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.

Answer (a) Amount of sweat evaporated/minute

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

Answer (c) According to the question given that $p V=$ 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).

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 $A B C D=$ area of rectangle $A B C D A$

$$ \begin{aligned} & =(A B) \times B C=(3 V_0-V_0) \times(2 p_0-p_0) \\ & =2 V_0 \times p_0=2 p_0 V_0 \end{aligned} $$

As the process is going anti-clockwise, hence there is a net compression in the gas. So, work done by the gas $=-2 p_0 V_0$.

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 $1 \rightarrow 2$ )

$ p_1 V_1 =p_2 V_2 $

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

$\Rightarrow p_2 =2 p_0$

For adiabatic process, (gas $B$ ) (during $1 \rightarrow 2$ ) $$ \begin{array}{rlrl} p_1 V _1 ^\gamma & =p _2 V _2 ^\gamma \\ \Rightarrow & p _0\left(2 V_0\right)^\gamma =p_2\left(V_0\right)^\gamma \\ \Rightarrow & p _2 =\left(\frac{2 V _0}{V _0}\right)^\gamma p_0=(2)^\gamma p _0 \end{array} $$

Hence, $\frac{\left(p_2\right) _B}{\left(p _2\right)_A}=$ Ratio of final pressure $=\frac{(2)^\gamma p_0}{2 p _0}=2^{\gamma-1}$ where, $\gamma$ is ratio of specific heat capacities for the gas.

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 $M_2$

$$ \begin{matrix} \Rightarrow & M_3 s(T_3-T)=M_1 s(T-T_1)+M_2 s(T-T_2) \\ \Rightarrow & \quad \text { (where, s is specific heat of the copper material) } \\ \Rightarrow & T[M_1+M_2+M_3]=M_3 T_3+M_1 T_1+M_2 T_2 \\ \Rightarrow & T=\frac{M_1 T_1+M_2 T_2+M_3 T_3}{M_1+M_2+M_3} \end{matrix} $$

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.

Answer $(a, d)$

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

We know that for an ideal gas $d U=$ change in internal energy $=n C_{V} d T=0$

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

$$ \begin{aligned} d Q & =d U+d W \\ & =0+d W \Rightarrow d Q=d W \end{aligned} $$

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 $B$

$$ d U_{A arrow B}=n C_{V} d T=n C_{V}(T_{B}-T_{A}) $$

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

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

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 $d U=0$

$\Rightarrow \quad d Q=d U+d W=0+d W=d W$

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.

Answer (a, c)

Consider the figure we can write

$$Q_1=W+Q_2$$

$$\Rightarrow W=Q _1-Q _2>0 \text { (By question) }$$

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

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

Thinking Process

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

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

For path 2,

$$ \begin{aligned} \text { Work done }(W_2) & =(W_1-100) J \\ \text { Heat given } Q_2 & =? \end{aligned} $$

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

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

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

In adiabatic compression, $\quad d Q=0$

$\therefore$ From first law of thermodynamics, $\quad d U=d Q-d W$

$d U=-d W$

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

Therefore, $\quad d U=$ 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 $T_1=500 K$

Temperature of the $sink T_2=300 K$

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

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

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

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

Thinking Process

Potential energy(PE) of an object at height $(h)$ is $m g h$. 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 $\frac{1}{2} m g h$.

Answer Given,

height of the stairs $=h=10 m$

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

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

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

Energy utilised in going up and down one time

$$ \begin{aligned} & =m g h+\frac{1}{2} m g h=\frac{3}{2} m g h \\ & =\frac{3}{2} \times 60 \times 10 \times 10 \\ & =9000 J=\frac{9000}{4.2}=\frac{3000}{1.4} cal \end{aligned} $$

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

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

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 $\Delta V$ and pressure is increased by $\Delta p$ by an stroke.

For just before and after an stroke, we can write

$$ \begin{matrix} p_1 V_1^{\gamma} & =p_2 V_2^{\gamma} & \\ \Rightarrow \quad p(V+\Delta V)^{\gamma} & =(p+\Delta p) V^{\gamma} \\ \Rightarrow \quad p V^{\gamma}(1+\frac{\Delta V}{V})^{\gamma} & =p(1+\frac{\Delta p}{p}) V^{\gamma} \\ \Rightarrow \quad p V^{\gamma}(1+\gamma \frac{\Delta V}{V}) & \approx p V^{\gamma}(1+\frac{\Delta p}{p}) \quad(\because \text { volume is fixed }) \\ \Rightarrow \quad \gamma \frac{\Delta V}{V} & =\frac{\Delta p}{p} \Rightarrow \Delta V=\frac{1}{\gamma} \frac{V}{p} \Delta p \\ d V & =\frac{1}{\gamma} \frac{V}{p} d p \end{matrix} $$

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

W = $\int_{p_1}^{p_2} p d V= \int_{p_1}^{p_2} p \times \frac{1}{\gamma} \frac{V}{p} d p $

= $\frac{V}{\gamma} \int _{p _ 1}^{p _ 2} d p= \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 $27^{\circ} C$

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

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

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

$\therefore$ Coefficient of performance of the refrigerator

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

Therefore, heat is taken out of the refrigerator at a rate of $19 kJ$ 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 $(\beta)=5$

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

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

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

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} p d V=K \int_{V_1}^{V_2} \frac{d V}{\sqrt{V}}=K[\frac{\sqrt{V}}{1 / 2}]_{V_1}^{V_2}=2 K(\sqrt{V_2}-\sqrt{V_1}) $

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

(b) From ideal gas equation,

$p V=n R T \Rightarrow T=\frac{p V}{n R}=\frac{p \sqrt{V} \sqrt{V}}{n R} $

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

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

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

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

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

This is the amount of heat supplied.

Answer (a) For the process $A B$,

$$ \begin{aligned} d V & =0 \Rightarrow d W=0 \quad(\because \text { volume is constant }) \\ \Rightarrow \quad & d Q=d U+d W=d U \\ d Q & =d U=\text { Change in internal energy. } \end{aligned} $$

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

(b) For the process $C D$, 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, $\quad W_{D A}=\frac{p_{A} V_{A}-p_{D} V_{D}}{1-\gamma} \quad[\because B C$ is adiabatic process $]$

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

$ \therefore \quad 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_{A B}+W_{B C}+W_{C D}+W_{D A}=W_{B C}+W_{D A} $

$ =\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}(2 V_{B})-p_{B} V_{B}+p_{A} V_{A}-2^{-\gamma} p_{B}(2 V_{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 $A B$,

Volume is constant, hence work done $d W=0$

Now, by first law of thermodynamics,

$$ \begin{aligned} d Q & =d U+d W=d U+0=d U \\ & =n C v d T=n C v(T_{B}-T_{A}) \\ & =\frac{3}{2} R(T_{B}-T_{A}) \\ & =\frac{3}{2}(R T_{B}-R T_{A})=\frac{3}{2}(p_{B} V_{B}-p_{A} V_{A}) \end{aligned} $$

Heat exchanged $=\frac{3}{2}(p_{B} V_{B}-p_{A} V_{A})$ (b) For process $B C$,

$$ \begin{aligned} p & =\text { constant } \\ d Q & =d U+d W=\frac{3}{2} R(T_{C}-T_{B})+p_{B}(V_{C}-V_{B}) \\ & =\frac{3}{2}(p_{C} V_{C}-p_{B} V_{B})+p_{B}(V_{C}-V_{B})=\frac{5}{2} p_{B}(V_{C}-V_{B}) \end{aligned} $$

Heat exchanged $=\frac{5}{2} p_{B}(V_{C}-V_{A})$

$(\because p_{B}=p_{C}.$ and $.p_{B}=V_{A})$

(c) For process $C D$, Because $C D$ is adiabatic, $d Q=$ Heat exchanged $=0$

(d) $D A$ 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,

$$ d Q=\frac{5}{2} p_{A}(V_{A}-V_{D}) $$

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

$\therefore \quad$ 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 $p=f(V)$

$ d Q=d U+d W=n C_{V} d T+p d V $

$\because p V=n R T \Rightarrow T=(\frac{1}{n R}) p V $

$\Rightarrow T=(\frac{1}{n R}) V f(V) $

$\Rightarrow d T=(\frac{1}{n R})[f(V)+V f^{\prime}(V)] d V$

Now from Eq. (i)

$$ \frac{d Q}{d V}=n C_{V} \frac{d T}{d V}+p \frac{d V}{d V}=n C_{V} \frac{d T}{d V}+p $$

$$ =\frac{n C_{V}}{n R} \times[f(V)+V f^{\prime}(V)]+p $$

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

$ \rightarrow [\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} $

$\because \quad C_{V} =\frac{R}{\gamma-1} \Rightarrow \frac{C_{V}}{R}=\frac{1}{\gamma-1}$

$\Rightarrow \quad[\frac{d Q}{d V}]_{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 $\frac{d Q}{d V}>0$, 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

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

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

Hence,

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

$$ \begin{aligned} \text { final pressure } & =p_{f}=p_{a}+k x \\ & =p_{a}+k \times(V_1-V_0) \end{aligned} $$

(c) From first law of thermodynamics $d Q=d U+d W$

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

$$ d U=C_{V}(T,-T_0)=C_{V}(T,-T_0) $$

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 $=p d V+$ increase in PE of the spring

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

Now, we can write $d Q=d U+d W$

$$ \begin{aligned} & =C_{V}(T-T_0)+p_{a}(V-V_0)+\frac{1}{2} k x^{2} \\ & =C_{V}(T-T_0)+p_{a}(V-V_0)+\frac{1}{2}(V_1-V_0)^{2} \end{aligned} $$

This is the required relation.