Numerical Problems-2 - Chapter 8 Heat and Thermodynamics 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
Copper wire of length 1 increase in length by $1 \%$ when heated from temperature $T_1$ to $T_2$. Find the percentage change in area when a copper plate of dimensions $21 \times 1$ is heated from $T_1$ to $T_2$
Answer:
From length expansion equation $\frac{\Delta \mathrm{L}}{\mathrm{L}}=\alpha_{\mathrm{L}} \Delta \mathrm{T}$
From Area expansion equation, $\left(\frac{\Delta \mathrm{V}}{\mathrm{V}_0}\right)=\alpha_{\mathrm{A}} \Delta \mathrm{T}$
But $\alpha_{\mathrm{A}}=2 \alpha_{\mathrm{L}}$, therefore, $\frac{\Delta \mathrm{A}}{\mathrm{A}_0}=2 \alpha_{\mathrm{A}} \Delta \mathrm{T}=2 \frac{\Delta \mathrm{L}}{\mathrm{L}}$

$
\begin{aligned}
& \text { Percentage change }=\frac{\Delta \mathrm{A}}{\mathrm{A}_0} \times 100=2\left(\frac{\Delta \mathrm{L}}{\mathrm{L}} \times 100\right)=2 \times 1 \\
& \text { Percentage change }=2 \%
\end{aligned}
$
Question 2.
A steel rod of length $25 \mathrm{~cm}$ has a cross-sectional area of $0.8 \mathrm{~cm}^2$. Find the force required to stretch this rod by the same amount as the expansion produced by heating it through $10^{\circ} \mathrm{C}$. (coefficient of linear expansion of steel is $10^{-5} /{ }^{\circ} \mathrm{C}$ and Young's modulus of steel is $2 \times 10^{10} \mathrm{Nm}^{-2}$ )
Answer:
The required force is given by $\mathrm{F}=\mathrm{YA} \alpha \Delta t=2 \times 10^{10} \times 0.8 \times 10^{-4} \times 10^{-5} \times 10=160 \mathrm{~N}$
Question 3.
The temperature at the bottom of a $40 \mathrm{~m}$ deap lake is $12^{\circ} \mathrm{C}$ and that at the surface is $35^{\circ} \mathrm{C}$. An air bubble of volume $1.0 \mathrm{~cm} 3$ rises from the bottom to the surface. Find its volume, (atmospheric pressure $=10 \mathrm{~m}$ of water)
Answer:

Let $P_1, V_1$ and $T_1$ be the pressure, bubble volume and absolute temperature at the bottom of the lake and let $P_2, V_2$ and $T_2$ be the corresponding quantities at the surface. Then
$
\begin{aligned}
\frac{\mathrm{P}_2 V_2}{T_2} & =\frac{P_1 V_1}{T_1} \Rightarrow V_2=\frac{P_1 T_2}{P_2 T_1} V_1 \\
V_2 & =\left(\frac{10+40}{10}\right)\left(\frac{273+35}{273+12}\right) \times 1=\frac{5 \times 308}{285} \\
V_2 & =5.4 \mathrm{~cm}^3
\end{aligned}
$
Question 4.
Ajar ' $\mathrm{A}$ ' is filled with an ideal gas characterised by parameters $\mathrm{P}, \mathrm{V}$ and $\mathrm{T}$ and another jar $\mathrm{B}$ is filled with an ideal gas with parameters $2 \mathrm{P}, \frac{\mathrm{V}}{4}$ and $2 \mathrm{~T}$. Find the ratio of the number of molecules in jar $\mathrm{A}$ and $\mathrm{B}$.
Answer:
$
\begin{aligned}
n_{\mathrm{A}} & =\frac{\mathrm{PV}}{\mathrm{RT}}, n_{\mathrm{B}}=\frac{(2 \mathrm{P})\left(\frac{\mathrm{V}}{4}\right)}{\mathrm{R}(2 \mathrm{~T})}=\frac{1}{4} n_{\mathrm{A}} ; n_{\mathrm{B}}=\frac{1}{4} n_{\mathrm{A}} \\
\therefore \quad n_{\mathrm{A}}: n_{\mathrm{B}} & =4: 1
\end{aligned}
$
Question 5.
By what percentage should the pressure of a given mass of a gas be increased so as to decrease its volume by $10 \%$ at a constant temperature?
Answer:

According to Boyle's law,
$
\begin{aligned}
(\mathrm{P}+\Delta \mathrm{P})(\mathrm{V}-\Delta \mathrm{V}) & =\mathrm{PV} \\
(\mathrm{P}+\Delta \mathrm{P})(\mathrm{V}-0.1 \mathrm{~V}) & =\mathrm{PV} \\
\mathrm{PV}+\mathrm{V} \Delta \mathrm{P}-0.1 \mathrm{PV}-0.1 \mathrm{~V} \Delta \mathrm{P} & =\mathrm{PV} \\
0.9 \mathrm{~V} \Delta \mathrm{P} & =0.1 \mathrm{PV} \\
\frac{\Delta \mathrm{P}}{\mathrm{P}} & =\frac{1}{9} \\
\text { Percentage increase } & =\frac{\Delta \mathrm{P}}{\mathrm{P}} \times 100=\frac{100}{9}=11.1 \\
\text { Percentage increase } & =11.1 \%
\end{aligned}
$
Question 6.
A flask is filled with $13 \mathrm{~g}$ of an ideal gas at $27^{\circ} \mathrm{C}$ and its temperature is raised to $52^{\circ} \mathrm{C}$. Find the mass of the gas that hias to be released to maintain the temperature of the gas in the flask at $52^{\circ} \mathrm{C}$ and the pressure is same as the initial pressure.
Answer:
Let $\mathrm{n}$ be the initial number of moles and ri be the final number of moles. Since pressure and volume remain the same, we have
$
\begin{aligned}
\qquad \mathrm{PV} & =n \mathrm{RT}=n \mathrm{R} \times 300=n^{\prime} \mathrm{R} \times 325 \\
\frac{n^{\prime}}{n} & =\frac{300}{325}=\frac{12}{13} \\
\frac{\text { Final mass }}{\text { Initial mass }} & =\frac{12}{13} \Rightarrow \text { Final mass }=\frac{12}{13} \times 13=12 \mathrm{~g} \\
\text { Mass to be released } & =13-12=1 \mathrm{~g}
\end{aligned}
$
Question 7.
A rod of metal-1 of length $50.0 \mathrm{~cm}$ elongates by $0.10 \mathrm{~cm}$ when it is heated from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. Another rod of metal- 2 of length $80.0 \mathrm{~cm}$ elongates by $0.08 \mathrm{~cm}$ for the same rise in temperature. A third rod of length $50.0 \mathrm{~cm}$, made by welding pieces of rod $1 \mathrm{rad}$ and $2 \mathrm{rad}$ placed end to end, elongates by $0.03 \mathrm{~cm}$ when it is heated from $0^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$. Then what is the length of metal-1 in the third rod at $0^{\circ} \mathrm{C}$ ?
Answer:
For metal-1, For metal-2,
$
\begin{array}{ll}
\alpha_1=\frac{\Delta l}{l \Delta t}=\frac{0.10}{50.0 \times 100} & \alpha_2=\frac{\Delta l}{l \Delta t}=\frac{0.08}{80.0 \times 100} \\
\alpha_1=2.0 \times 10^{-5} /{ }^{\circ} \mathrm{C} & \alpha_2=1.0 \times 10^{-5} /{ }^{\circ} \mathrm{C}
\end{array}
$

Let the lengths of metal-1 and metal- 2 in the third rod at $0^{\circ} \mathrm{C}$ be $1_1$ and $1_2$, respectively.
Then $\quad l_1+l_2=50.0$
When this rod is heated to $50^{\circ} \mathrm{C}$, then
and
$
\begin{aligned}
l_1^{\prime} & =l_1\left(1+\alpha_1 50\right) ; l_2^{\prime}=l_2\left(1+\alpha_2 50\right) \\
l_1^{\prime}+l_2^{\prime} & =l_1+l_2+\left(l_1 \alpha_1+l_2 \alpha_2\right) 50 \\
50.03 & =50.0+\left(2 l_1+l_2\right) \times 50 \times 10^{-5} \\
2 l_1+l_2 & =60
\end{aligned}
$
Solving (1) and (2), $\quad l_1=10.0 \mathrm{~cm}, l_2=40.0 \mathrm{~cm}$
Question 8.
A balloon is filled at $27^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure by $500 \mathrm{~m}^3 \mathrm{He}$. Then find the volume of $\mathrm{He}$ at $-3^{\circ} \mathrm{C}$ and $0.5 \mathrm{~mm}$ Hg pressure.
Answer:
$
\begin{aligned}
\frac{P_2 V_2}{T_2} & =\frac{P_1 V_1}{T_1} \\
V_2 & =\frac{P_1 V_1 T_2}{P_2 T_1} \frac{1 \times 500 \times(273-3)}{0.5 \times(273+27)}=\frac{500 \times 270}{0.5 \times 300} \Rightarrow V_2=900 \mathrm{~m}^3
\end{aligned}
$
Question 9.
During an experiment an ideal gas is found to obey an additional law $\mathrm{VP}^2=$ constant. The gas is initially at temperature $\mathrm{T}$ and Volume $V$. Find the resulting temperature when it expands to volume $2 \mathrm{~V}$.
Answer:
$
\begin{aligned}
\mathrm{PV} & =\mathrm{RT} \Rightarrow \mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}} \\
\mathrm{VP}^2 & =\text { constant } \\
\therefore \quad \mathrm{V}\left(\frac{\mathrm{RT}}{\mathrm{V}}\right)^2 & =\text { constant (or) } \frac{\mathrm{R}^2 \mathrm{~T}^2}{\mathrm{~V}}=\text { constant } \\
\frac{\mathrm{T}^2}{\mathrm{~V}} & =\text { constant } \\
\therefore \quad \frac{\mathrm{T}^{\prime 2}}{2 \mathrm{~V}} & =\frac{\mathrm{T}^2}{\mathrm{~V}} \Rightarrow \mathrm{T}^{\prime}=\sqrt{2} \mathrm{~T}
\end{aligned}
$

Question 12.
A mass of ideal gas at pressure $P$ is expanded isothermally to four times the original volume and then slowly compressed adiabatically to its original volume. Assuming $\gamma\left(=\mathrm{C}_{\mathrm{P}} / \mathrm{C}_{\mathrm{V}}\right)$ to be 1.5 , find the new pressure of the gas.
Answer:
Let the initial volume be V. After isothermal expansion, pressure $=\frac{P}{4}$, volume $=4 \mathrm{~V}$
Let $\mathrm{P}^{\prime}$ is the pressure after adiabatic compression. The volume then is $\mathrm{V}$. Therefore,
$
\begin{aligned}
\mathrm{P}^{\prime} \mathrm{V}^\gamma & =\frac{\mathrm{P}}{4}(4 \mathrm{~V})^\gamma ; \mathrm{P}^{\prime}=\frac{\mathrm{P}}{4}(4)^{\frac{3}{2}} \\
\mathrm{P}^{\prime} & =2 \mathrm{P}
\end{aligned}
$
Question 13.
A carnot engine operating between temperature $T_1$ and $T_2$ has efficiency $\frac{1}{6}$. When $T_2$ is lowered by $62 \mathrm{~K}$, its efficiency increases to $\frac{1}{3}$. Then find the values of $T_1$ and $T_2$.
Answer:
Efficiency
$
\eta=1-\frac{T_2}{T_1}
$
In the first case:
$
\begin{aligned}
1-\frac{T_2}{T_1} & =\frac{1}{6} \Rightarrow \frac{T_2}{T_1}=\frac{5}{6} \\
T_1 & =\frac{6}{5} \mathrm{~T}_2
\end{aligned}
$
In the second case:
$
\begin{aligned}
& 1-\frac{\left(T_2-62\right)}{T_1}=\frac{1}{3} ; \frac{T_2-62}{T_1}=\frac{2}{3} \\
& \frac{T_2-62}{\left(\frac{6}{5}\right) T_2}=\frac{2}{3}
\end{aligned}
$

$
\begin{aligned}
& \mathrm{T}_2-62=\frac{2}{3} \times \frac{6}{5} \mathrm{~T}_2=\frac{4}{5} \mathrm{~T}_2 \Rightarrow \mathrm{T}_2=310 \mathrm{~K} \\
& \mathrm{~T}_1=\frac{6}{5} \times 310=372 \mathrm{~K} \\
& \mathrm{~T}_1=372 \mathrm{~K} \text { and } \mathrm{T}_2=310 \mathrm{~K}
\end{aligned}
$
Question 14.
A perfect gas goes from state A to state B by absorbing $8 \times 10^5 \mathrm{~J}$ of heat and doing $6.5 \times 10^5 \mathrm{~J}$ of external work. It is now transferred between the same two states in another process in which it absorbs $10 \mathrm{~s} \mathrm{~J}$ of heat. In second process. Find the work done in the second process.
Answer:
According to the first law of thermodynamics
$\Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}$
In the first process $\Delta \mathrm{U}=8 \times 10^5-6.5 \times 10^5=1.5 \times 10^5 \mathrm{~J}$
Now $\Delta \mathrm{U}$, being a state function, remains the same in the second process,
$
\Delta \mathrm{W}=\Delta \mathrm{Q}-\Delta \mathrm{U}=1 \times 10^5-1.5 \times 10^5
$
$
\Delta \mathrm{W}=-0.5 \times 10^5 \mathrm{~J}
$
The negative sign shows that work is done on the gas.
Question 15.
A carnot engine, having efficiency of $\eta=\frac{1}{10}$ as heat engine, is used as a refrigerator. If the work done on the system is $10 \mathrm{~J}$, then find the amount of energy absorbed from the reservoir at lower temperature.
Answer:
As heat engine, let $\mathrm{Q}_1$ be the heat energy absorbed from the source and $\mathrm{Q}_2$ be the energy rejected to the sink. Then

Given :
$
\begin{aligned}
Q_1 & =W_1+Q_2=10 \\
\eta & =1-\frac{Q_2}{Q_1}=\frac{1}{10} \\
\frac{Q_2}{Q_1} & =1-\frac{1}{10}=\frac{9}{10} \\
Q_1 & =\frac{10}{9} Q_2
\end{aligned}
$
From equation (1) and (2)
$
\begin{aligned}
10+Q_2 & =\frac{10}{} \\
10 & =\left(\frac{10}{9}-1\right) Q_2 \\
\therefore \quad Q_2 & =90 \mathrm{~J}
\end{aligned}
$
Question 16.
An ideal gas compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas?
Answer:
Work done on the gas $=$ Area under the curve

$\mathrm{W}_{\text {adiabatic }}>\mathrm{W}_{\text {isothermal }}>\mathrm{W}_{\text {isobaric }}$
$\mathrm{w}_{\text {isochoric }}$ is obviously zero because in an isochoric process there is no change in Volume.