Numerical Problems-1 - Chapter 8 - Heat and Thermodynamics - 11th Science Guide Samacheer Kalvi Solutions

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Numerical Problems
Question 1.
Calculate the number of moles of air is in the inflated balloon at room temperature as shown in the figure.

The radius of the balloon is $10 \mathrm{~cm}$, and pressure inside the balloon is $180 \mathrm{kPa}$

Answer:
$
\begin{aligned}
& \mathrm{P}=180 \mathrm{kPa}=180 \times 10^3 \mathrm{~Pa}, \mathrm{~V}=\frac{4}{3} \pi r^3, r=10 \mathrm{~cm}, \mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{k}^{-1}, \\
& \mathrm{~T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}
\end{aligned}
$
From ideal gas equation of state
$
\begin{aligned}
\mathrm{PV} & =\mu \mathrm{RT} \\
\mu & =\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{180 \times 10^3 \times \frac{4}{3} \times 3.14 \times\left(10 \times 10^{-2}\right)^3}{8.314 \times 300} \\
& =\frac{0.7536 \times 10^3}{2494.2}=3.021 \times 10^{-4} \times 10^3 \\
\mu & \cong 0.3 \mathrm{~mol}
\end{aligned}
$
Question 2.
In the planet Mars, the average temperature is around $-53^{\circ} \mathrm{C}$ and atmospheric pressure is $0.9 \mathrm{kPa}$. Calculate the number of moles of the molecules in unit volume in the planet Mars? Is this greater than that in earth?
Answer:
$
\begin{aligned}
& \mathrm{T}=-53^{\circ} \mathrm{C}=220 \mathrm{~K} \\
& \mathrm{P}=0.9 \times 10^3 \mathrm{~Pa} \\
& \mathrm{~V}=1 \mathrm{~m}^3
\end{aligned}
$
Number of molecules $\mu=\frac{\mathrm{VP}}{\mathrm{RT}}=\frac{0.9 \times 10^3}{8.314 \times 220}=0.00049 \times 10^3 ; \mu=0.49 \mathrm{~mol}$
Question 3.
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $V_1$ and contains ideal gas at pressure $P_1$ and temperature $T_1$. The other chamber has volume $V_2$ and contains ideal gas at pressure $P_2$ and temperature $T_2$. If the partition is removed without doing any work on the gases, calculate the final equilibrium temperature of the container.
Answer:

Let $\mathrm{T}$ be the equilibrium temperature and let $\mathrm{n}_1$ and $\mathrm{n}_2$ be the number of moles in vessels 1 and 2 respectively. As there is no loss of energy,
$
\begin{aligned}
n_1\left(\frac{3}{2} \mathrm{RT}_1\right)+n_2\left(\frac{3}{2} \mathrm{RT}_2\right) & =\left(n_1+n_2\right)\left(\frac{3}{2} \mathrm{RT}\right) \\
n_1 \mathrm{~T}_1+n_2 \mathrm{~T}_2 & =\left(n_1+n_2\right) \mathrm{T} \\
\mathrm{T} & =\frac{n_1 \mathrm{~T}_1+n_2 \mathrm{~T}_2}{n_1+n_2} \\
n_1 & =\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{RT}_1}, n_2=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{RT}_2}
\end{aligned}
$
Substituting $n_1$ and $n_2$ values and solving, we get
$
T=\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_2+P_2 V_2 T_1}
$
Question 4.
The temperature of a uniform rod of length $L$ having a coefficient of linear expansion $\alpha_{\mathrm{L}}$ is changed by $\Delta \mathrm{T}$. Calculate the new moment of inertia of the uniform rod about axis passing through its center and perpendicular to an axis of the rod.
Answer:
Moment of inertia of a uniform rod of mass and length 1 about its perpendicular bisector. Moment of inertia of the rod

$
I=\frac{1}{12} M L^2
$
Increase in length of the rod when temperature is increased by $\Delta T$, is given by $\mathrm{L}^{\prime}=\mathrm{L}\left(1+\alpha_{\mathrm{L}} \Delta \mathrm{T}\right)$
New moment of inertia of the rod
$
\begin{aligned}
& I^{\prime}=\frac{M L^{\prime 2}}{12}=\frac{M}{12} L^2\left(1+\alpha_L \Delta T\right)^2 \\
& I^{\prime}=I\left(1+\alpha_L \Delta T\right)^2
\end{aligned}
$
Question 5.
Draw the TP diagram ( $\mathrm{P}-\mathrm{x}$ axis, $\mathrm{T}-\mathrm{y}$ axis), $V T(T-x$ axis, $V-y$ axis) diagram for
(a) Isochoric process
(b) Isothermal process
(c) Isobaric process
Answer:
(a) Isochoric process : $\mathrm{PV}_o=n \mathrm{RT}$

(b) Isothermal process: $\mathrm{PV}=n \mathrm{RT}_{\mathrm{o}}$

(c) Isobaric process : $\mathrm{P}_0 \mathrm{~V}=n \mathrm{RT}$

Question 6.
A man starts bicycling in the morning at a temperature around $25^{\circ} \mathrm{C}$, he checked the pressure of tire which is equal to be $500 \mathrm{kPa}$. Afternoon he found that the absolute pressure in the tyre is increased to $520 \mathrm{kPa}$. By assuming the expansion of tyre is negligible, what is the temperature of tyre at afternoon?
Answer:
For ideal gas equation of state
$\mathrm{PV}=\mathrm{nRT}$
$
\mathrm{P}_1=500 \mathrm{kPa}, \mathrm{T}_1=25^{\circ} \mathrm{C}=25+273=298 \mathrm{~K}, \mathrm{P}_2=520 \mathrm{kPa}, \mathrm{T}_2=?
$
Expansion of tyre is negligible $\left(\mathrm{V}_{\text {constant }}\right)$
$
\begin{aligned}
\frac{\mathrm{P}_1 V_1}{\mathrm{~T}_1} & =\frac{\mathrm{P}_2 V_2}{\mathrm{~T}_2} \\
\therefore \quad \mathrm{T}_2 & =\left(\frac{\mathrm{P}_2}{\mathrm{P}_1}\right) \mathrm{T}_1=\frac{520}{500} \times 298=\frac{154960}{500}=309.92 \mathrm{~K} \\
\mathrm{~T}_2 & =309.92-273 ; \mathrm{T}_2=36.9^{\circ} \mathrm{C}
\end{aligned}
$

Question 7.
Normal human body of the temperature is $98.6^{\circ} \mathrm{F}$. During high fever if the temperature increases to $104^{\circ} \mathrm{F}$, what is the change in peak wavelength that emitted by our body? (Assume human body is a black body)
Answer:
Normal human body temperature $(\mathrm{T})=98.6^{\circ} \mathrm{F}$
Convert Fahrenheit into Kelvin,
$
\frac{\mathrm{F}-32}{180}=\frac{\mathrm{K}-273}{100}
$
So, $\mathrm{T}=98.6^{\circ} \mathrm{F}=310 \mathrm{~K}$
From Wien's displacement law
Maximum wavelength, $\lambda_{\max }=\frac{b}{\mathrm{~T}}=\frac{2.898 \times 10^{-3}}{310}=9348 \times 10^{-9} \mathrm{~m}$
$
\lambda_{\max }=9348 \mathrm{~nm} \text { (at } 98.6^{\circ} \mathrm{F} \text { ) }
$
During high fever, human body temperature, $\mathrm{T}=104^{\circ} \mathrm{F}=313 \mathrm{~K}$
Peak wavelength
$
\begin{aligned}
& \lambda_{\max }=\frac{b}{\mathrm{~T}}=\frac{2.898 \times 10^{-3}}{313}=9259 \times 10^{-9} \mathrm{~m} \\
& \lambda_{\max }=9259 \mathrm{~nm} \quad\left(\text { at } 104^{\circ} \mathrm{F}\right)
\end{aligned}
$
Question 8.
In an adiabatic expansion of the air, the volume is increased by $4 \%$, what is percentage change in pressure? (For air $\gamma=1.4$ )
Answer:

From equation for adiabatic process,
$
\mathrm{PV}^\gamma=\text { constant }
$
Using differentiation, we get
$
\begin{aligned}
\mathrm{P} \gamma \mathrm{V}^{\gamma-1} d \mathrm{~V}+d \mathrm{P}^\gamma & =0 \\
\frac{d \mathrm{P}}{\mathrm{P}} & =-\gamma \frac{d \mathrm{~V}}{\mathrm{~V}}
\end{aligned}
$
Volume ' $\mathrm{V}$ ' is increased. by $4 \%$ and $\gamma=1.4$
$
\frac{d \mathrm{P}}{\mathrm{P}} \times 100=-\gamma\left(\frac{d \mathrm{~V}}{\mathrm{~V}} \times 100\right)=-1.4 \times 4=-5.6
$
Pressure is decreased. by $5.6 \%$
Question 9.
In a petrol engine, (internal combustion engine) air at atmospheric pressure and temperature of $20^{\circ} \mathrm{C}$ is compressed in the cylinder by the piston to $1 / 8$ of its original volume. Calculate the temperature of the compressed air. (For air $\gamma=1.4$ ) Answer:
$
\mathrm{T}_1=20^{\circ} \mathrm{C}=20+273=293 \mathrm{~K}, \mathrm{~V}_1=1 \mathrm{~m}^3, \mathrm{~V}_2=\frac{1}{8} \mathrm{~V}_1 m^3, \gamma=1.4
$
From equation of adiabatic process
$
\begin{aligned}
\mathrm{TV}^{\gamma-1} & =\text { constant } \\
\mathrm{T}_1 \mathrm{~V}_1^{\gamma-1} & =\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\
\mathrm{~T}_2 & =\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1} \mathrm{~T}_1 \\
\mathrm{~T}_2 & =\left(\frac{1}{\left(\frac{1}{8}\right)}\right)^{1.4-1} \times 293=(8)^{0.4} \times 293 \\
& =673.1 \mathrm{~K}=673.1-273 ; \mathrm{T}_2 \cong 400^{\circ} \mathrm{C}
\end{aligned}
$
Question 10
Consider the following cyclic process consist of isotherm, isochoric and isobar which is given in the figure.

Draw the same cyclic process qualitatively in the $\mathrm{V}-\mathrm{T}$ diagram where $\mathrm{T}$ is taken along $\mathrm{x}$ - direction and $\mathrm{V}$ is taken along $\mathrm{y}$ - direction. Analyze the nature of heat exchange in each process.
Answer:
Process 1 to $2=$ increase in volume. So heat must be added.
Process 2 to $3=$ Volume remains constant. Increase in temperature.
The given heat is used to increase the internal energy.

Process 3 to 1 : Pressure remains constant. Volume and Temperature are reduced. Heat flows out of the system. It is an isobaric compression where the work is done on the system.
Question 11.
An ideal gas is taken in a cyclic process as shown in the figure.

Calculate
(a) work done by the gas.
(b) work done on the gas
(c) Net work done in the process
Answer:
(a) Work done by the gas (along $\mathrm{AB}$ )
$
\mathrm{W}=\mathrm{P} \times \Delta \mathrm{V}=600 \times 3=1800 \mathrm{~J} ; \mathrm{W}=1.8 \mathrm{~kJ}
$
(b) Work is done on the gas (along $\mathrm{BC}$ )
$
\begin{aligned}
& \mathrm{W}=-\mathrm{P} \times \Delta \mathrm{V}=-400 \times 3=-1200 \mathrm{~J} \\
& \mathrm{~W}=-1.2 \mathrm{~kJ}
\end{aligned}
$

(c) Net work done in the process = Area under the curve $A B$ $=$ Rectangle area + triangle area
$
\begin{aligned}
& =(l \times b)+\left(\frac{1}{2} \times b \times h\right)=(400 \times 3)+\left(\frac{1}{2} \times 3 \times 200\right) \\
& =1200+300=1500 \mathrm{~J} ; \mathrm{W}=1.5 \mathrm{~kJ}
\end{aligned}
$
Question 12.
For a given ideal gas $6 \times 10^5 \mathrm{~J}$ heat energy is supplied and the volume of gas is increased from $4 \mathrm{~m}^3$ to $6 \mathrm{~m}^3$ at atmospheric pressure. Calculate
(a) the work done by the gas
(b) change $i n$ internal energy of the gas
(c) graph this process $; \mathrm{n} \mathrm{PV}$ and TV diagram.
Answer:
Heat energy supplied to gas $Q=6 \times 10^5 \mathrm{~J}$
Change in volume $\Delta \mathrm{V}=(6-4)=2 \mathrm{~m}^3$
$1 \mathrm{~atm}=1.013 \times 10^5 \mathrm{Nm}^{-2}$
(a) Work done by the gas $\mathrm{W}=\mathrm{P} \times \Delta \mathrm{V}=1.013 \times 10^5 \times 2=2.026 \times 10^5$ $\mathrm{W}=202.6 \mathrm{~kJ} 1$
(b) Change in internal energy of the gas
$
\begin{aligned}
& \Delta \mathrm{U}=\mathrm{Q}-\mathrm{P} \cdot \Delta \mathrm{V}=6 \times 10^5-\left(202.6 \times 10^3\right) \\
&=6 \times 10^5-2.026 \times 10^5=3.974 \times 10^5 \\
& \Delta \mathrm{U}=397.4 \mathrm{~kJ}
\end{aligned}
$

Question 13.
Suppose a person wants to increase the efficiency of the reversible heat engine that is operating between $100^{\circ} \mathrm{C}$ and $300^{\circ} \mathrm{C}$. He had two ways to increase the efficiency,
(a) By decreasing the cold reservoir temperature from $100^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ and keeping the hot reservoir temperature constant
(b) by increasing the temperature of the hot reservoir from $300^{\circ} \mathrm{C}$ to $350^{\circ} \mathrm{C}$ by keeping the cold reservoir temperature constant. Which is the suitable method?
Answer:
Heat engine operates at initial temperature $=100^{\circ} \mathrm{C}+273=373 \mathrm{~K}$
Final temperature $=300^{\circ} \mathrm{C}+273=573 \mathrm{~K}$
At melting point $=273 \mathrm{~K}$
Efficiency $\quad \eta=1-\frac{T_2}{T_1}=1-\frac{273}{573}=0.3491 ; \eta=34.9 \%$
(a) By decreasing the cold reservoir, efficiency
$
\begin{gathered}
\mathrm{T}_1=350^{\circ} \mathrm{C}+273=623 \mathrm{~K}, \quad \mathrm{~T}_2=100^{\circ} \mathrm{C}+273=373 \mathrm{~K} \\
\eta=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1}=1-\frac{323}{573}=0.436 \\
\eta=43.6 \%
\end{gathered}
$
(b) By increasing the temperature of hot reservoir, efficiency,
$
\begin{aligned}
\mathrm{T}_1=350^{\circ} \mathrm{C}+273 & =623 \mathrm{~K}, \mathrm{~T}_2=100^{\circ} \mathrm{C}+273=373 \mathrm{~K} \\
\eta & =1-\frac{\mathrm{T}_2}{\mathrm{~T}_1}=1-\frac{373}{623}=0.401 \\
\eta & =40.12 \%
\end{aligned}
$
Method (a) More efficiency than method (b).
Question 14 .
A Carnot engine whose efficiency is $45 \%$ takes heat from a source maintained at a temperature of $327^{\circ} \mathrm{C}$. To have an engine of efficiency $60 \%$ what must be the intake temperature for the same exhaust (sink) temperature?
Answer:
Efficiency of Carnot engine $\left(\eta_1\right)=45 \%=0.45$
Initial intake temperature $\left(\mathrm{T}_1\right)=327^{\circ} \mathrm{C}=600 \mathrm{~K}$
New efficiency $\left(\eta_2\right)=60 \%=0.6$
Efficiency of Carnot engine is given by
$
\eta=1-\frac{T_2}{T_1}
$

$T_1$ is temperature of source ; $T_2$ is temperature of sink
1st Case: $\quad T_2=(1-\eta) T_1=(1-0.45) \times 600 \Rightarrow T_2=330 \mathrm{~K}$
$
\text { 2nd Case: } \quad \begin{aligned}
\frac{T_2}{T_1} & =1-\eta \\
T_1 & =\frac{T_2}{1-\eta}=\frac{330}{1-0.6}=\frac{330}{0.4} \\
T_1 & =825 \mathrm{~K}=825-273 ; \mathrm{T}_1=552^{\circ} \mathrm{C}
\end{aligned}
$
Question 15.
An ideal refrigerator keeps its content at $0^{\circ} \mathrm{C}$ while the room temperature is $27^{\circ} \mathrm{C}$. Calculate its coefficient of performance.
Answer:
Content placed at temperature $\quad \mathrm{T}_{\mathrm{L}}=0{ }^{\circ} \mathrm{C}=0+273$
$
\mathrm{T}_{\mathrm{L}}=273 \mathrm{~K}
$
Refrigerator placed at temperature $\mathrm{T}_{\mathrm{H}}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$
Coefficient of performance
$
(\beta)=\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}=\frac{273}{300-273}=\frac{273}{27} ; \beta=10.11
$