Past NEET Papers

Hint

Given, mass of water, \(m=1 \mathrm{~g}\)
Volume of \(1 \mathrm{~g}\) of water \(=1 \mathrm{~cm}^3=10^{-8} \mathrm{~m}^3\)
Volume of \(1 \mathrm{~g}\) of steam \(=1671 \mathrm{~cm}^3\)
\(
=1671 \times 10^{-8} \mathrm{~m}^3
\)
Pressure, \(p=1 \times 10^5 \mathrm{~Pa}\)
Latent heat of vaporization of water,
\(
L=2256 \mathrm{~J} / \mathrm{g}
\)
Change in volume,
\(
\begin{aligned}
\Delta V & =(1671-1) \times 10^{-8} \mathrm{~m}^3 \\
& =1670 \times 10^{-8} \mathrm{~m}^3 \dots(i)
\end{aligned}
\)
Heat supplied,
\(
\Delta Q=m L=1 \times 2256=2256 \mathrm{~J} \dots(ii)
\)
As the steam expands, so the work done in expansion is
\(
\begin{array}{rlr}
\Delta W & =p \Delta V \\
& =1 \times 10^5 \times 1670 \times 10^{-8} & \text { [from Eq. (i)] } \\
& =167 \mathrm{~J} & \text {…(iii) }
\end{array}
\)
According to the first law of thermodynamics,
\(
\begin{aligned}
& \Delta Q=\Delta U+\Delta W \\
\Rightarrow \quad & \Delta U=\Delta Q-\Delta W \\
= & 2256-167 \quad \text { [from Eq. (ii) and (iii)] } \\
= & 2089 \mathrm{~J}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \Delta \mathrm{Q}=54 \text { cal }=54 \times 4.18 \text { joule }=225.72 \text { joule } \\
& \left.\Delta \mathrm{W}=\mathrm{P} [\mathrm{V}_{\text {steam }}-\mathrm{V}_{\text {water }}\right][\text { For water } 0.1 \text { gram }=0.1 \mathrm{cc}] \\
& =1.013 \times 10^5\left[167.1 \times 10^{-6}-0.1 \times 10^{-6}\right] \text { joule } \\
& =1.013 \times 167 \times 10^{-1}=16.917 \text { joule } \\
& \text { By FLirst Law Of Thermodynamics } \\
& \Rightarrow \Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}=225.72-16.917 \\
& \quad \Delta \mathrm{U}=208.8 \text { joule }
\end{aligned}
\)

Hint

In an adiabatic process, the system is completely insulated from the surroundings. Thus, heat is neither absorbed nor released by the system to the surroundings. So, \(\Delta Q=0\).
Sudden processes are adiabatic like bursting of cycle tyre, etc.
If the pressure of gas is kept constant, then the process is called isobaric, i.e. \(\Delta p=0\).
If the temperature of the system remains constant, then it is called an isothermal process, i.e. \(\Delta T=0\).
If the volume of gas is constant in a system, then it is called an isochoric process, i.e. \(\Delta V=0\).

Hint

From the \(p V\) diagram, it is clear that the pressure of ideal gas in the piston cylinder is constant during the thermodynamic process. Hence, this process is isobaric.

Hint

Since the entire system is thermally insulated. Thus, there will be no transfer of heat between the system and the surrounding. Hence, when the stop cock is suddenly opened, a sudden process of expansion will take place with no heat transfer. Thus, the given process will be adiabatic, as it is a process that occurs without transferring heat or mass between a thermodynamic system and its surroundings.
Hence, the correct option is (a).

Hint

Efficiency of Carnot engine is given as
\(
\eta=1-\frac{T_2}{T_1}
\)
where \(T_2=\) temperature of sink
and \(T_1=\) temperature of source.
Hence, \(\eta\) depends upon the temperature of source and sink both.

Hint

According to the given graph, Volume \((V) \propto\) Temperature \((T)\) or \(\frac{V}{T}=\) constant Thus, the process is isobaric.

\(\therefore\) Work done, \(\Delta W=p \Delta V\)
\(
=n R \Delta T=n R\left(T_B-T_A\right) \ldots(i)
\)
Heat absorbed \(\Delta O=n C_p \Delta T\)
\(
=n C_p\left(T_B-T_A\right) \dots(ii)
\)
As, \(\quad C_\rho=\frac{\gamma R}{\gamma-1}, \quad\left(\right.\) where, \(\left.\gamma=1+\frac{2}{f}\right)\)
For a monoatomic gas, \(f=3\)
\(
\Rightarrow \quad C_p=\left(R+\frac{3}{2} R\right)=\frac{5}{2} R
\)
Substituting the value of \(C_p\) in Eq. (ii), we get
\(
\Delta Q=n\left(\frac{5}{2} R\right)\left(T_B-T_A\right)
\)
Hence, \(\frac{\Delta W}{\Delta 0}=\frac{n R\left(T_B-T_A\right)}{n\left(\frac{5}{2} R\right)\left(T_B-T_A\right)}=\frac{2}{5}\)

Alternate:

Gas is monatomic, so \(\mathrm{C}_{\mathrm{p}}=\frac{5}{2} R\)
\(
\begin{aligned}
& \begin{aligned}
& \text { Given process } \\
& \therefore \quad \mathrm{dQ}=\mathrm{n} \mathrm{C}_{\mathrm{p}} \mathrm{dT}
\end{aligned} \\
& \Rightarrow \mathrm{dQ}=n\left(\frac{5}{2} R\right) d T \\
& \mathrm{dW}=\mathrm{P} \mathrm{dV}=\mathrm{n} \mathrm{RdT} \\
& \therefore \quad \text { Required ratio }=\frac{d W}{d Q}=\frac{n R d T}{n\left(\frac{5}{2} R\right) d T}=\frac{2}{5} \\
&
\end{aligned}
\)

Hint

Efficiency ofideal heat engine, \(\eta=\left(1-\frac{T_2}{T_1}\right)\)
Sink temperature, \(\mathrm{T}_2=100^{\circ} \mathrm{C}=100+273=373 \mathrm{~K}\)
Source temperature, \(T_1=0^{\circ} \mathrm{C}=0+273=273 \mathrm{~K}\)
Percentage efficiency, \(\% \eta=\left(1-\frac{T_2}{T_1}\right) \times 100\) \(\quad=\left(1-\frac{273}{373}\right) \times 100=\left(\frac{100}{373}\right) \times 100=26.8 \%\)

Hint

In the isochoric process, the curve is parallel to \(y\)-axis because the volume is constant. Isobaric is parallel to \(x\)-axis because the pressure is constant. Along the curve, it will be isothermal because the temperature is constant.
So, \(\mathrm{P} \rightarrow \mathrm{c} \Rightarrow \mathrm{Q} \rightarrow \mathrm{a} \Rightarrow \mathrm{R} \rightarrow \mathrm{d} \Rightarrow \mathrm{S} \rightarrow \mathrm{b}\)

Hint

Given, efficiency of engine, \(\eta=\frac{1}{10}\) work done on system \(\mathrm{W}=10 \mathrm{~J}\) Coefficient of performance of refrigerator \(\beta=\frac{\mathrm{Q}_2}{\mathrm{~W}}=\frac{1-\eta}{\eta}=\frac{1-\frac{1}{10}}{\frac{1}{10}}=\frac{\frac{9}{10}}{\frac{1}{10}}=9\)

Energy absorbed from reservoir \(\mathrm{Q}_2=\beta \mathrm{W}, \quad \mathrm{Q}_2=9 \times 10=90 \mathrm{~J}\)
A refrigerator works along the reverse direction of a heat engine.

Hint

The solution to this question can be understood by plotting a p-V graph for the compression of a gas isothermally and adiabatically simultaneously to half of its initial volume. i.e.

Since the isothermal curve is less steeper than the adiabatic curve. So, area under the p-V curve for the adiabatic process has more magnitude than an isothermal curve. Hence, work done in the adiabatic process will be more than in the isothermal process.

Hint

Given: \(\mathrm{T}_2=4^{\circ} \mathrm{C}=4+273=277 \mathrm{~K}\)
\(
\mathrm{T}_1=30^{\circ} \mathrm{C}=30+273=303 \mathrm{~K}
\)
\(
\begin{aligned}
& \beta=\frac{Q_2}{W}=\frac{T_2}{T_1-T_2} \\
& \therefore \quad \beta=\frac{600 \times 4.2}{W}=\frac{277}{303-277} \\
& \Rightarrow \quad W=236.5 \text { joule }
\end{aligned}
\)
\(
\text { Power } \mathrm{P}=\frac{\mathrm{W}}{\mathrm{t}}=\frac{236.5 \text { joule }}{1 \mathrm{sec}}=236.5 \mathrm{watt} \text {. }
\)

Hint

For a refrigerator, we know that
\(
\frac{Q_1}{W}=\frac{Q_1}{Q_1-Q_2}=\frac{T_1}{T_1-T_2}
\)
where,
\(Q_1=\) amount of heat delivered to the
room
\(W=\) electrical energy consumed
\(T_1=\) room temperature \(=t_1+273\)
\(T_2=\) temperature of sink \(=t_2+273\)
\(
\begin{aligned}
& \therefore \quad \frac{Q_1}{1}=\frac{t_1+273}{t_1+273-\left(t_2+273\right)} \\
& \Rightarrow \quad Q_1=\frac{t_1+273}{t_1-t_2}
\end{aligned}
\)

Hint

In cyclic process \(\mathrm{ABCA}\)
\(
\begin{aligned}
& \mathrm{Q}_{\text {cycle }}=\mathrm{W}_{\text {cycle }} \\
& \mathrm{Q}_{\mathrm{AB}}+\mathrm{Q}_{\mathrm{BC}}+\mathrm{Q}_{\mathrm{CA}}=\text { Area of } \Delta \mathrm{ABC} \\
& 400+100+\mathrm{Q}_{\mathrm{C} \rightarrow \mathrm{A}}=\frac{1}{2}\left(2 \times 10^{-3}\right)\left(4 \times 10^4\right) \\
& \Rightarrow \mathrm{Q}_{\mathrm{C} \rightarrow \mathrm{A}}=-460 \mathrm{~J} \quad \Rightarrow \mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{C}}=+460 \mathrm{~J}
\end{aligned}
\)

Hint

\(
\text { For a diatomic gas, }
\)
\(
C_v=\frac{5}{2} R
\)
The change in internal energy of gas in the transition from \(A\) to \(B\) is
\(
\begin{aligned}
\Delta U & =n C_V d T=n\left(\frac{5 R}{2}\right)\left(T_B-T_A\right) \\
& =n R \frac{5}{2}\left(\frac{p_B V_B}{n R}-\frac{p_A V_A}{n R}\right) \\
& =\frac{5}{2}\left(2 \times 10^3 \times 6-5 \times 10^3 \times 4\right) \\
& =\frac{5}{2} \times\left(-8 \times 10^3\right)=-\frac{4 \times 10^4}{2}=-20 \mathrm{~kJ}
\end{aligned}
\)

Hint

Given, ideal gas is compressed to half its initial volume i.e.
\(
V_0=\frac{V}{2}
\)

The isochoric process is one in which volume is kept constant, meaning that work done by the system will be zero. i.e.
\(
W_{\text {isocharic }}=0
\)
As we know, work done on the gas = Area under curve, i.e.
\(
W_{\text {adiabatic }}>W_{\text {isothermal }}>W_{\text {isobaric }}
\)

Alternate:

Since area under the curve is maximum for adiabatic process so, work done \((\mathrm{W}=\mathrm{PdV})\) on the gas will be maximum for adiabatic process.

Hint

Key Concept Coefficient of performance \((\beta)\) of a refrigerator is defined as the ratio of quantity of heat removed per cycle \(\left(O_2\right)\) to the work done on the working substance per cycle to remove this heat.
Given, coefficient of performance of a refrigerator, \(\beta=5\)
Temperature of surface, i.e. inside freezer,
\(
T_2=-20^{\circ} \mathrm{C}=-20+273=253 \mathrm{~K}
\)
Temperature of surrounding, i.e. heat rejected outside \(T_1=\) ?
So, \(\quad \beta=\frac{T_2}{T_1-T_2} \Rightarrow 5=\frac{253}{T_1-253}\)
\(
\Rightarrow 5 T_1-1265=253
\)
\(
\begin{aligned}
5 T_1 & =1518 \\
T_1 & =\frac{1518}{5}=303.6 \mathrm{~K} \\
T_1 & =303.6-273=30.6 \cong 31^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

For isothermal process \(\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2\)
\(
\Rightarrow \quad \mathrm{PV}=\mathrm{P}_2(2 \mathrm{~V}) \Rightarrow \mathrm{P}_2=\frac{\mathrm{P}}{2}
\)
For adiabatic process
\(
\mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_3 \mathrm{~V}_3^\gamma
\)
\(
\left(\frac{\mathrm{P}}{2}\right)(2 \mathrm{v})^\gamma=\mathbf{P}_3(16 \mathrm{v})^\gamma
\)
\(
\mathrm{P}_3=\frac{3}{2}\left(\frac{1}{8}\right)^{5 / 3}=\frac{\mathrm{P}}{64}
\)

Hint

Work done in the cyclic process \(=\) Area bounded by the closed configuration
\(=\) Area of closed configuration
\(
=\frac{1}{2} \times V_0 \times p_0-\frac{1}{2} \times V_0 \times p_0=0 \text { (zero). }
\)

Hint

According to the question,
Net work done \(=\) Area enclosed in \(p V\) curve i.e. \(\triangle A B C\)
Area of \(\triangle A B C\)
\(
\begin{aligned}
& =\frac{1}{2} \times 5 \times 10^{-3} \times 4 \times 10^5 \mathrm{~J} \\
& =10^3 \mathrm{~J}=1000 \mathrm{~J}
\end{aligned}
\)

Hint

From first law of thermodynamics
\(
\begin{aligned}
& Q_{\mathrm{adc}}=\Delta U_{a d c}+W_{a d c} \\
& 50 \mathrm{~J}=\Delta U_{a d c}+20 \mathrm{~J} \\
& \Delta U_{a d c}=30 \mathrm{~J}
\end{aligned}
\)
\(
\text { Again, } \quad \begin{aligned}
Q_{a b c} & =\Delta U_{a b c}+W_{a b c} \\
W_{a b c} & =Q_{a b c}-\Delta U_{a b c} \\
& =Q_{a b c}-\Delta U_{a d c} \\
& =36 \mathrm{~J}-30 \mathrm{~J} \\
& =6 \mathrm{~J}
\end{aligned}
\)

Hint

According to question \(\mathrm{P} \propto \mathrm{T}^3\) But as we know for an adiabatic process the pressure \(\mathrm{P} \propto \mathrm{T}^{\frac{\gamma}{\gamma-1}}\).
So, \(\frac{\gamma}{\gamma-1}=3 \Rightarrow \gamma=\frac{3}{2}\) or, \(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{3}{2}\)

Hint

Adiabatic equations of state are
\(P V^\gamma=\mathrm{constant} \quad T V^{\gamma-1}=\) constant,  \(P^{1-\gamma} T^\gamma=\) constant
When a thermodynamic system undergoes a change in such a way that to exchange of heat takes place between the system and surroundings, then the process is an adiabatic process. In this process \(P, V\) and \(T\) change but \(\Delta Q=0\).

Hint

Efficiency of engine \(A, \eta_1=1-\frac{T}{T_1}\), Efficiency of engine \(B, \eta_2=1-\frac{T_2}{T}\) Here, \(\eta_1=\eta_2\)
\(
\therefore \frac{T}{T_1}=\frac{T_2}{T} \Rightarrow T=\sqrt{T_1 T_2}
\)

Hint

Internal energy is the state function.
In cyclic process; \(\Delta U=0\)
According to 1st law of thermodynamics \(\Delta Q=\Delta U+W\)
So heat absorbed
\(\Delta Q=W=\) Area under the curve \(=-(2 \mathrm{~V})(\mathrm{P})=-2 \mathrm{PV}\)
So heat rejected \(=2 \mathrm{PV}\)

Hint

(d) 1 st process is an isothermal expansion which is only correctly shown in option (d)
2nd process is isobaric compression which is correctly shown in option (d)

Note:

According to the question, firstly gas expands from volume \(V\) to \(3 \mathrm{~V}\) and after this volume is reduced from \(3 \mathrm{~V}\) to \(V\) at constant pressure. In isothermal expansion, \(p-V\) curve is a rectangular hyperbola.

Hint

Concept Apply first law of thermodynamics.
According to first law of thermodynamics
\(
\Delta U=\Delta Q+\Delta W
\)
For isothermal process, \(\Delta U=0\)
So, \(\Delta Q=-\Delta W\)
Given, \(\quad \Delta W=-150 \mathrm{~J}\)
So, \(\Delta Q=+150 \mathrm{~J}\)
When \(Q\) is positive, the heat is added to the gas.

Hint

\(
\begin{aligned}
& \text { (c) } \mathrm{T}_1=273+27=300 \mathrm{~K} \\
& \mathrm{~T}_2=273+927=1200 \mathrm{~K}
\end{aligned}
\)
For adiabatic process,
\(
\begin{aligned}
& \mathrm{P}^{1-\gamma} \mathrm{T}^\gamma=\text { constant } \\
& \Rightarrow \mathrm{P}_1^{1-\gamma} \mathrm{T}_1^\gamma=\mathrm{P}_2{ }^{1-\gamma} \mathrm{T}_2^\gamma \\
& \Rightarrow\left(\frac{P_2}{P_1}\right)^{1-\gamma}=\left(\frac{T_1}{T_2}\right)^\gamma \Rightarrow\left(\frac{P_1}{T_2}\right)^{1-\gamma}=\left(\frac{T_2}{T_1}\right)^\gamma \\
& \Rightarrow\left(\frac{P_1}{P_2}\right)^{1-1.4}=\left(\frac{1200}{300}\right)^{1.4} \Rightarrow\left(\frac{P_1}{P_2}\right)^{-0.4}=(4)^{1.4}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow\left(\frac{\mathrm{P}_2}{\mathrm{P}_1}\right)^{0.4}=4^{1.4} \\
& \Rightarrow \mathrm{P}_2=\mathrm{P}_1 4^{\left(\frac{1.4}{0.4}\right)}=\mathrm{P}_1 4^{\left(\frac{7}{2}\right)} \\
& \Rightarrow \mathrm{P}_1\left(2^7\right)=2 \times 128=256 \mathrm{~atm}
\end{aligned}
\)

Hint

Change in entropy is given by
\(
\begin{aligned}
& \Delta \mathrm{S}=\frac{\mathrm{dQ}}{\mathrm{T}} \text { or } \Delta \mathrm{S}=\frac{\Delta \mathrm{Q}}{\mathrm{T}}=\frac{\mathrm{mL}_{\mathrm{f}}}{273} \\
& \Delta \mathrm{S}=\frac{1000 \times 80}{273}=293 \mathrm{cal} / \mathrm{K}
\end{aligned}
\)

Hint

From first law of thermodynamics
\(
\Delta Q=\Delta U+\Delta W
\)
For adiabatic process, \(\Delta Q=0\)
\(\because \quad \Delta Q=0\)
So, \(\quad \Delta U=-\Delta W\)

Hint

According to first law of thermodynamics
\(
\begin{aligned}
& \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} \\
& \Delta \mathrm{U}=\mathrm{Q}-\mathrm{W} \\
& =2 \times 4.2 \times 1000-500=8400-500 \\
& =7900 \mathrm{~J}
\end{aligned}
\)

Hint

In an isochoric process, volume remains constant whereas pressure remains constant in the isobaric process.

Explanation:

For an adiabatic process, there should not be any exchange of heat between the system and its surroundings. All walls of the container must be perfectly insulated. In adiabatic changes, gases obey Poisson’s law, i.e., \(p V^\gamma=\) constant. In an isochoric process, volume remains constant and in an isobaric process, pressure remains constant.

Hint

For a cyclic process, \(\Delta U=0\) or \(E=0\)

In a cyclic process, the initial state coincides with the final state. Hence, the change in internal energy is zero, as it depends only on the initial and final states.
In case of cyclic process, \(\mathrm{u}_f=\mathrm{u}_i \therefore \Delta \mathrm{u}=\mathrm{u}_f-\mathrm{u}_i=0\) And also \(\Delta \mathrm{u} \propto \Delta T \therefore \Delta T=0\)

Hint

Since efficiency of engine is \(\eta=1-\frac{T_2}{T_1}\)
According to problem, \(\frac{1}{6}=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1} \dots(1)\)
When the temperature of the sink is reduced by \(62^{\circ} \mathrm{C}\), its efficiency is doubled
\(
2\left(\frac{1}{6}\right)=1-\frac{\mathrm{T}_2-62}{\mathrm{~T}_1} \dots(2)
\)
Solving (1) and (2) \(\quad \mathrm{T}_2=372 \mathrm{~K}\) \(\mathrm{T}_1=99^{\circ} \mathrm{C}=\) Temperature of the source.

Hint

We know that efficiency of Carnot Engine
\(
\eta=\frac{T_1-T_2}{T_1}
\)
where, \(T_1\) is temp. of source and \(T_2\) is temp. of sink
\(
\therefore \quad 0.40=\frac{T_1-300}{T_1} \Rightarrow T_1-300=0.40 T_1
\)
\(
0.6 T_1=300 \Rightarrow T_1=\frac{300}{.6}=\frac{3000}{6}=500 K
\)
Now efficiency to be increased by \(50 \%\)
\(
\begin{aligned}
& \therefore \quad 0.60=\frac{T_1-300}{T_1} \Rightarrow T_1-300=0.6 T_1 \\
& 0.4 T_1=300 \Rightarrow T_1=\frac{300}{4}=\frac{300 \times 10}{4}=750 \\
& \text { Increase in temp }=750-500=250 \mathrm{~K}
\end{aligned}
\)

Hint

The isothermal process occurs very slowly so it is quasi-static and hence it is reversible.
The conditions for reversibility

• Must take place infinitely slowly
• The temperature of the system must not differ appreciably
• There must be complete absence of dissipative forces such as friction, viscosity, electric resistance, etc.

Hint

We know that efficiency of Carnot engine
\(
\begin{aligned}
& =1-\frac{T_2}{T_1}=1-\frac{400}{500}=\frac{1}{5} \\
& {\left[\because T_1=(273+227) \mathrm{K}=500 \mathrm{~K}\right.} \\
& \text { and } \left.T_2=(273+127) \mathrm{K}=400 \mathrm{~K}\right]
\end{aligned}
\)
Efficiency of Heat engine \(=\frac{\text { Work output }}{\text { Heat input }}\)
or, \(\frac{1}{5}=\frac{\text { work output }}{6 \times 10^4}\)
\(\Rightarrow\) work output \(=1.2 \times 10^4 \mathrm{cal}\)

Hint

\(
T_1=T, W=6 R \text { joules, } \gamma=\frac{5}{3}
\)
\(
\begin{aligned}
W & =\frac{P_1 V_1-P_2 V_2}{\gamma-1}=\frac{n R T_1-n R T_2}{\gamma-1} \\
& =\frac{n R\left(T_1-T_2\right)}{\gamma-1}
\end{aligned}
\)
\(
\begin{aligned}
& n=1, T_1=T \Rightarrow \frac{R\left(T-T_2\right)}{5 / 3-1}=6 R \\
& \Rightarrow T_2=(T-4) \mathrm{K}
\end{aligned}
\)

Hint

The efficiency of heat engine is \(\eta=1-\frac{T_2}{T_1}\) or \(\frac{W}{0_1}=1-\frac{T_2}{T_1}\)
\(T_2=\) Temperature of sink
\(T_1=\) Temperature of source
\(W=\) Work done
Given, \(Q_1=6 \mathrm{kcal}\) = heat absorbed from the source
\(T_1=227+273=500 \mathrm{~K}\)
and \(T_2=127+273=400 \mathrm{~K}\)
Hence, \(\frac{W}{6}=1-\frac{400}{500}\)
or \(\frac{W}{6}=\frac{100}{500}\)
or \(\quad W=1.2 \mathrm{kcal}\)
Thus, the amount of heat converted into work is \(1.2 \mathrm{kcal}\).

Hint

\(\eta=1-\frac{T_2}{T_1}\) or \(\frac{50}{100}=1-\frac{500}{T_1}\)
\(\Rightarrow T_1=1000 \mathrm{~K}\)
Also, \(\quad \frac{60}{100}=1-\frac{T_2}{1000} \Rightarrow T_2=400 \mathrm{~K}\)

Hint

Efficiency of heat engine is,
\(\eta=1-\frac{T_2}{T_1}\) or \(\eta=\frac{T_1-T_2}{T_1}\)
\(T_2=\) temperature of sink
\(T_1=\) temperature of source
Given, \(T_1=273+127=400 \mathrm{~K}\)
\(T_2=273+27=300 \mathrm{~K}\)
\(\therefore \quad \eta=\frac{400-300}{400}=\frac{100}{400}=0.25=25 \%\)
Hence, \(26 \%\) efficiency is impossible for a given heat engine.

Hint

Initially the efficiency of the engine was \(\frac{1}{6}\) which increases to \(\frac{1}{3}\) when the sink temperature reduces by \(62^{\circ} \mathrm{C}\). \(\eta=\frac{1}{6}=1-\frac{T_2}{T_1}\), when \(T_2=\operatorname{sink}\) temperature \(T_1=\) source temperature \(\Rightarrow T_2=\frac{5}{6} T_1\)
Secondly,
\(
\begin{aligned}
& \frac{1}{3}=1-\frac{T_2-62}{T_1}=1-\frac{T_2}{T_1}+\frac{62}{T_1}=1-\frac{5}{6}+\frac{62}{T_1} \\
& \text { or, } T_1=62 \times 6=372 \mathrm{~K}=372-273=99^{\circ} \mathrm{C}
\end{aligned}
\)
\(\& T_2=\frac{5}{6} \times 372=310 \mathrm{~K}=310-273=37^{\circ} \mathrm{C}\)

Hint

\(\mathrm{T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}\)
\(
\gamma=\frac{5}{3} ; \quad V_2=\frac{8}{27} V_1 ; \quad \frac{V_1}{V_2}=\frac{27}{8}
\)
From adiabatic process we know that
\(
\begin{aligned}
& T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1} \\
& \frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^{\gamma-1}=\left(\frac{27}{8}\right)^{\frac{5}{3}-1} \\
& \frac{T_2}{T_1}=\frac{9}{4} \Rightarrow T_2=\frac{9}{4} \times T_1=\frac{9}{4} \times 300=675 \mathrm{~K} \\
& T_2=675-273^{\circ} \mathrm{C}=402^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

Change in internal energy is equal to work done in adiabatic system \(\Delta W=-\Delta U\) (Expansion in the system)
\(
\begin{aligned}
& =-\frac{1}{\gamma-1}\left(P_1 V_1-P_2 V_2\right) \\
& \Delta U=\frac{1}{1-\gamma}\left(P_2 V_2-P_1 V_1\right)
\end{aligned}
\)
Here, \(V_1=V, V_2=2 V\)
\(
\begin{aligned}
& \therefore \Delta U=\frac{1}{1-\gamma}[P \times 2 V-P V]=\frac{P V}{1-\gamma} \\
& \Rightarrow \Delta U=-\frac{P V}{\gamma-1}
\end{aligned}
\)

Hint

In thermodynamics for the same change in volume, the work done is maximum in the isobaric process because, in \(P-V\) graph, the area enclosed by the curve and volume axis is maximum in the isobaric process.

Explanation:

In thermodynamics, for some change in volume, the work done is maximum for the curve having maximum area enclosed with the volume axis.
Area enclosed by the curve \(\propto\) (Slope of curve)

\(
\text { (slope })_{\text {sobaric }}<(\text { slope })_{\text {sothermal }}<(\text { slope })_{\text {adiabatic }}
\)

\(
\begin{aligned}
& \Rightarrow(\text { Area })_{\text {isobaric }}>(\text { Area })_{\text {isothermal }} \\
& >(\text { Area })_{\text {adiabatic }} \\
&
\end{aligned}
\)

\(
\begin{aligned}
& (\text { Slope })_{\text {aclateresc }}=-\gamma\left(\frac{p}{V}\right) \\
& \text { and }(\text { Slope })_{\text {sactsomal }}=-\frac{p}{V} \\
& \left.\therefore \quad(\text { Slope })_{\text {adidarc }}=\gamma \times \text { (slope }\right)_{\text {karacrmal }} \\
&
\end{aligned}
\)
Slope of adiabatic curve is always steeper than that of isothermal curve.

Hint

Efficiency of Carnot engine is given by
\(
\eta=1-\frac{T_2}{T_1}=\frac{T_1-T_2}{T_1} \dots(i)
\)
Given, \(T_1=\) temperature of reservoir
\(
=100+273=373 \mathrm{~K}
\)
\(T_2=\) temperature of sink
\(
=-23+273=250 \mathrm{~K}
\)
Substituting in Eq. (i), we get
\(
\begin{aligned}
\therefore \quad \eta & =\frac{373-250}{373}=\frac{123}{373} \\
& =\frac{100+23}{373}
\end{aligned}
\)

Hint

We know that in the adiabatic process, \(P V^{\gamma}= \dots(1)\) constant
From the ideal gas equation, we know that \(P V=n R T\)
\(V=\frac{n R T}{P} \dots(2)\)
Puttingt the value from equation (2) in equation (1),
\(
\begin{gathered}
P\left(\frac{n R T}{P}\right)^\gamma=\text { constant } \\
P^{(1-\gamma)} T^{\gamma}=\text { constant }
\end{gathered}
\)

Hint

Initial temperature \(\left(T_1\right)=18^{\circ} \mathrm{C}=291 \mathrm{~K}\)
Let Initial volume \(\left(V_1\right)=V\)
Final volume \(\left(V_2\right)=\frac{V}{8}\)
According to adiabatic process, \(T V^{-1}=\) constant
According to question, \(T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}\)
\(
\begin{aligned}
& \Rightarrow T_2=293\left(\frac{V_1}{V_2}\right)^{\gamma-1} \\
& \Rightarrow T_2=293(8)^{\frac{7}{5}-1}=293 \times 2.297=668.4 \mathrm{~K} \\
& {\left[\text { For diatomic gas } \gamma=\frac{C_p}{C_v}=\frac{7}{5}\right]} \\
&
\end{aligned}
\)

Hint

Efficiency of carnot engine \(\left(\eta_1\right)=40 \%\) \(=0.4\); Initial intake temperature \(\left(T_1\right)=500 \mathrm{~K}\) and new efficiency \(\left(\eta_2\right)=50 \%=0.5\).
Efficiency \((\eta)=1-\frac{T_2}{T_1}\) or \(\frac{T_2}{T_1}=1-\eta\).
Therefore in first case, \(\frac{T_2}{500}=1-0.4=0.6\). \(\Rightarrow T_2=0.6 \times 500=300 \mathrm{~K}\)
And in second case, \(\frac{300}{T_1}=1-0.5=0.5\)
\(
\Rightarrow T_1=\frac{300}{0.5}=600 \mathrm{~K}
\)

Hint

Concept Apply the first law of thermodynamics to calculate the required work done.

From the first law of thermodynamics
\(
\Delta 0=\Delta U+\Delta W
\)
where, \(\Delta \emptyset=\) heat given
\(\Delta U=\) change in internal energy
\(\Delta W=\) work done
Here, \(\Delta Q=110 \mathrm{~J}\)
\(
\begin{aligned}
\Delta U & =40 \mathrm{~J} \\
\therefore \quad \Delta W & =\Delta Q-\Delta U=110-40=70 \mathrm{~J}
\end{aligned}
\)

Hint

The thermodynamic state of a homogeneous system may be represented by certain specific thermodynamic variables such as pressure \(p\), volume \(V\), temperature \(T\) and entropy S. Out of these four variables, any two are independent and when they are known the others may be determined. Thus, there are only two independent variables and the others may be considered their functions. For complete knowledge of the system certain relations are required and for this purpose we introduce some functions of variables \(p_1, T\) and \(S\) known as thermodynamic functions. There are four principal thermodynamic functions
(i) Internal energy (U)
(ii) Helmholtz function (F)
(iii) Enthalpy (H)
(iv) Gibb’s energy (G)
Hence, work done is not thermodynamic function.

Hint

A change in the pressure and volume of a gas without any change in its temperature is called an isothermal change. In such a change, there is a free exchange of heat between the gas and its surroundings.
\(\therefore \quad T=\) constant, \(\Delta T=0\)
So, internalenergy \((U)\) remains constant 0.

Note: Under isothermal conditions, there is no change in internal energy.

Hint

\(
\text { Work done }=\text { Area under curve } A C B D A
\)

Hint

\(B C\) is isobaric process,
\(
\begin{aligned}
& \therefore W=P_B \times\left(V_D-V_A\right)=240 \mathrm{~J} \\
& \therefore Q=600+200=800 \mathrm{~J}
\end{aligned}
\)
Using \(\Delta Q=\Delta U+\Delta W\)
\(
\Rightarrow \Delta U=\Delta Q-\Delta W=800-240=560 \mathrm{~J}
\)
Since \(A B\) is an isochoric process i.e., \(\Delta V=0\) so, no work is done.

Alternate:

For path \(A B\), applying first law of thermodynamics
\(
d Q=d U+d W \text { or } d Q=U_B-U_A+d W
\)
or \(\quad 600=U_B-U_A+0\)
(for isochoric process \(d V=0\) so, \(d W=0\) )
\(
U_B-U_A=600 \dots(i)
\)
\(
\begin{aligned}
& \text { For path } B C \text {, } \\
& d Q=d U+d W \\
& \therefore \quad 200=U_C-U_B+p_B\left(V_C-V_B\right) \\
& \text { (as } B C \text { is isobaric process) } \\
& \therefore 200=U_c-U_B+8 \times 10^4 \\
& \left(5 \times 10^{-3}-2 \times 10^{-3}\right) \\
& \text { or } 200=U_C-U_B+240 \\
& \therefore \quad U_c-U_B=-40 \mathrm{~J} \dots(ii) \\
&
\end{aligned}
\)
For \(A C\), change in internal energy can be calculated by adding Eqs. (i) and (ii),
\(
\begin{aligned}
& =U_C-U_A \\
& =U_C-U_B+U_B-U_A \\
& =-40+600=560 \mathrm{~J}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& T_1^\gamma P_1^{1 -\gamma}=T_2^\gamma P_2^{1 -\gamma} \\
& \Rightarrow\left(\frac{T_2}{T_1}\right)^\gamma=\left(\frac{P_1}{P_2}\right)^{1-\gamma} \\
& \Rightarrow T_2=T_1 \cdot\left(\frac{P_1}{P_2}\right)^{\frac{1-\gamma}{\gamma}}=300 \times(8)^{-2 / 5}=142^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

According to the first law of thermodynamics, when some quantity of heat \((d Q)\) is supplied to a system capable of doing external work, then the quantity of heat absorbed by the system \((\mathrm{d Q}\) is equal to the sum of the increase in the internal energy of the system (dU) due to rise in temperature and the external work done by the system \((\mathrm{dW})\) in expansion,
i.e. \(d Q=d U+d W\)
This law, which is basically the law of conservation of energy applies to every process in nature.

Hint

\(
\begin{aligned}
& \left(\frac{d P}{d V}\right)_{\text {adiabatic }}=-\gamma P \\
& \left(\frac{d P}{d V}\right)_{\text {isothermal }}=-P \\
& \left(\frac{d P}{d V}\right)_{\text {adiabatic }}>\left(\frac{d P}{d V}\right)_{\text {isothermal }}
\end{aligned}
\)

Hint

From ideal gas equation
\(
P V=n R T \quad\left[n=\frac{\text { mass of water }}{\text { mol. wt. }}=\frac{4.5 \times 10^3}{18}\right]
\)
\(
V=\frac{n R T}{P}
\)
At. STP \(\Rightarrow T=273 \mathrm{~K}\)
\(
\begin{aligned}
& P=10^5 \mathrm{~N} / \mathrm{m}^2 \\
& V=\frac{4.5 \times 10^3}{18} \times \frac{8.3 \times 273}{10^5}=5.66 \mathrm{~m}^3
\end{aligned}
\)

Hint

Efficiency \(\eta=\frac{50}{100}=\frac{1}{2}\)
Efficiency of Carnot engine
\(
\begin{aligned}
& \eta=1-\frac{T_2}{T_1} \\
& \eta=1-\frac{T_2}{600} \\
& \frac{1}{2}=1-\frac{T_2}{600} \\
& \frac{T_2}{600}=\frac{1}{2} \Rightarrow T_2=300 \mathrm{~K} \\
& T_2=300-273=27^{\circ} \mathrm{C}
\end{aligned}
\)