Additional Numerical Problems - Chapter 9 - Semiconductor Electronics - 12th Physics Guide Samacheer Kalvi Solutions

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Additional Numerical Problems
Question 1.
If the rms speed of hydrogen molecules at $300 \mathrm{~K}$ is $1930 \mathrm{~ms}^{-1}$. Then what is the rms speed of oxygen molecules
Answer:
$
\text { From rms speed, } \begin{aligned}
v_{r m s} & =\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \\
v_{\mathrm{O}_2} & =\sqrt{\frac{3 \mathrm{RT}_{\mathrm{O}_2}}{\mathrm{M}_{\mathrm{O}_2}}} ; v_{\mathrm{H}_2}=\sqrt{\frac{3 \mathrm{RT}_{\mathrm{H}_2}}{\mathrm{M}_{\mathrm{H}_2}}} \\
\frac{v_{\mathrm{O}_2}}{v_{\mathrm{H}_2}} & =\sqrt{\frac{\mathrm{T}_{\mathrm{O}_2}}{\mathrm{~T}_{\mathrm{H}_2}} \cdot \frac{\mathrm{M}_{\mathrm{H}_2}}{\mathrm{M}_{\mathrm{O}_2}}} \\
v_{\mathrm{O}_2} & =\left(\sqrt{\frac{1200}{300} \times \frac{2}{32}}\right) \times 1930=\frac{1930}{2} \\
v_{\mathrm{O}_2} & =965 \mathrm{~ms}^{-1}
\end{aligned}
$
Question 2.
The rms velocity of the molecules in a sample of helium is $5 / 7$ times that of molecules in a sample of hydrogen. If the temperature of hydrogen is $0^{\circ} \mathrm{C}$. Then, what is the temperature of helium?
Answer:
From RMS speed, $\sqrt{\frac{3 \mathrm{RT}_{\mathrm{He}}}{\mathrm{M}_{\mathrm{He}}}}=\frac{5}{7}\left(\sqrt{\frac{3 \mathrm{RT}_{\mathrm{H}_2}}{\mathrm{M}_{\mathrm{H}_2}}}\right)$

$
\begin{aligned}
\therefore \quad \mathrm{T}_{\mathrm{He}} & =\frac{25}{49}\left(\frac{\mathrm{M}_{\mathrm{He}}}{\mathrm{M}_{\mathrm{H}_2}}\right) \mathrm{T}_{\mathrm{H}_2}=\frac{25}{49} \times \frac{4}{2} \times 273 \approx 273 \mathrm{~K} \\
\mathrm{~T}_{\mathrm{He}} & =0^{\circ} \mathrm{C}
\end{aligned}
$
Question 3.
A cylinder of fixed capacity 44.8 litres contain helium gas at standard pressure and temperature. What is the amount of heat needed to raise the temperature of the gas by $15^{\circ} \mathrm{C} ?\left(\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1}\right.$ $\mathrm{K}^{-1}$ )
Answer:
Volume of 1 mole of He at STP $=22.4$ litres
Total volume of $\mathrm{He}$ at $\mathrm{STP}=44.8$ litres
$
\therefore \quad \text { No. of moles of } \mathrm{He}, n=\frac{44.8}{22.4}=2
$
Molar specific heat of $\mathrm{He}$ (monoatomic gas) at constant volume,
$
\begin{aligned}
\mathrm{C}_{\mathrm{V}} & =\frac{3}{2} \mathrm{R}=\frac{3}{2} \times 8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\
\Delta \mathrm{T} & =15^{\circ} \mathrm{C} \\
\mathrm{Q} & =n \mathrm{C}_{\mathrm{V}} \Delta \mathrm{T}=2 \times \frac{3}{2} \times 8.31 \times 15 \\
\mathrm{Q} & =373.95 \mathrm{~J}
\end{aligned}
$

Question 4.
An insulated container containing monoatomic gas of molar mass $\mathrm{m}$ is moving with a velocity $\mathrm{v}_0$. If the container is suddenly stopped. Find the change in temperature.
Answer:
Suppose the container has $n$ moles of the monoatomic gas. Then the loss in K.E. of the gas
$
\Delta \mathrm{E}=\frac{1}{2}(m n) v_o^2
$
If the temperature of the gas changes by $\Delta \mathrm{T}$, then heat gained by the gas,
Now,
$
\begin{aligned}
\Delta \mathrm{Q} & =n \mathrm{C}_{\mathrm{V}} \Delta \mathrm{T} \\
\Delta \mathrm{Q} & =\frac{3}{2} n \mathrm{R} \Delta \mathrm{T} \\
\Delta \mathrm{Q} & =\Delta \mathrm{E} \\
\frac{3}{2} n \mathrm{R} \Delta \mathrm{T} & =\frac{1}{2}(m n) v_o^2 \\
\Delta \mathrm{T} & =\frac{m v_o^2}{3 \mathrm{R}}
\end{aligned}
$
Question 5.
Estimate the total number of molecules inclusive of oxygen, nitrogen, water vapour and other constituents in a room of capacity $25.0 \mathrm{~m}^3$ at a temperature of $27^{\circ} \mathrm{C}$ and 1 atmospheric pressure. $\left(\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{JK}^{-1}\right)$
Solution:
At Boltzmann's constant, $k_{\mathrm{B}}=\frac{\mathrm{R}}{\mathrm{N}} ; \mathrm{R}=k_{\mathrm{B}} \mathrm{N}$
Now, $\quad \mathrm{PV}=n \mathrm{RT}=n k_{\mathrm{B}} \mathrm{NT}$
$\therefore$ The number of molecules in the room, $n \mathrm{~N}=\frac{\mathrm{PV}}{k_{\mathrm{B}} \mathrm{T}}=\frac{1.013 \times 10^5 \times 25}{300 \times 1.38 \times 10^{-23}}$
The number of molecules in the room $=6.117 \times 10^{26}$

Question 6.
Estimate the average energy of a helium atom at
(i) room temperature $\left(27^{\circ} \mathrm{C}\right)$
(ii) the temperature on the surface of the sun $(6000 \mathrm{~K})$ and
(iii) the temperature of $10^7 \mathrm{~K} .\left(\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{JK}^{-1}\right)$
Answer:
Average K.E. per molecule is $\overline{\mathrm{E}}=\frac{3}{2} k_{\mathrm{B}} \mathrm{T}$
(i) $\mathrm{T}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$
$
\overline{\mathrm{E}}=\frac{3}{2} \times 1.38 \times 10^{-23} \times 300=6.21 \times 10^{-21} \mathrm{~J}
$
(ii) $\mathrm{T}=6000 \mathrm{~K}$
$
\overline{\mathrm{E}}=\frac{3}{2} \times 1.38 \times 10^{-23} \times 6000=1.242 \times 10^{-19} \mathrm{~J}
$
(iii) $\mathrm{T}=10^7 \mathrm{~K}$
$
\overline{\mathrm{E}}=\frac{3}{2} \times 1.38 \times 10^{-23} \times 10^7=2.07 \times 10^{-16} \mathrm{~J}
$
Question 7.
The molecules of a given mass of a gas have rms velocity of $200 \mathrm{~ms} 1$ at $27^{\circ} \mathrm{C}$ and $1.0 \times 10 \mathrm{~s} \mathrm{Nm}^{-2}$ pressure. When the temperature and pressure of the gas are respectively. $127^{\circ} \mathrm{C}$ and $0.05 \times 10 \mathrm{~s}$ $\mathrm{Nm}^{-2}$.
Find the rms velocity of its molecules in $\mathrm{ms}^{-1}$
Answer:
$
\begin{aligned}
& \mathrm{T}_1^*=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}, \mathrm{~T}_2=127^{\circ} \mathrm{C}+273=400 \mathrm{~K} \\
& r m s \text { speed of molecules is } v_{r m s}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}
\end{aligned}
$
So it depends only on temperature,
$
\begin{aligned}
& v_{r m s} \propto \sqrt{\mathrm{T}} \\
& \frac{v_1}{v_2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}} \Rightarrow \frac{200}{\mathrm{~V}_2}=\sqrt{\frac{300}{400}} ; \frac{200}{v_2}=\frac{\sqrt{3}}{2} \\
& v_2=\frac{400}{\sqrt{3}} \\
& v_2=231 \mathrm{~ms}^{-1}
\end{aligned}
$

Question 8 .
At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Mass of oxygen molecule $(\mathrm{m})=2.76 \times 10^{-26} \mathrm{~kg}$ Boltzmann's constant $\left.\left(\mathrm{k}_{\mathrm{B}}\right)=1.38 \times 10^{-23} \mathrm{JK}^{-1}\right)$
Answer:
$
v_{\text {escape }}=11200 \mathrm{~ms}^{-1}
$
Say at temperature $\mathrm{T}$ it attains $v_{\text {escape }}$
$
\begin{aligned}
v_{\text {escape }} & =\sqrt{\frac{3 k_{\mathrm{B}} \mathrm{T}}{m_{\mathrm{O}_2}}} \\
\mathrm{~T} & =\frac{\left(v_{\text {escape }}\right)^2 \times m_{0_2}}{3 k_{\mathrm{B}}}=\frac{(11200)^2 \times 2.76 \times 10^{-26}}{3 \times 1.38 \times 10^{-23}} \\
& =\frac{3.46 \times 10^8 \times 10^{-26}}{4.14 \times 10^{-23}}=0.8357 \times 10^5 \\
\mathrm{~T} & =8.357 \times 10^4 \mathrm{~K}
\end{aligned}
$
Question 9.
The temperature of a gas is raised from $27^{\circ} \mathrm{C}$ to $927^{\circ} \mathrm{C}$. What is the rms molecular speed?
Answer:
$
\begin{aligned}
& v_{r m s} \propto \sqrt{\mathrm{T}} \\
& \frac{v_{r m s} \text { at }\left(927^{\circ} \mathrm{C}\right)}{v_{r m s} \text { at }\left(27^{\circ} \mathrm{C}\right)}=\sqrt{\frac{927+273}{27+273}}=\sqrt{\frac{1200}{300}}=2 \\
& \therefore \quad v_{r m s} \text { at }\left(927^{\circ} \mathrm{C}\right)=2 v_{r m s} \text { at }\left(27^{\circ} \mathrm{C}\right)
\end{aligned}
$
Question 10.
A gaseous mixture consists of $16 \mathrm{~g}$ of helium and $16 \mathrm{~g}$ of oxygen. Find the ratio $\frac{C_P}{C_V}$ of the mixture.

Answer:
Number of moles of helium $(n)=\frac{16}{4}=4$; Number of moles of oxygen $\left(n^{\prime}\right)=\frac{16}{32}=\frac{1}{2}$ For the monoatomic helium gas, degrees of freedom, $f=3$
So
$
\mathrm{C}_{\mathrm{V}}=\frac{f}{2} \mathrm{R}=\frac{3}{2} \mathrm{R}
$
For the diatomic oxygen gas, $f=5$
$
\begin{gathered}
\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R} \\
\mathrm{C}_{\mathrm{V}}(\text { mixture })=\frac{n \mathrm{C}_{\mathrm{V}}+n^{\prime} \mathrm{C}_{\mathrm{V}}}{n+n^{\prime}}=\frac{4 \times \frac{3}{2} \mathrm{R}+\frac{1}{2} \times \frac{5}{2} \mathrm{R}}{\left(4+\frac{1}{2}\right)}=\frac{29}{18} \mathrm{R} \\
\gamma(\text { mixture })=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1+\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}(\text { mixture })}=1+\frac{\mathrm{R}}{\left(\frac{29}{18} \mathrm{R}\right)}=1+\frac{18}{29}=\frac{47}{29} \\
\gamma(\text { mixture })=1.62
\end{gathered}
$