Exemplar MCQs

Hint

The answer is option (c) 4 is the isobaric process, 1 is isochoric. Out of 3 and 2,3 has a smaller
slope and hence is isothermal. Remaining process 2 is adiabatic.

Hint

(a) Rate of bum calories is equivalent to sweat produced. Then, Amount of sweat evaporated/ minute
\(
\begin{aligned}
& =\frac{\text { Sweat produced } / \text { minute }}{\text { Number of calories required for evaporation } / \mathrm{kg}} \\
& =\frac{\text { Calories produced (heat produced) per minute }}{\text { Latent heat (in cal } / \mathrm{kg} \text { ) }} \\
& =\frac{14.5 \times 10^3}{580 \times 10^3}=\frac{145}{580}=0.25 \mathrm{~kg}
\end{aligned}
\)

Hint

i.e., We know \(P V=\) 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., \(P_2<P_1\). So, we have to keep in mind while drawing the T-P graph, that temperature \((T)\) is constant and pressure at point 2 is greater than the pressure at 1, which corresponds to diagram (iii).

Hint

The answer is the option (b). Since the direction of arrows is anticlockwise, so work done is negative equal to the area of the loop \(=-2 P_0 V_0\)

Hint

(a) According to the P-V diagram shown for the container A (which is going through the isothermal process) and for container B (which is going through the adiabatic process).

Both the process involves compression of the gas.
(i) Isothermal compression (gas A) (during \(1 \rightarrow 2\) )
\(
\begin{array}{lrl}
& P_1 V_1=P_2 V_2 \\
\Rightarrow & P_0\left(2 V_0\right)^\gamma=P_2\left(V_0\right)^\gamma \\
\Rightarrow & P_0\left(2 V_0\right)=P_2\left(V_0\right)
\end{array}
\)
(ii) Adiabatic compression, (gas \(B\) ) (during \(1 \rightarrow 2\) )
\(
\begin{array}{rlr}
& P_1 V_1^\gamma=P_2 V_2^\gamma \\
\Rightarrow & P_0\left(2 V_0\right)^\gamma=P_2\left(V_0\right)^\gamma \\
\Rightarrow & P_2=\left(\frac{2 V_0}{V_0}\right)^\gamma, P_0=(2)^\gamma P_0
\end{array}
\)
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.

Hint

(b) According to question, since there is no net loss to the surroundings and the equilibrium temperature of the system is \(T\).
Let us assume that \(T_1, T_2<T<T_3\).
Heat lost by \(M_3=\) Heat gained by \(M_1+\) Heat gained by \(M_2\)
\(
\Rightarrow \quad M_3 s\left(T_3-T\right)=M_1 s\left(T-T_1\right)+M_2 s\left(T-T_2\right)
\)
(where \(s\) is the specific heat of the copper material)
\(
\begin{aligned}
& \Rightarrow \quad T\left[M_1+M_2+M_3\right]=M_3 T_3+M_1 T_1+M_2 T_2 \\
& \Rightarrow \quad T=\frac{M_1 T_1+M_2 T_2+M_3 T_3}{M_1+M_2+M_3}
\end{aligned}
\)

Hint

\((a, b, d)\)
Key concept: Reversible process: A reversible process is one which can be reversed in such a way that all changes occurring in the direct process are exactly repeated in the opposite order and inverse sense and no change is left in any of the bodies taking part in the process or in the surroundings.
The conditions for reversibility are:
– There must be complete absence of dissipative forces such as friction, viscosity, electric resistance etc.
– The direct and reverse processes must take place infinitely slowly.
– The temperature of the system must not differ appreciably from its surroundings.
Irreversible process: Any process which is not reversible exactly is an irreversible process. All natural
processes such as conduction, radiation, radioactive decay, etc. are irreversible. All practical processes such
as free expansion, Joule-Thomson expansion, and electrical heating of a wire are also irreversible.
(a) In this case internal energy of the rod is increased from external work done by the hammer which in turn increases its temperature. So, 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 a temperature \(T_2\).
(c) In a quasi-static isothermal expansion, the gas is ideal, this process is reversible because the cylinder is fitted with a frictionless piston.
(d) As the weight is added to the cylinder arrangement in the form of external pressure hence, it cannot be reversed back itself.

Hint

(a, d)
Key concept: First Law of Thermodynamics:
It is a statement of conservation of energy in a thermodynamical process.
According to it heat given to a system \((\Delta Q\) ) is equal to the sum of the increase in its internal energy (AIT) and the work done (AW) by the system against the surroundings.
\(\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}\)
According to the first law of thermodynamics.
\(\Delta \mathrm{U} \propto \Delta \mathrm{T}\)
\(\Delta \mathrm{U}=0[\mathrm{As} \Delta \mathrm{T}=0]\)
\(\Delta Q=\Delta W\), i.e., heat supplied in an isothermal change is used to do work against external surroundings.
or if the work is done on the system then an equal amount of heat energy will be liberated by the system

Hint

(b,c)
Key concept: Internal energy \((U)\): Internal energy of a system is the energy possessed by the system due to molecular motion and molecular configuration.
The energy due to molecular motion is called internal kinetic energy \(U_K\) and that due to the molecular configuration is called internal potential energyUp.
i.e., Total internal energy \(U=U_K+U_P\)
(i) For an ideal gas, as there is no molecular attraction \(U_p=0\)
i.e., the internal energy of an ideal gas is totally kinetic and is given by
\(
U=U_K=3 / 2 R T
\)

and change in internal energy \(\Delta U=\frac{3}{2} \mu R \Delta T\)
(ii) In case of gases whatever be the process
\(
\begin{aligned}
\Delta U & =\mu \frac{f}{2} R \Delta T=\mu C_V \Delta T \\
& =\mu \frac{R}{(\gamma-1)} \Delta T=\frac{\mu R\left(T_f-T_i\right)}{\gamma-1}=\frac{\mu R T_f-\mu R T_i}{\gamma-1} \\
& =\frac{\left(P_f V_f-P_i V_i\right)}{\gamma-1}
\end{aligned}
\)
(iii) Change in internal energy in a cyclic process is always zero as for cyclic process \(U_f=U_i\)
So \(\Delta U=U_f-U_i=0\)
Change in internal energy does not depend on the path of the process. So it is called a point function, i.e. it depends only on the initial and final states ( \(A\) and \(B\) ) of the system, i.e. \(\Delta U=U f-U_i\) Hence internal energy is same for all four paths I, II, III and IV.
The work done by an ideal gas is equal to the area bounded between \(\mathrm{P}-\mathrm{V}\) curve. Work done from \(A\) to \(B, \Delta W_{A \rightarrow B}=\) Area under the \(P-V\) curve which is maximum for the path \(\mathrm{I}\).

Hint

(a, c)
(a) The given process is a cyclic process, i.e. it returns to the original state 1. And change in internal energy in a cyclic process is always zero as for cyclic process \(U_f=U_i S o, \Delta U=U_f-U_i=0\) Hence, total heat is completely converted to mechanical energy. Such a process is not possible by the second law of thermodynamics.
(c) Here, two curves are intersecting, 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. So, we conclude that such a process does not exist because curves representing two adiabatic processes do not intersect.

Hint

Key concept: Refrigerator or Heat Pump:
A refrigerator or heat pump is basically a heat engine run in the reverse direction. It essentially consists of three parts:
Source: At higher temperature \(T_1\)
Working substance: It is called refrigerant liquid ammonia and freon works as a working substance.

Sink: At lower temperature \(T_2\).
The working substance takes heat \(\mathrm{Q}_2\) from a sink (contents of refrigerator) at lower temperature, has a net amount of work done W on it by an external agent (usually compressor of refrigerator) and gives out a larger amount of heat \(Q_1\), to a hot body at temperature \(T_1\) (usually atmosphere). Thus, it transfers heat from a cold body to a hot body at the expense of mechanical energy supplied to it by an external agent. The cold body is thus cooled more and more.
We know that the diagram represents the working of a refrigerator. So, we can write \(Q_1=W+Q_2\)
According to the problem, \(W>0\), then \(\Rightarrow W=Q_1-Q_2>0\)
So there are two possibilities:
(a) If both \(Q_1\) and \(Q_2\) are positive,
\(
\Rightarrow Q_1>Q_2>0
\)
(c) If both \(Q_1\) and \(Q_2\) are negative,
\(
Q_2<Q_1<0
\)

Hint

Yes, this is possible when the entire heat supplied to the system is utilised in expansion.
As \(\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}\) and \(\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}\)
\(\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}+\Delta \mathrm{W}\)
If temperature remains constant, then \(\Delta T=0\), this implies \(\Delta Q=\Delta W\). This implies that heat supplied should perform work against the surroundings.

Hint

According to the first law of thermodynamics,
\(\Delta \mathrm{Q}=\mathrm{AU}+\Delta \mathrm{W}\). Let us apply this for each path.
For path 1: Heat given \(Q_1=+1000 \mathrm{~J}\)
Let work done for path \(1=W_1\)
For path 2 :
Work done \(\left(\mathrm{W}_2\right)=\left(\mathrm{W}_1-100\right) \mathrm{J}\)
Heat given \(Q_2\) ?
As change in internal energy between two states for different path is same.
\(
\begin{aligned}
& \Delta U=Q_i-W_1=Q_2-W_2 \\
& 1000-W_1=Q_2-\left(W_1-100\right) \\
& \Rightarrow Q_2=1000-100=900 \mathrm{~J}
\end{aligned}
\)

Hint

Sol: Yes, it is possible to increase the temperature of a gas without adding heat to it, during adiabatic compression the temperature of a gas increases while no heat is given to it.
For an adiabatic compression, no heat is given or taken out in adiabatic process.
Therefore, \(\Delta Q=0\)
According to the first law of thermodynamics,
\(
\begin{aligned}
& \Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W} \\
& \Delta \mathrm{U}=-\Delta \mathrm{W}(\Delta \mathrm{Q}=0)
\end{aligned}
\)
In compression work is done on the gas, i.e. work done is negative. Therefore, \(\Delta \mathrm{U}=\) Positive
Hence, internal energy of the gas increases due to which its temperature increases.

Hint

The volume of a car tyre is fixed. During driving, the 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, the pressure of gas increases.

Hint

Key concept: Carnot theorem: The efficiency of Carnot’s heat engine depends only on the temperature of source \(\left(T_1\right)\) and temperature of \(\operatorname{sink}\left(T_2\right)\), and heat supplied \(\left(Q_1\right)\) i.e., \(\eta=W / Q_1=1-T_2 / T_1\)
(The efficiency of engine is defined as the ratio of work done to the heat supplied.)
Carnot stated that no heat engine working between two given temperatures of source and sink can be more efficient than a perfectly reversible engine (Carnot engine) working between the same two temperatures.
Carnot’s reversible engine working between two given temperatures is considered to be the most efficient
engine.
According to the problem, temperature of the source \(T_1=500 \mathrm{~K}\), Temperature of the sink \(T_2=300 \mathrm{~K}\),
Work done per cycle \(W=1 \mathrm{~kJ}=1000 \mathrm{~J}\)
Heat transferred to the engine per cycle \(Q_1=\) ?
Efficiency of a Carnot engine
\(
(\eta)=1-\frac{T_2}{T_1}=1-\frac{300}{500}=\frac{200}{500}=\frac{2}{5}
\)
and \(\eta=\frac{W}{Q_1} \Rightarrow Q_1=\frac{W}{\eta}=\frac{1000}{(2 / 5)}=2500 \mathrm{~J}\)
As \(\quad Q_1-Q_2=W, Q_2=Q_1-W\)
\(
=2500 \mathrm{~J}-1000 \mathrm{~J}=1500 \mathrm{~J}
\)

Hint

The energy produced by \(1 \mathrm{~kg}\) fat \(=7000 \mathrm{k} \mathrm{cal}\)
Energy produced by \(5 \mathrm{~kg}\) fat \(=35000 \mathrm{k} \mathrm{cal}=35 \times 10^6 \mathrm{cal}\)
Energy consumed to go up one time=mgh
Energy consumed to go up and down one time \(=m g h+0.5 \mathrm{mgh}\)
\(
E=\frac{3}{2} m g h=\frac{3}{2} \times 60 \times 10 \times 10=9000 \mathrm{~J}=\frac{9000}{4.2} \mathrm{cal}
\)
Let us assume that he goes up and down \(n\) times, then
\(
\begin{aligned}
& n \times \frac{9000}{4.2}=35 \times 10^6 \\
& n=35 \times 10^6 \times \frac{4.2}{9000}=16.3 \times 10^3
\end{aligned}
\)

Hint

Since the process is adiabatic, there is no exchange of heat in the process, Let, pressure is increased by
\(\mathrm{AP}\) and volume is increased by AV at each stroke.
For just before and after an stroke, we can write
For just before and after an stroke, we can write
\(
\begin{aligned}
& P_1 V_1^\gamma=P_2 V_2^\gamma \\
\Rightarrow & P(V+\Delta V)^\gamma=(P+\Delta P) V^\gamma \quad(\because \text { volume is fixed }) \\
\Rightarrow & P V^\gamma\left(1+\frac{\Delta V}{V}\right)^\gamma=P\left(1+\frac{\Delta P}{P}\right) V^\gamma
\end{aligned}
\)
As \(\Delta V<<V\), so by using binomial approximation we get
\(
\begin{aligned}
& \Rightarrow \quad P V^\gamma\left(1+\gamma \frac{\Delta V}{V}\right) \approx P V^\gamma\left(1+\frac{\Delta P}{P}\right) \\
& \Rightarrow \quad \gamma \frac{\Delta V}{V}=\frac{\Delta P}{P}
\end{aligned}
\)
\(
\Rightarrow \quad P \Delta V=\frac{V}{\gamma} \Delta P
\)
Hence, work done is increasing the pressure from \(P_1\) to \(P_2\) is
\(
\begin{aligned}
W & =\int P \Delta V=\frac{V}{\gamma} \int_P^{P_2} \Delta P=\frac{V}{\gamma}|P|_{P_1}^{P_2} \\
& =\frac{\left(P_2-P_1\right) V}{\gamma}
\end{aligned}
\)
Important note: As \(\Delta V\) is very small so we can also write this as \(d V\), then the work done will be same as
\(
W=\int_{R_1}^{P_2} P d V=\int_{P_1}^{P_2} P \times \frac{1}{\gamma} \frac{V}{P} d P=\frac{\left(P_2-P_1\right)}{\gamma} V
\)

Hint

Carnot designed a theoretical engine which is free from all the defects of a practical engine. The Carnot engine is the most efficient heat engine operating between two given temperatures. The efficiency of Carnot engine is
\(
\eta=1-\frac{T_2}{T_1}
\)
According to the problem, efficiency of a perfect engine working between \(-3^{\circ} \mathrm{C}\) and \(27^{\circ} \mathrm{C}\) (i.e., \(T_2=-3+273=270 \mathrm{~K}\) and \(T_1=27+273=300 \mathrm{~K}\) ).
Thus according to the Carnot theorem \(\eta_{\text {engine }}=1-\frac{T_2}{T_1}=1-\frac{270 \mathrm{~K}}{300 \mathrm{~K}}=0.1\)
Since, efficiency of the refrigerator \(\left(\eta_{\text {ref }}\right)\) is \(50 \%\) of \(\eta_{\text {engine }}\)
\(
\therefore \quad \eta_{\text {ref }}=0.5 \eta_{\text {engine }}=0.05
\)
If \(Q_1\) is the heat transferred per second at higher temperature by doing work \(W\), then
\(
\begin{aligned}
& \eta_{\text {ref. }}=\frac{W}{Q_1} \text { or } Q_1=\frac{W}{\eta_{\text {ret }}}=\frac{1 \mathrm{~kJ}}{0.05}=20 \mathrm{~kJ} \\
& \text { (as } W=1 \mathrm{~kW} \times 1 \mathrm{~s}=1 \mathrm{~kJ} \text { ) }
\end{aligned}
\)
Since \(\eta_{\text {ref }}\) is \(0.05\), heat removed from the refrigerator per second, i.e.,
\(
\begin{aligned}
Q_2 & =Q_1-\eta_{\text {ref }} Q_1=Q_1\left(1-\eta_{\text {ref }}\right) \\
& =20 \mathrm{~kJ}(1-0.05)=19 \mathrm{~kJ}
\end{aligned}
\)
Therefore, heat is taken out of the refrigerator at a rate of \(19 \mathrm{~kJ}\) per second.

Hint

Key concept: The performance of a refrigerator is expressed by means of “coefficient of performance” \(\beta\)
which is defined as the ratio of the heat extracted from the cold body to the work needed to transfer it to the hot body.
i.e., \(\quad \beta=\frac{\text { Heat extracted }}{\text { Work done }}=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2} \quad \therefore \beta=\frac{Q_2}{Q_1-Q_2}\)
A perfect refrigerator is the one which transfers heat from cold to hot body without doing work.
i.e., \(W=0\) so that \(Q_1=Q_2\) and hence \(\beta=\infty\)
According to the problem, coefficient of performance \((\omega)=5\)
\(
T_1=(27+273) \mathrm{K}=300 \mathrm{~K}
\)
Coefficient of performance \((\omega)=\frac{T_2}{T_1-T_2}\)
\(
\begin{aligned}
& 5=\frac{T_2}{300-T_2} \Rightarrow 1500-5 T_2=T_2 \\
\Rightarrow & 6 T_2=1500 \Rightarrow T_2=250 \mathrm{~K} \\
\Rightarrow & T_2=(250-273)^{\circ} \mathrm{C}=-23^{\circ} \mathrm{C}
\end{aligned}
\)