Thinking Process

The slope of the curve for the adiabatic process will be more that is the curve will be steeper. Slope of $p-V$ curve in adiabatic process $=\gamma (p/V)$ where as slope of is otemal process $=-p/v$

Answer (c) For the curve 4 pressure is constant, so this is an isobaric process.

For the curve 1, volume is constant, so it is isochoric process. Between curves 3 and 2 , curve 2 is steeper, so it is adiabatic and 3 is isothermal.

Note We should be careful while deciding between isothermal and adiabatic curves because these curves look similar.

• For option (a) 4: The pressure is constant, so this is an isobaric process.
• For option (b) 3: The curve is less steep compared to curve 2, indicating it is an isothermal process.
• For option (d) 1: The volume is constant, so it is an isochoric process.

Answer (a) Amount of sweat evaporated/minute

$$\begin{array}{rl}& =\frac{\text{ Sweat produced / minute }}{\text{ Number of calories required for evaporation }/kg}\\ & =\frac{\text{ Amount of heat produced per minute in jogging }}{\text{ Latent heat (in cal }/kg\text{ ) }}\\ & =\frac{14.5\times {10}^3}{580\times {10}^3}=\frac{145}{580}=0.25kg\end{array}$$

• Option (b) $2.25kg$: This option is incorrect because it suggests a much higher amount of sweat evaporated per minute than calculated. Given the heat produced per minute and the latent heat of evaporation, the correct calculation yields 0.25 kg, not 2.25 kg. The value 2.25 kg would imply an unrealistic amount of sweat evaporation for the given energy expenditure.

• Option (c) $0.05kg$: This option is incorrect because it underestimates the amount of sweat evaporated per minute. The correct calculation shows that 0.25 kg of sweat is evaporated per minute, not 0.05 kg. This value would imply that the energy expenditure is much lower than given.

• Option (d) $0.20kg$: This option is incorrect because it slightly underestimates the amount of sweat evaporated per minute. The correct calculation based on the given energy expenditure and latent heat of evaporation results in 0.25 kg, not 0.20 kg. This discrepancy indicates a miscalculation or incorrect assumption about the energy required for evaporation.

Answer (c) According to the question given that $pV=$ constant Hence, we can say that the gas is going through an isothermal process.

Clearly, from the graph that between process 1 and 2 temperature is constant and the gas expands and pressure decreases i.e., $p_2<p_1$ which corresponds to diagram (iii).

• Option (i): This diagram shows a linear relationship between temperature (T) and pressure (p), which is characteristic of an isochoric process (constant volume). However, the given process is isothermal (constant temperature), so this option is incorrect.

• Option (ii): This diagram shows a direct proportionality between temperature (T) and pressure (p), which is characteristic of an isochoric process (constant volume). Since the process is isothermal, this option does not represent the correct relationship.

• Option (iv): This diagram shows an inverse relationship between temperature (T) and pressure (p), which is characteristic of an adiabatic process (no heat exchange). The given process is isothermal, so this option is incorrect.

Thinking Process

Work done in a process by which a gas is going through can be calculated by area of the $p$ - $V$ diagram.

Answer (b) Consider the $p$ - $V$ diagram given in the question.

Work done in the process $ABCD=$ area of rectangle $ABCDA$

$$\begin{array}{rl}& =(AB)\times BC=(3V_0-V_0)\times (2p_0-p_0)\\ & =2V_0\times p_0=2p_0{V}_0\end{array}$$

As the process is going anti-clockwise, hence there is a net compression in the gas. So, work done by the gas $=-2p_0{V}_0$.

• Option (a) $6p_0{V}_0$: This option is incorrect because the area of the rectangle $ABCDA$ is calculated as $2p_0{V}_0$, not $6p_0{V}_0$. The dimensions of the rectangle are $(3V_0-V_0)$ in volume and $(2p_0-p_0)$ in pressure, which results in an area of $2p_0{V}_0$.

• Option (c) $+2p_0{V}_0$: This option is incorrect because although the magnitude of the work done is $2p_0{V}_0$, the process is going anti-clockwise, indicating a net compression. Therefore, the work done by the gas is negative, not positive.

• Option (d) $+4p_0{V}_0$: This option is incorrect because the area of the rectangle $ABCDA$ is $2p_0{V}_0$, not $4p_0{V}_0$. Additionally, the process is anti-clockwise, indicating a net compression, so the work done by the gas should be negative, not positive.

Answer (a) Consider the $p-V$ diagram shown for the container $A$ (isothermal) and for container $B$ (adiabatic).

Container $A$ (Isothermal)

Container $B$ (Adiabatic)

Both the process involving compression of the gas.

For isothermal process (gas $A$ ) (during $1\to 2$ )

$p_1{V}_1=p_2{V}_2$

$\Rightarrow p_0(2V_0)=p_2(V_0)$

$\Rightarrow p_2=2p_0$

For adiabatic process, (gas $B$ ) (during $1\to 2$ ) $$\begin{array}{rl}{p}_1{V}_1^{\gamma }& =p_2{V}_2^{\gamma }\\ \Rightarrow & p_0{\left(2V_0\right)}^{\gamma }=p_2{\left(V_0\right)}^{\gamma }\\ \Rightarrow & p_2={\left(\frac{2V_0}{V_0}\right)}^{\gamma }{p}_0=(2{)}^{\gamma }{p}_0\end{array}$$

Hence, $\frac{{\left(p_2\right)}_B}{{\left(p_2\right)}_A}=$ Ratio of final pressure $=\frac{(2{)}^{\gamma }{p}_0}{2p_0}=2^{\gamma -1}$ where, $\gamma$ is ratio of specific heat capacities for the gas.

• Option (b) $(\frac{1}{2}{)}^{\gamma -1}$: This option is incorrect because it suggests that the final pressure of gas in container B is less than that in container A by a factor of $(\frac{1}{2}{)}^{\gamma -1}$. However, the adiabatic process results in a higher final pressure compared to the isothermal process due to the additional work done on the gas without heat exchange. Therefore, the ratio should be greater than 1, not less than 1.

• Option (c) $(\frac{1}{1-\gamma }{)}^2$: This option is incorrect because it does not correctly represent the relationship between the pressures in the adiabatic and isothermal processes. The expression $(\frac{1}{1-\gamma }{)}^2$ does not align with the physical principles governing the adiabatic compression, where the pressure increases by a factor of $2^{\gamma }$ compared to the initial pressure.

• Option (d) $(\frac{1}{\gamma -1}{)}^2$: This option is incorrect because it incorrectly represents the ratio of the final pressures. The correct ratio involves the factor $2^{\gamma -1}$, which directly comes from the relationship between the pressures in the adiabatic and isothermal processes. The expression $(\frac{1}{\gamma -1}{)}^2$ does not correctly capture the effect of the adiabatic compression on the pressure.

Answer (b) Let the equilibrium temperature of the system is $T$.

Let us assume that $T_1,T_2<T<T_3$.

According to question, there is no net loss to the surroundings.

Heat lost by $M_3=$ Heat gained by $M_1+$ Heat gained by $M_2$

$$\begin{array}{cc}\Rightarrow & M_{3}s(T_3-T)=M_{1}s(T-T_1)+M_{2}s(T-T_2)\\ \Rightarrow & \text{ (where, s is specific heat of the copper material) }\\ \Rightarrow & T[M_1+M_2+M_3]=M_3{T}_3+M_1{T}_1+M_2{T}_2\\ \Rightarrow & T=\frac{M_1{T}_1+M_2{T}_2+M_3{T}_3}{M_1+M_2+M_3}\end{array}$$

• Option (a) is incorrect because it assumes that the equilibrium temperature is simply the arithmetic mean of the initial temperatures of the three blocks. This does not take into account the different masses of the blocks, which affect the amount of heat each block can absorb or release.

• Option (c) is incorrect because it divides the weighted sum of the initial temperatures by three times the total mass, which does not correctly account for the conservation of energy. The correct formula should only divide by the total mass, not three times the total mass.

• Option (d) is incorrect because it introduces an unnecessary multiplication of the specific heat (s) with the temperatures in the numerator. Since the specific heat (s) is a constant and cancels out in the equation, it should not appear in the final expression for the equilibrium temperature.

Thinking Process

If any process can be returned back such that both, the system and the surroundings return to their original states, with no other change anywhere else in the universe, then this process is called reversible process.

Answer $(a,b,d)$

(a) When the rod is hammered, the external work is done on the rod which increases its temperature. The process cannot be retraced itself.

(b) In this process energy in the form of heat is transferred to the gas in the small container by big reservoir at temperature $T_2$.

(d) As the weight is added to the cylinder arrangement in the form of external pressure hence, it cannot be reversed back itself.

• (c) A quasi-static isothermal expansion of an ideal gas in a cylinder fitted with a frictionless piston is a reversible process because it can be reversed by infinitesimally small changes in the external conditions, and the system remains in thermodynamic equilibrium throughout the process.

Answer $(a,d)$

For an isothermal process change in temperature of the system $dT=0\Rightarrow T=$ constant.

We know that for an ideal gas $dU=$ change in internal energy $=n{C}_{V}dT=0$

[where, $n$ is number of moles and $C_V$ is specific heat capacity at constant volume] From first law of thermodynamics,

$$\begin{array}{rl}dQ& =dU+dW\\ & =0+dW\Rightarrow dQ=dW\end{array}$$

• (b) (d Q=0): This option is incorrect because, in an isothermal process, heat is exchanged with the surroundings to maintain a constant temperature. Therefore, (d Q) is not zero; instead, it is equal to the work done by or on the system.

• (c) (d Q=d U): This option is incorrect because, in an isothermal process for an ideal gas, the change in internal energy (d U) is zero (since internal energy is a function of temperature, which remains constant). Therefore, (d Q) cannot be equal to (d U); instead, (d Q) is equal to the work done (d W).

Thinking Process

Internal energy is a state function and work done by the gas is a path dependent function. The work done in a thermodynamical process is equal to the area bounded between $p$ - $V$ curve.

Answer (b, c)

Change in internal energy for the process $A$ to $B$

$$d{U}_{AarrowB}=n{C}_{V}dT=n{C}_V(T_B-T_A)$$

which depends only on temperatures at $A$ and $B$.

Work done for $A$ to $B,d{W}_{AarrowB}=$ Area under the $p-V$ curve which is maximum for the path $I$.

• Option (a): Change in internal energy is a state function and depends only on the initial and final states, not on the path taken. Therefore, the change in internal energy is the same for all paths (I, II, III, and IV) from A to B. Hence, it is incorrect to say that the change in internal energy is the same in IV and III but not in I and II.

• Option (d): The work done is represented by the area under the $p-V$ curve. Since path II is a straight vertical line, it represents an isochoric process (constant volume), and no work is done in an isochoric process. Therefore, the work done in case II is zero, which is indeed the minimum possible. However, this option is not selected because the correct answer already includes the maximum work done in case I, which is more relevant to the question.

Answer $(a,c)$

(a) The given process is a cyclic process i.e., it returns to the original state 1 .

Hence, change in internal energy $dU=0$

$\Rightarrow dQ=dU+dW=0+dW=dW$

Hence, total heat supplied is converted to work done by the gas (mechanical energy) which is not possible by second law of thermodynamics.

(c) When the gas expands adiabatically from 2 to 3 . It is not possible to return to the same state without being heat supplied, hence the process 3 to 1 cannot be adiabatic.

• (b) Mechanical energy is not completely converted to heat in this process. In a cyclic process, the net work done over one complete cycle is equal to the net heat added to the system. Therefore, the statement that mechanical energy is completely converted to heat is incorrect.

• (d) Curves representing an adiabatic process and an isothermal process can intersect. An isothermal process occurs at a constant temperature, while an adiabatic process occurs without heat exchange. These processes can intersect on a PV diagram, as they represent different paths between the same states.

Answer (a, c)

Consider the figure we can write

$$Q_1=W+Q_2$$

$$\Rightarrow W=Q_1-Q_2>0\text{ (By question) }$$

$$\Rightarrow Q_1>Q_2>0(\text{ If both }{Q}_1\text{ and }{Q}_2\text{ are positive })$$

$$\text{ We can also, write }{Q}_2<Q_1<0(\text{ If both }{Q}_1\text{ and }{Q}_2\text{ are negative }).$$

• Option (b) $Q_2>Q_1>0$ is incorrect because if $Q_2$ were greater than $Q_1$ and both were positive, then $W=Q_1-Q_2$ would be negative, which contradicts the given condition that $W>0$.

• Option (d) $Q_1<0,Q_2>0$ is incorrect because if $Q_1$ were negative and $Q_2$ were positive, then $W=Q_1-Q_2$ would be negative, which again contradicts the given condition that $W>0$.

Thinking Process

We have to apply first law of thermodynamics for each path.

Answer For path 1, Heat given $Q_1=+1000J$ Work done $=W_1$ (let)

For path 2,

$$\begin{array}{rl}\text{ Work done }(W_2)& =(W_1-100)J\\ \text{ Heat given }{Q}_2& =?\end{array}$$

As change in internal energy between two states for different path is same.

$$\begin{array}{ccc}\therefore & \mathrm{\Delta }U& =Q_1-W_1=Q_2-W_2\\ & \Rightarrow & 1000-W_1& =Q_2-(W_1-100)\\ & Q_2& =1000-100=900J\end{array}$$

Answer Yes, during adiabatic compression the temperature of a gas increases while no heat is given to it.

In adiabatic compression, $dQ=0$

$\therefore$ From first law of thermodynamics, $dU=dQ-dW$

$dU=-dW$

In compression work is done on the gas i.e., work done is negative.

Therefore, $dU=$ Positive

Hence, internal energy of the gas increases due to which its temperature increases.

Answer During, driving, temperature of the gas increases while its volume remains constant.

So, according to Charle’s law, at constant volume $(V)$,

Pressure $(p)\propto$ Temperature $(T)$

Therefore, pressure of gas increases.

Thinking Process

The efficiency of a Carnot’s engine is $\eta =1-\frac{T_2}{T_1^{\prime }}$

where, $T_2$ is temperature of the sink and $T_1$ is temperature of the source.

Answer Given, temperature of the source $T_1=500K$

Temperature of the $sink{T}_2=300K$

Work done per cycle $W=1kJ=1000J$

Heat transferred to the engine per cycle $Q_1=$ ?

Efficiency of a Carnot engine $(\eta )=1-\frac{T_2}{T_1}=1-\frac{300}{500}=\frac{200}{500}=\frac{2}{5}$

$$\begin{array}{cc}\text{ and }& \eta =\frac{W}{Q_1}\\ \Rightarrow & Q_1=\frac{W}{\eta }=\frac{1000}{(2/5)}=2500J\end{array}$$

Thinking Process

Potential energy(PE) of an object at height $(h)$ is $mgh$. The energy in the form offat will be utilised to increase PE of the person. Thus, the calorie consumed by the person in going up is mgh, then according to problem calorie consumed by the person in coming down is $\frac{1}{2}mgh$.

Answer Given,

height of the stairs $=h=10m$

Energy produced by burning $1kg$ of fat $=7000kcal$

$\therefore$ Energy produced by burning $5kg$ of fat $=5\times 7000=35000kcal$

$$=35\times {10}^{6}cal$$

Energy utilised in going up and down one time

$$\begin{array}{rl}& =mgh+\frac{1}{2}mgh=\frac{3}{2}mgh\\ & =\frac{3}{2}\times 60\times 10\times 10\\ & =9000J=\frac{9000}{4.2}=\frac{3000}{1.4}cal\end{array}$$

$\therefore$ Number of times, the person has to go up and down the stairs

$$\begin{array}{rl}& =\frac{35\times {10}^6}{(3000/1.4)}=\frac{35\times 1.4\times {10}^6}{3000}\\ & =16.3\times {10}^3\text{ times }\end{array}$$

Thinking Process

There is no exchange of heat in the process, hence this can be considered as an adiabatic process.

Answer Let, volume is increased by $\mathrm{\Delta }V$ and pressure is increased by $\mathrm{\Delta }p$ by an stroke.

For just before and after an stroke, we can write

$$\begin{array}{ccc}{p}_1{V}_1^{\gamma }& =p_2{V}_2^{\gamma }& \\ \Rightarrow p(V+\mathrm{\Delta }V{)}^{\gamma }& =(p+\mathrm{\Delta }p)V^{\gamma }\\ \Rightarrow p{V}^{\gamma }(1+\frac{\mathrm{\Delta }V}{V}{)}^{\gamma }& =p(1+\frac{\mathrm{\Delta }p}{p})V^{\gamma }\\ \Rightarrow p{V}^{\gamma }(1+\gamma \frac{\mathrm{\Delta }V}{V})& \approx p{V}^{\gamma }(1+\frac{\mathrm{\Delta }p}{p})(\because \text{ volume is fixed })\\ \Rightarrow \gamma \frac{\mathrm{\Delta }V}{V}& =\frac{\mathrm{\Delta }p}{p}\Rightarrow \mathrm{\Delta }V=\frac{1}{\gamma }\frac{V}{p}\mathrm{\Delta }p\\ dV& =\frac{1}{\gamma }\frac{V}{p}dp\end{array}$$

Hence, work done is increasing the pressure from $p_1$ to $p_2$

W = ${\int }_{p_1}^{p_2}pdV={\int }_{p_1}^{p_2}p\times \frac{1}{\gamma }\frac{V}{p}dp$

= $\frac{V}{\gamma }{\int }_{p_1}^{p_2}dp=\frac{V}{\gamma }(p_2-p_1)$

$W=\frac{(p_2-p_1)}{\gamma }V$

Thinking Process

The Carnot engine is the most efficient heat engine operating between two given temperature. This is why it is known as perfect engine. The efficiency of Carnot engine is

$$\eta =1-\frac{T_2}{T_1}$$

Answer Given, temperature of the source is ${27}^{\circ }C$

$$\begin{array}{cc}\Rightarrow & T_1=(27+273)K=300K\\ \text{ Temperature of sink }& T_2=(-3+273)K=270K\end{array}$$

Efficiency of a perfect heat engine is given by

$$\eta =1-\frac{T_2}{T_1}=1-\frac{270}{300}=\frac{1}{10}$$

Efficiency of refrigerator is $50$ of a perfect engine

$\therefore {\eta }^{\prime }=0.5\times \eta =\frac{1}{2}\eta =\frac{1}{20}$

$\therefore$ Coefficient of performance of the refrigerator

$$\begin{array}{rl}\beta & =\frac{Q_2}{W}=\frac{1-{\eta }^{\prime }}{{\eta }^{\prime }}\\ & =\frac{1-(1/20)}{(1/20)}=\frac{19/20}{1/20}=19\\ \Rightarrow Q_2& =\beta W=19W\\ & =19\times (1kW)=19kW=19kJ/s.\end{array}(\because \beta =\frac{Q_2}{W})$$

Therefore, heat is taken out of the refrigerator at a rate of $19kJ$ per second.

Thinking Process

Coefficient of performance $(\beta )$ of a refrigerator is ratio of quantity of heat removed per cycle $(Q_2)$ to the amount of work done on the refrigerator.

Answer Given, coefficient of performace $(\beta )=5$

$$T_1=(27+273)K=300K,T_2=?$$

Coefficient of performance $(\beta )=\frac{T_2}{T_1-T_2}$

$$\begin{array}{rlrl}& & 5& =\frac{T_2}{300-T_2}\Rightarrow 1500-5T_2=T_2\\ \Rightarrow & & 6T_2& =1500\Rightarrow T_2=250K\\ \Rightarrow & & T_2& =(250-273{)}^{\circ }C=-{23}^{\circ }C\end{array}$$

Answer Consider the diagram $p-V$, where variation is shown for each process.

Process 1 is isobaric and process 2 is isothermal.

Since, work done $=$ area under the $p-V$ curve. Here, area under the $p-V$ curve 1 is more. So, work done is more when the gas expands in isobaric process.

Answer Let $p{V}^{1/2}=$ Constant $=K,p=\frac{K}{\sqrt{V}}$

(a) Work done for the process 1 to 2 ,

W = ${\int }_{V_1}^{V_2}pdV=K{\int }_{V_1}^{V_2}\frac{dV}{\sqrt{V}}=K[\frac{\sqrt{V}}{1/2}{]}_{V_1}^{V_2}=2K(\sqrt{V_2}-\sqrt{V_1})$

$=2p_1{V}_1^{1/2}(\sqrt{V_2}-\sqrt{V_1})=2p_2{V}_2^{1/2}(\sqrt{V_2}-\sqrt{V_1})$

(b) From ideal gas equation,

$pV=nRT\Rightarrow T=\frac{pV}{nR}=\frac{p\sqrt{V}\sqrt{V}}{nR}$

$\text{ Hence, }T=\frac{K\sqrt{V}}{nR}(\text{ As, }p\sqrt{V}=K)$

$\Rightarrow T_1=\frac{K\sqrt{V_1}}{nR}\Rightarrow T_2=\frac{K\sqrt{V_2}}{nR}$

$\frac{T_1}{T_2}=\frac{\frac{K\sqrt{V_1}}{nR}}{\frac{K\sqrt{V_2}}{nR}}=\sqrt{\frac{V_1}{V_2}}=\sqrt{\frac{V_1}{2V_1}}=\frac{1}{\sqrt{2}}(\because V_2=2V_1)$

(c) Given, internal energy of the gas $=U=(\frac{3}{2})RT$

$$\begin{array}{rl}\mathrm{\Delta }U& =U_2-U_1=\frac{3}{2}R(T_2-T_1)\\ & =\frac{3}{2}R{T}_1(\sqrt{V}-1)[\because T_2=\sqrt{2}{T}_1\text{ from (b) }]\\ \mathrm{\Delta }W& =2p_1{V}_1^{1/2}(\sqrt{V_2}-\sqrt{V_1})\\ & =2p_1{V}_1^{1/2}(\sqrt{2}\times \sqrt{V_1}-\sqrt{V_1})\\ & =2p_1{V}_1(\sqrt{2}-1)=2R{T}_1(\sqrt{2}-1)\\ & =\frac{3}{2}R{T}_1(\sqrt{2}-1)+2R{T}_1(\sqrt{2}-1)\\ & =(\sqrt{2}-1)R{T}_1(2+3/2)\\ & =(\frac{7}{2})R{T}_1(\sqrt{2}-1)\end{array}$$

This is the amount of heat supplied.

Answer (a) For the process $AB$,

$$\begin{array}{rl}dV& =0\Rightarrow dW=0(\because \text{ volume is constant })\\ \Rightarrow & dQ=dU+dW=dU\\ dQ& =dU=\text{ Change in internal energy. }\end{array}$$

Hence, in this process heat supplied is utilised to increase, internal energy of the system. Since, $p=(\frac{nR}{V})T$, in isochoric process, $T\propto p$. So temperature increases with increases of pressure in process $AB$ which inturn increases internal energy of the system i.e., $dU>0$. This imply that $dQ>0$. So heat is supplied to the system in process $AB$.

(b) For the process $CD$, volume is constant but pressure decreases.

Hence, temperature also decreases so heat is given to surroundings.

(c) To calculate work done by the engine in one cycle, we calculate work done in each part separately.

Similarly, $W_{DA}=\frac{p_A{V}_A-p_D{V}_D}{1-\gamma }[\because BC$ is adiabatic process $]$

$\because B$ and $C$ lies on adiabatic curve $BC$.

$\therefore p_B{V}_B^{\gamma }=p_C{V}_C^{\gamma }$

$p_C=p_B(\frac{V_B}{V_C}{)}^{\gamma }=p_B(\frac{1}{2}{)}^{\gamma }=2^{-\gamma }{p}_B$ $p_D=2^{-\gamma }{p}_A$

$W=W_{AB}+W_{BC}+W_{CD}+W_{DA}=W_{BC}+W_{DA}$

$=\frac{(p_C{V}_C-p_B{V}_B)}{1-\gamma }+\frac{(p_A{V}_A-p_D{V}_D)}{1-\gamma }$

$W=\frac{1}{1-\gamma }[2^{-\gamma }{p}_B(2V_B)-p_B{V}_B+p_A{V}_A-2^{-\gamma }{p}_B(2V_B)]$

$=\frac{1}{1-\gamma }[p_B{V}_B(2^{-\gamma +1}-1)-p_A{V}_A(2^{-\gamma +1}-1).$

$=\frac{1}{1-\gamma }(2^{1-\gamma }-1)(p_B-p_A)V_A$

$=\frac{3}{2}[1-(\frac{1}{2}{)}^{(}2/3](p_B-p_A)V_A$

Thinking Process

Find amount of heat associated with each process by using first law of thermodynamics.

Answer (a) For process $AB$,

Volume is constant, hence work done $dW=0$

Now, by first law of thermodynamics,

$$\begin{array}{rl}dQ& =dU+dW=dU+0=dU\\ & =nCvdT=nCv(T_B-T_A)\\ & =\frac{3}{2}R(T_B-T_A)\\ & =\frac{3}{2}(R{T}_B-R{T}_A)=\frac{3}{2}(p_B{V}_B-p_A{V}_A)\end{array}$$

Heat exchanged $=\frac{3}{2}(p_B{V}_B-p_A{V}_A)$ (b) For process $BC$,

$$\begin{array}{rl}p& =\text{ constant }\\ dQ& =dU+dW=\frac{3}{2}R(T_C-T_B)+p_B(V_C-V_B)\\ & =\frac{3}{2}(p_C{V}_C-p_B{V}_B)+p_B(V_C-V_B)=\frac{5}{2}{p}_B(V_C-V_B)\end{array}$$

Heat exchanged $=\frac{5}{2}{p}_B(V_C-V_A)$

$(\because p_B=p_C.$ and $.p_B=V_A)$

(c) For process $CD$, Because $CD$ is adiabatic, $dQ=$ Heat exchanged $=0$

(d) $DA$ involves compression of gas from $V_D$ to $V_A$ at constant pressure $p_A$.

$\therefore$ Heat exchanged can be calculated by similar way as $B{C}_1$,

Hence,

$$dQ=\frac{5}{2}{p}_A(V_A-V_D)$$

Answer According to question, slope of the curve $=f(V)$, where $V$ is volume.

$\therefore$ Slope of $p=f(V)$, curve at $(V_0,p_0)=f(V_0)$

Slope of adiabatic at $(V_0,p_0)=k(-\gamma )V_0^{-1-\gamma }=-\gamma p_0/V_0$

Now heat absorbed in the process $p=f(V)$

$dQ=dU+dW=n{C}_{V}dT+pdV$

$\because pV=nRT\Rightarrow T=(\frac{1}{nR})pV$

$\Rightarrow T=(\frac{1}{nR})Vf(V)$

$\Rightarrow dT=(\frac{1}{nR})[f(V)+V{f}^{\prime }(V)]dV$

Now from Eq. (i)

$$\frac{dQ}{dV}=n{C}_V\frac{dT}{dV}+p\frac{dV}{dV}=n{C}_V\frac{dT}{dV}+p$$

$$=\frac{n{C}_V}{nR}\times [f(V)+V{f}^{\prime }(V)]+p$$

$$=\frac{C_V}{R}[f(V)+V{f}^{\prime }(V)]+f(V)$$

$\to [\frac{dQ}{dV}{]}_{V=V_0}$=

$\frac{C_V}{R}$ $[f(V_0)+(V_0)f^{\prime }]$ $(V_0)+f(V_0)$

$f(V_0)[\frac{C_V}{R}+1]+V_0{f}^{\prime }(V_0)\frac{C_V}{R}$

$\because C_V=\frac{R}{\gamma -1}\Rightarrow \frac{C_V}{R}=\frac{1}{\gamma -1}$

$\Rightarrow [\frac{dQ}{dV}{]}_{V=V_0}=[\frac{1}{\gamma -1}+1]f(V_0)+\frac{V_0{f}^{\prime }(V_0)}{\gamma -1}$

$=\frac{\gamma }{\gamma -1}{p}_0+\frac{V_0}{\gamma -1}{f}^{\prime }(V_0)$

Heat is absorbed where $\frac{dQ}{dV}>0$, when gas expands

Hence, $\gamma p_0+V_0{f}^{\prime }(V_0)>0$ or $f^{\prime }(V_0)>(-\gamma \frac{p_0}{V_0})$

Thinking Process

We will assume the piston is massless, hence, at equilibrium atmospheric pressure and inside pressure will be same.

Answer (a) Initially the piston is in equilibrium hence, $p_i=p_a$

(b) On supplying heat, the gas expands from $V_0$ to $V_1$

$\therefore$ Increase in volume of the gas $=V_1-V_0$

As the piston is of unit cross-sectional area hence, extension in the spring

$$x=\frac{V_1-V_0}{\text{ Area }}=V_1-V_0$$

$\therefore$ Force exerted by the spring on the piston

Hence,

$$=F=kx=k(V_1-V_0)$$

$$\begin{array}{rl}\text{ final pressure }& =p_f=p_a+kx\\ & =p_a+k\times (V_1-V_0)\end{array}$$

(c) From first law of thermodynamics $dQ=dU+dW$

If $T$ is final temperature of the gas, then increase in internal energy

$$dU=C_V(T,-T_0)=C_V(T,-T_0)$$

We can write,

$$T=\frac{p_f{V}_1}{R}=[\frac{p_a+k(V_1-V_0)}{R}]\frac{V_1}{R}$$

Work done by the gas $=pdV+$ increase in PE of the spring

$$=p_a(V_1-V_0)+\frac{1}{2}k{x}^2$$

Now, we can write $dQ=dU+dW$

$$\begin{array}{rl}& =C_V(T-T_0)+p_a(V-V_0)+\frac{1}{2}k{x}^2\\ & =C_V(T-T_0)+p_a(V-V_0)+\frac{1}{2}(V_1-V_0{)}^2\end{array}$$

This is the required relation.