work equation:
$10 \, J = W_{12} + 0 \, J + (-20 \, J)$
$10 \, J = W_{12} - 20 \, J$
$W_{12} = 30 \, J$
For an isothermal process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta U + W$. Therefore,for the isothermal process $(1 \rightarrow 2)$:
$\Delta Q_{12} = 0 + W_{12} = 30 \, J$.
Thus,the heat added to the system in the isothermal process is $30 \, J$.

(B) The work done by a gas during an expansion process is equal to the area under the $p-V$ curve.
For an ideal gas expanding from volume $V$ to $2V$,the isothermal process is represented by the curve $AB$ and the adiabatic process is represented by the curve $AC$ on the $p-V$ diagram.
Since the adiabatic curve is steeper than the isothermal curve,the area under the isothermal curve $AB$ is greater than the area under the adiabatic curve $AC$ for the same change in volume.
Therefore,the work done in the isothermal process $(W_i)$ is greater than the work done in the adiabatic process $(W_a)$.
Since the gas is expanding,the work done is positive in both cases.
Thus,$W_i > W_a > 0$.

(D) Given,internal energy $U = a V^3$.
For an ideal gas,$U = \frac{f}{2} n R T$. Since $n = 1$ and assuming a monatomic gas $(f = 3)$,we have $U = \frac{3}{2} R T$.
Equating the two expressions: $\frac{3}{2} R T = a V^3$.
Using the ideal gas law $P V = R T$,we substitute $R T = P V$:
$\frac{3}{2} P V = a V^3 \Rightarrow P = \frac{2 a}{3} V^2$.
The work done $W$ is given by $\int_{V_i}^{V_f} P dV$.
Here,$V_i = V$ and $V_f = 2V$.
$W = \int_{V}^{2V} \frac{2 a}{3} V^2 dV = \frac{2 a}{3} \left[ \frac{V^3}{3} \right]_{V}^{2V} = \frac{2 a}{9} (8V^3 - V^3) = \frac{2 a}{9} (7V^3) = \frac{14 a V^3}{9}$.
Since $a V^3 = \frac{3}{2} R T$ at the initial state,we substitute $a V^3 = \frac{3}{2} R T$:
$W = \frac{14}{9} \times \left( \frac{3}{2} R T \right) = \frac{7}{3} R T$.

(B) The cycle consists of a straight line path from $A$ to $B$ and an adiabatic path from $B$ to $A$.
Along the straight line path from $A$ to $B$,the gas undergoes expansion,and the temperature changes. Since the path is a straight line,the heat exchange varies. Specifically,heat is absorbed by the gas during the expansion phase along the straight line.
Along the curved path from $B$ to $A$,the process is adiabatic,meaning no heat is exchanged $(dQ = 0)$.
However,in a complete cycle,for work to be done by the gas,heat must be absorbed from a source and some must be rejected to a sink. In this specific cycle,heat is absorbed during the straight-line expansion and rejected during the adiabatic compression (if we consider the cycle direction). Therefore,option $B$ correctly describes the heat exchange characteristics.

(D) For the cyclic process $A \rightarrow B \rightarrow C \rightarrow A$,the change in internal energy $\Delta U = 0$. By the first law of thermodynamics,$\Delta Q = \Delta W$,where $\Delta W$ is the area enclosed by the triangle $ABC$ in the $p-V$ diagram.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (V_2 - V_1)(p_2 - p_1)$.
For process $AB$ (isobaric,$p = p_1$):
$\Delta Q_{AB} = n C_p \Delta T = n (\frac{5}{2} R) \Delta T = \frac{5}{2} p_1 (V_1 - V_2)$.
For process $BC$ (isochoric,$V = V_2$):
$\Delta Q_{BC} = n C_V \Delta T = n (\frac{3}{2} R) \Delta T = \frac{3}{2} V_2 (p_2 - p_1)$.
Using $\Delta Q_{total} = \Delta Q_{AB} + \Delta Q_{BC} + \Delta Q_{AC} = \Delta W$,we find:
$\Delta Q_{AC} = \Delta W - \Delta Q_{AB} - \Delta Q_{BC}$.
Substituting the values:
$\Delta Q_{AC} = \frac{1}{2}(V_2 - V_1)(p_2 - p_1) - [\frac{5}{2} p_1 (V_1 - V_2) + \frac{3}{2} V_2 (p_2 - p_1)]$.
Simplifying this expression yields:
$\Delta Q_{AC} = 2(p_2 V_2 - p_1 V_1) + \frac{1}{2}(p_1 V_2 - p_2 V_1)$.

(A) For the adiabatic processes $BC$ and $DA$:
$P_{12} V_2^\gamma = P_{34} V_3^\gamma \quad \dots(i)$
$P_{34} V_4^\gamma = P_{12} V_1^\gamma \quad \dots(ii)$
Multiplying $(i)$ and $(ii)$:
$P_{12} P_{34} (V_2 V_4)^\gamma = P_{12} P_{34} (V_1 V_3)^\gamma$
$V_2 V_4 = V_1 V_3 \Rightarrow \frac{V_2}{V_3} = \frac{V_1}{V_4}$
Using the ideal gas law $PV = nRT$ for isobaric processes:
$T_1 = \frac{P_{12} V_1}{nR}, T_2 = \frac{P_{12} V_2}{nR}, T_3 = \frac{P_{34} V_3}{nR}, T_4 = \frac{P_{34} V_4}{nR}$
Efficiency $\eta = 1 - \frac{Q_{rejected}}{Q_{supplied}} = 1 - \frac{n C_p (T_3 - T_4)}{n C_p (T_2 - T_1)} = 1 - \frac{T_3 - T_4}{T_2 - T_1}$
Substituting $T$ values:
$\eta = 1 - \frac{\frac{P_{34}}{nR}(V_3 - V_4)}{\frac{P_{12}}{nR}(V_2 - V_1)} = 1 - \frac{P_{34}}{P_{12}} \left( \frac{V_3 - V_4}{V_2 - V_1} \right)$
Since $\frac{V_2}{V_3} = \frac{V_1}{V_4} = k$,then $V_2 = k V_3$ and $V_1 = k V_4$.
$\eta = 1 - \frac{P_{34}}{P_{12}} \left( \frac{V_3 - V_4}{k(V_3 - V_4)} \right) = 1 - \frac{P_{34}}{P_{12} k} = 1 - \frac{P_{34}}{P_{12}} \cdot \frac{V_3}{V_2} = 1 - \frac{T_3}{T_2} = 1 - \frac{T_4}{T_1}$

(C) The efficiency of the cycle is given by $\eta = \frac{W_{\text{net}}}{Q_{\text{supplied}}}$.
The net work done $W_{\text{net}}$ is the area enclosed by the cycle on the $P-V$ diagram. The area of a quarter ellipse with semi-axes $a = \frac{V_{0}}{2}$ and $b = \frac{P_{0}}{2}$ is $W_{\text{net}} = \frac{1}{4} \pi a b = \frac{1}{4} \pi \left(\frac{V_{0}}{2}\right) \left(\frac{P_{0}}{2}\right) = \frac{\pi P_{0} V_{0}}{16}$.
Heat is supplied to the gas only during the process $CA$. For process $CA$,the change in internal energy is $\Delta U_{CA} = n C_{V} \Delta T = \frac{3}{2} R (T_{A} - T_{C})$. Using $PV = nRT$,we have $T_{A} = \frac{(3/2 P_{0}) V_{0}}{R} = \frac{3 P_{0} V_{0}}{2R}$ and $T_{C} = \frac{P_{0} (V_{0}/2)}{R} = \frac{P_{0} V_{0}}{2R}$.
Thus,$\Delta U_{CA} = \frac{3}{2} R \left(\frac{3 P_{0} V_{0}}{2R} - \frac{P_{0} V_{0}}{2R}\right) = \frac{3}{2} P_{0} V_{0}$.
The work done during the path $CA$ is the area under the curve $CA$,which is the area of the rectangle plus the area of the quarter ellipse: $W_{CA} = P_{0} \left(\frac{V_{0}}{2}\right) + \frac{\pi P_{0} V_{0}}{16} = \frac{P_{0} V_{0}}{2} + \frac{\pi P_{0} V_{0}}{16}$.
Total heat supplied $Q_{\text{supplied}} = W_{CA} + \Delta U_{CA} = \left(\frac{P_{0} V_{0}}{2} + \frac{\pi P_{0} V_{0}}{16}\right) + \frac{3}{2} P_{0} V_{0} = 2 P_{0} V_{0} + \frac{\pi P_{0} V_{0}}{16} = P_{0} V_{0} \left(2 + \frac{\pi}{16}\right) = P_{0} V_{0} \left(\frac{32 + \pi}{16}\right)$.
Efficiency $\eta = \frac{\pi P_{0} V_{0} / 16}{P_{0} V_{0} (32 + \pi) / 16} = \frac{\pi}{32 + \pi}$.

(D) The correct answer is $D$.
$1$. Internal energy $(U)$ is a state function, meaning its change $(\Delta U)$ depends only on the initial and final states of the system, not on the path taken.
$2$. Heat $(Q)$ and work $(W)$ are path functions. Their values depend on the specific process or path followed to transition between two states.
$3$. Since statements $B$ and $C$ are correct, and statement $A$ is often interpreted in the context of path dependence (as the values of $Q$ and $W$ are determined by the path between states), all statements are considered correct in the context of thermodynamics.

(D) In a $P-T$ graph:
$1.$ Process $A$ is a vertical line, meaning temperature $T$ is constant. This represents an isothermal process.
$2.$ Process $B$ is a horizontal line, meaning pressure $P$ is constant. This represents an isobaric process.
$3.$ Process $C$ is a straight line passing through the origin $(0, 0)$, meaning $P \propto T$. From the ideal gas law $PV = nRT$, if $P/T = nR/V$ is constant, then volume $V$ must be constant. This represents an isochoric process.
$4.$ Process $D$ is a curve where the slope $dP/dT$ increases with temperature. For an adiabatic process, $P^{1-\gamma} T^{\gamma} = \text{constant}$, which leads to a non-linear relationship in the $P-T$ plane. Thus, $D$ represents an adiabatic process.
Comparing these with the options, the correct alternative is $D$ - Adiabatic process.

(B) In the given $P-V$ diagram:
$1$. Process $A \longrightarrow B$: The pressure $P$ is constant while the volume $V$ increases. This is an isobaric expansion.
$2$. Process $B \longrightarrow C$: This is an isothermal process ($T$ is constant). For an ideal gas,$PV = nRT$,so $P \propto 1/V$. The curve is a rectangular hyperbola.
$3$. Process $C \longrightarrow D$: The volume $V$ is constant while the pressure $P$ decreases. This is an isochoric cooling process.
$4$. Process $D \longrightarrow A$: This is an isothermal process ($T$ is constant). Similar to $B \longrightarrow C$,$P \propto 1/V$.
Now,let's analyze the $P-T$ graph:
- For an isobaric process $(A \longrightarrow B)$,$V \propto T$. Since $V$ increases,$T$ must increase at constant $P$.
- For an isothermal process $(B \longrightarrow C)$,$T$ is constant,so the graph is a vertical line segment.
- For an isochoric process $(C \longrightarrow D)$,$P \propto T$. This is a straight line passing through the origin $(0,0)$.
- For an isothermal process $(D \longrightarrow A)$,$T$ is constant,so the graph is a vertical line segment.
Comparing these with the given options,the correct $P-T$ graph is represented by option $B$.

(A) The process is isothermal,so the change in internal energy $\Delta U$ is equal to the heat supplied $Q$ because the work done by the gas is not requested,but rather the energy required to change the state of the gas.
Initial internal energy of $20$ moles of diatomic gas: $U_{\text{initial}} = n_1 \times C_{v1} \times T = 20 \times \frac{5}{2} R T = 50 R T$.
After dissociation,$8$ moles of diatomic gas become $16$ moles of monatomic gas,and $12$ moles of diatomic gas remain.
Final internal energy: $U_{\text{final}} = (12 \times \frac{5}{2} R T) + (16 \times \frac{3}{2} R T) = 30 R T + 24 R T = 54 R T$.
Heat energy supplied $Q = U_{\text{final}} - U_{\text{initial}} = 54 R T - 50 R T = 4 R T$.

(B) For a cyclic process,the change in internal energy $\Delta U = 0$.
According to the first law of thermodynamics,$Q = \Delta U + W$. Since $\Delta U = 0$,we have $Q = W$.
The work done in a cyclic process is equal to the area enclosed by the $PV$ diagram.
Since the cycle is traced in a clockwise direction,the net work done $W$ by the gas is positive $(W > 0)$.
Therefore,the net heat energy $Q$ supplied to the system must be positive $(Q > 0)$,meaning heat is supplied to the system.
Thus,option $(b)$ is correct.

(A) For an isothermal process,$P_1 V_1 = P_2 V_2$. Given $V_2 = 2V_1$,we have $P_1 V_1 = P_2 (2V_1)$,which gives $P_2 = P_1 / 2$.
Let the pressure be further decreased by a factor such that the new pressure is $P'$. We want to reach the original state $(P_1, V_1)$ from $(P', 2V_1)$ via an adiabatic process.
For an adiabatic process,$P' (2V_1)^\gamma = P_1 (V_1)^\gamma$,where $\gamma = 5/3$ for a monoatomic gas like neon.
$P' (2)^\gamma = P_1 \implies P' = P_1 (2)^{-\gamma} = P_1 (2)^{-5/3}$.
The pressure at the end of the isothermal expansion was $P_2 = P_1 / 2 = P_1 (2)^{-1}$.
The fractional decrease in pressure required is $\frac{P_2 - P'}{P_2} = \frac{P_1 (2)^{-1} - P_1 (2)^{-5/3}}{P_1 (2)^{-1}} = 1 - \frac{(2)^{-5/3}}{(2)^{-1}} = 1 - 2^{-5/3 + 1} = 1 - 2^{-2/3}$.

(A) According to the first law of thermodynamics: $\Delta Q = \Delta U + \Delta W$.
Given that the work done by the gas is equal to the increase in internal energy,we have $\Delta W = \Delta U$.
Substituting this into the first law equation: $\Delta Q = \Delta U + \Delta U = 2\Delta U$.
We know that $\Delta Q = nC\Delta T$ and $\Delta U = nC_v\Delta T$,where $C$ is the molar heat capacity for the process and $C_v$ is the molar heat capacity at constant volume.
Thus,$nC\Delta T = 2nC_v\Delta T$,which simplifies to $C = 2C_v$.
For an ideal gas,$C_v = \frac{R}{\gamma - 1}$.
Given $\gamma = 1.5$,we have $C_v = \frac{R}{1.5 - 1} = \frac{R}{0.5} = 2R$.
Therefore,$C = 2 \times (2R) = 4R$.

(A) The work done during the phase change is given by $W = P \Delta V$.
Given $P = 3 \times 10^5\,Pa$ and $\Delta V = 1600\,cm^3 = 1600 \times 10^{-6}\,m^3 = 1.6 \times 10^{-3}\,m^3$.
So,$W = (3 \times 10^5) \times (1.6 \times 10^{-3}) = 480\,J$.
It is given that $10\%$ of the total heat supplied $(Q)$ is used for this work.
Therefore,$0.10 \times Q = 480\,J$,which implies $Q = 4800\,J$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$,where $\Delta U$ is the change in internal energy.
Thus,$\Delta U = \Delta Q - W = 4800\,J - 480\,J = 4320\,J$.
Alternatively,since $10\%$ of heat is used for work,$90\%$ of the heat is used to increase the internal energy.
$\Delta U = 0.90 \times 4800\,J = 4320\,J$.

(B) The change in internal energy is given by $\Delta U = nC_{v} \Delta T$.
For an isothermal process, the temperature $T$ is constant, so $\Delta T = 0$, which implies $\Delta U = 0$. Thus, $A \longrightarrow II$.
For an adiabatic process, the heat exchange $\Delta Q = 0$. According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$, so $\Delta U = -\Delta W$. If work is done by the gas, $\Delta W > 0$, then $\Delta U < 0$, meaning internal energy decreases. Thus, $B \longrightarrow I$.
For an isochoric process, the volume $V$ is constant, so $\Delta V = 0$. Since work done $\Delta W = P \Delta V$, no work is done on or by the gas. Thus, $C \longrightarrow IV$.
For an isobaric process, the pressure $P$ is constant. The heat absorbed $\Delta Q$ is used to both increase the internal energy $\Delta U$ and perform work $\Delta W = P \Delta V$. Thus, $D \longrightarrow III$.
Therefore, the correct matching is $A-II, B-I, C-IV, D-III$.

(A) For a diatomic gas where molecules rotate but do not oscillate,the degrees of freedom $f = 5$ ($3$ translational + $2$ rotational).
The molar heat capacity at constant pressure is $C_p = \frac{f+2}{2}R = \frac{5+2}{2}R = \frac{7}{2}R$.
The molar heat capacity at constant volume is $C_v = \frac{f}{2}R = \frac{5}{2}R$.
The heat supplied at constant pressure is $Q = nC_p\Delta T = 735\,J$.
So,$n\left(\frac{7}{2}R\right)\Delta T = 735$,which gives $nR\Delta T = 735 \times \frac{2}{7} = 210\,J$.
The increase in internal energy is given by $\Delta U = nC_v\Delta T = n\left(\frac{5}{2}R\right)\Delta T$.
Substituting the value of $nR\Delta T$,we get $\Delta U = \frac{5}{2} \times 210 = 5 \times 105 = 525\,J$.

(D) Statement $I:$ $\Delta Q > 0$. According to the $1^{\text{st}}$ law of thermodynamics,$\Delta Q = \Delta U + W$. If heat is added to a system,it is possible for the system to perform work $W$ such that $W > \Delta Q$,which would result in $\Delta U < 0$ (a decrease in internal energy and temperature). Thus,Statement $I$ is false.
Statement $II:$ Work done by a system is given by $W = \int P \, dV$. If $W > 0$,then $\int P \, dV > 0$. Since pressure $P$ is always positive for a gas,the change in volume $dV$ must be positive,meaning the volume of the system must increase. Thus,Statement $II$ is true.

(D) For the isothermal process in container $A$,the temperature $T$ remains constant.
Using Boyle's Law: $P_1 V_1 = P_2 V_2$.
Given $P_1 = P$,$V_1 = V$,and $V_2 = V/8$.
$P \cdot V = P_A \cdot (V/8) \implies P_A = 8P$.
For the adiabatic process in container $B$,the relation is $P_1 V_1^\gamma = P_2 V_2^\gamma$,where $\gamma = 5/3$ for a monoatomic gas.
$P \cdot V^{5/3} = P_B \cdot (V/8)^{5/3}$.
$P_B = P \cdot (V / (V/8))^{5/3} = P \cdot (8)^{5/3}$.
Since $8 = 2^3$,$8^{5/3} = (2^3)^{5/3} = 2^5 = 32$.
So,$P_B = 32P$.
The ratio of the final pressure of gas in $B$ to that of gas in $A$ is $\frac{P_B}{P_A} = \frac{32P}{8P} = 4$.

(A) In a $P-V$ diagram,the slope of an adiabatic process is $\gamma$ times the slope of an isothermal process. Therefore,the steeper curve represents an adiabatic process.
$1$. Curve $B$ is steeper than curve $A$,so process $B$ is adiabatic. The equation for an adiabatic process is $PV^\gamma = k$.
$2$. Curve $A$ is less steep,so process $A$ is isothermal. The equation for an isothermal process is $PV = k$.
Comparing this with the given options,option $A$ correctly identifies process $B$ as adiabatic $(PV^\gamma = k)$ and process $A$ as isothermal $(PV = k)$.

(D) From the first law of thermodynamics,$dQ = du + dW$.
For a process,$C dT = C_V dT + P dV$,where $C$ is the molar heat capacity.
Dividing by $dT$,we get $C = C_V + P \frac{dV}{dT}$.
Given the equation of state $PV^2 = RT$.
Differentiating with respect to $T$ at constant pressure $P$:
$P(2V) \frac{dV}{dT} = R$.
Therefore,$P \frac{dV}{dT} = \frac{R}{2}$.
Substituting this into the expression for $C$:
$C = C_V + \frac{R}{2}$.

(B) For path $A$ to $B$, the process follows $PV^3 = RT$. Given $T = 300 \, K$ (constant temperature process is not implied, but $T$ is constant at $300 \, K$ for the state $A$ and $B$ as per the equation $PV^3 = RT$ at $T=300 \, K$).
$W_{AB} = \int_{V_A}^{V_B} P \, dV = \int_{2}^{4} \frac{RT}{V^3} \, dV = RT \int_{2}^{4} V^{-3} \, dV$
$W_{AB} = 8 \times 300 \times \left[ \frac{V^{-2}}{-2} \right]_{2}^{4} = 2400 \times \left( -\frac{1}{2} \right) \left( \frac{1}{16} - \frac{1}{4} \right) = -1200 \times \left( \frac{1-4}{16} \right) = -1200 \times \left( -\frac{3}{16} \right) = 75 \times 3 = 225 \, J$.
For path $B$ to $C$, it is an isobaric process at $P = 10 \, Pa$ from $V = 4 \, m^3$ to $V = 2 \, m^3$.
$W_{BC} = P(V_C - V_B) = 10(2 - 4) = -20 \, J$.
For path $C$ to $A$, it is an isochoric process at $V = 2 \, m^3$, so $W_{CA} = 0 \, J$.
Total work $W_{net} = W_{AB} + W_{BC} + W_{CA} = 225 - 20 + 0 = 205 \, J$.