Thermodynamics Practice Questions

For complete cyclic process $\Delta{U}=0$ $\therefore$ from $\Delta{Q}=\Delta {U}+{W}$ $=0+{W}$ $\Delta{Q}={W}$ $=$ Area $=\pi {r}_{1} \.{r}_{2}$ $=\pi \\times (10 \\times 10^{3}) \\times (10 \\times 10^{-3})$ $\Delta{Q}=100 \pi$

${PV}={nRT}$ ${PV} \propto{T}$ Straight line with positive slope (nR)

Given that, $r_{1}$ and $r_{2}$ be the radii of inner and outer shells. A material of thermal conductivity (K) is filled between region. Consider an elementary sphere of radius r and thickness dr. Now, thermal resistance of elementary sphere $\text{dR}=\frac{\text{dr}}{\text{K} 4 \pi r^{2}}$ $[\because inner surface area,$\text{A}=4 \pi r^{2}$]$ So, total thermal resistance, $R=\int \text{dR}=\int ^{r_2}_{r_1} \frac{d r}{4 \pi K r^{2}}=-\frac{1}{4 \pi \text{K}}[\frac{1}{r}]_{r_{1}}^{r_{2}}$ $\text{R}=\frac{r_{2}-r_{1}}{4 \pi \text{K} r_{1} r_{2}}$ Hence, heat current (or the rate of flow of heat) $\frac{\text{dQ}}{d t}=\frac{\Delta\theta }{\text{R}}={(\theta _{2}-\theta _{1})}/{\frac{(r_{2}-r_{1})}{4 \pi \text{K} r_{1} r_{2}}}=\frac{4 \pi \text{K} r_{1} r_{2}(\theta _{2}-\theta _{1})}{r_{2}-r_{1}}$

The expression of the root mean square velocity of the gas molecules, $V_{ {ms}}=\sqrt{\;\frac{3 R T}{M}}$ The expression of the average velocity of the gas molecules, $v_{a w} \;\; =\sqrt{\;\frac{8 R T}{\pi M}}$ $\;{v_{ {rms}}}/{v_{ {av}}} \;\; =\;{\sqrt{\frac{3 R T}{M}}}/{\sqrt{\;\frac{8 R T}{\pi M}}}=\sqrt{\;\frac{3 \pi }{8}}$

Let $p_{i}, p_{f}, V_{i}$ and $V_{f}$ be the initial and final pressure and volume. Given, $A B$ is isothermal $(\Delta T=0)$, $B C$ is isochoric $(\Delta V=0)$ and $C A$ is adiabatic $(\Delta Q=0)$ Since, isothermal work $(W_{A B})=p_{1} V_{1} \ln \;\frac{V_{f}}{V_{i}}$where, $V_{i}$ and $V_{f}$ are volume at $A$ and $B$, respectively. $\therefore \;\; W_{A B}=p_{1} V_{1} \ln \;\frac{2 V_{1}}{V_{1}}=p_{1} V_{1} \ln 2$ Since, at constant volume, work done is zero. $\therefore \;\; W_{B C}=0$ Since, $W_{C A}$ is an adiabatic work done, i.e. $W_{C A} \;\; =\;\frac{1}{1-\gamma }(p_{f} V_{f}-p_{i} V_{i})$ $\Rightarrow \;\; W_{C A} \;\; =\;\frac{1}{1-\gamma }(p_{1} V_{1}-\;\frac{p_{1}}{4} \times 2 V_{1})$ $\;\; =\;\frac{1}{1-\gamma }(p_{1} V_{1}-p_{1} V_{1} / 2)=\;\frac{1}{1-\gamma } \;\frac{p_{1} V_{1}}{2}$ $\therefore$ Net work done, $W_{\;\text{net }\;}=W_{A B}+W_{B C}+W_{C A}$ $\;\; =p_{1} V_{1} \ln 2+0+\;\frac{1}{1-\gamma } \;\frac{p_{1} V_{1}}{2}$ $\;\; =p_{1} V_{1}[\ln 2+1 / 2(1-\gamma )]$ From ideal gas law, $p V=n R T$ $\therefore W_{\;\text{net }\;}=R T[\ln 2-1 / 2(\gamma -1)]$ $( \because n=1)$

Elastic energy $=\frac{Y}{2}(\text{ strain })^{2} \\times$ Area $\\times$ length \Rightarrow Elastic energy per unit length $=\frac{Y}{2}(\text{ strain})^{2} \\times$ Area $($ strain $=\frac{\Deltal}{l}=\alpha \Delta{T}=10^{-5} \\times 10=10^{-4})$ $=\frac{10^{11}}{2} \\times (10^{-4})^{2} \\times 10^{-2}=5 {J} / {m}$

In graph (A) vertical line is isochoric process, so the graph (A) is incorrect. In graph (B) horizontal line is isobaric process, so the graph (B) is incorrect. In graph (C) and (D) vertical line is isothermal process and the curve line is adiabatic process. So, the correct representation of thermodynamic process graph is (C) and (D). So, the correct option is (b)

Given, initial load, $\text{M}_{1}=185 {\text{kg}}$ Initial pressure, $\text{p}_{1}=76 {\text{cm}}$ of Hg Initial temperature, $\text{T}_{1}=27° {\text{C}}=(273+27) {\text{K}}=300 {\text{K}}$ Final pressure at height, $\text{p}_{2}=45 {\text{cm}}$ of Hg Final temperature, $\text{T}_{2}=-7° {\text{C}}=[273+(-7)] {\text{K}}=266 {\text{K}}$ Volume of balloon is constant, $\text{V}_{1}=\text{V}_{2}=\text{V}$ We know that, for air, $\text{p}=\rho \text{RT}$ Now, for initial condition, we can write $\rho _{1}=\rho _{1} \text{RT}_{1}$ ...(i) Similarly, for final condition, we can write $\text{P}_{2}=\rho _{2} \text{RT}_{2}$ ...(ii) Dividing Eq. (i) by Eq. (ii), we get $\frac{\text{p}_{1}}{\text{p}_{2}}=\frac{\rho _{1} \text{T}_{1}}{\rho _{2} \text{T}_{2}}$ Substituting the given values in above expression, we get ${76 {\text{cm}} { \text{of} } {\text{Hg}}}/{45 {\text{cm}} { \text{of} } {\text{Hg}}}=\frac{\rho _{1}}{\rho _{2}}({300 {\text{K}}}/{266 {\text{K}}})$ $\Rightarrow \frac{\rho _{1}}{\rho _{2}}=\frac{76 \times 266}{45 \times 300}$ We know that, density of balloon can be written as $\rho \frac{\text{M}}{\text{V}}$ As, volume is constant ρ ∝ M We can write, $\frac{\rho _{1}}{\rho _{2}}=\frac{\text{M}_{1}}{\text{M}_{2}}=\frac{76 \times 266}{45 \times 300}$ $\Rightarrow \frac{\text{M}_{1}}{\text{M}_{2}}=\frac{20216}{13500}$ Substituting the value of $\text{M}_{1}$ in above expression, we get $\text{M}_{2}=185 \times \frac{13500}{20216}=123.54 {\text{kg}}$ Thus, the load carried by balloon will be 123.54 kg.

$\gamma =1+\frac{2}{{f}}$ ${f}=\frac{2}{\gamma -1}$

$\Delta{Q}=\Delta{U}+\Delta{W}$ Here $\Delta{W}=0$ $\DeltaQ=\DeltaU=n C_{V} \DeltaT$ $\Delta{Q}=4 \\times \frac{5 {R}}{2}(50)=500 {R}$ Hence option (4).

Given, temperature of hot reservoir $=T_{1}$ Temperature of cold reservoir, $T_{2}=400 {K}$Heat of hot reservoir, $Q_{1}=300 {J}$Heat of cold reservoir, $Q_{2}=240 {J}$Since, efficiency $(η)=1-\;\frac{T_{2}}{T_{1}}=\;\frac{Q_{1}-Q_{2}}{Q_{1}}$ $1-\;\frac{T_{2}}{T_{1}}=1-\;\frac{Q_{2}}{Q_{1}}$ $\Rightarrow \;\; 1-\;\frac{400}{T_{1}}=1-\;\frac{240}{300}$ $\Rightarrow \;\; \;\frac{400}{T_{1}}=\;\frac{4}{5} \Rightarrow T_{1}=500 {K}$

Given, $C_{p}=7 / 2 R, C_{V}=5 / 2 R$ Since, change in internal energy $(d U)=n C_{V} d T$ Heat change $(d Q)=n C_{p} d T$ Work done $(d W)=n R d T$ $\therefore \;\; d U: d Q: d W \;\; =n C_{V} d T: n C_{p} d T: n R d T$ $\;\; =C_{V}: C_{p}: R=\;\frac{5}{2} R: \;\frac{7}{2} R: R$ $\;\; =5 R: 7 R: 2 R$ $\;\; =5: 7: 2$

From ideal gas equation, $p V=(n_{1}+n_{2}+n_{3}) R T$ where, $n_{1}=$ number of moles of oxygen $=\;\frac{16}{32}$ $n_{2}=$ number of moles of nitrogen $=\;\frac{28}{28}$ $n_{3}=$ number of moles of carbon dioxide $=\;\frac{44}{44}$ $\Rightarrow \;\; p V=[\;\frac{16}{32}+\;\frac{28}{28}+\;\frac{44}{44}] R T=[\;\frac{1}{2}+1+1]_{R T}$ $=\;\frac{5}{2} R T \Rightarrow p=\;\frac{5}{2} \;\frac{R T}{V}$ This is the required pressure of the mixture of the gases.

Since, each vibrational mode, corresponds to two degrees of freedom, hence, f = 3 (trans.) + 3(rot.) + 8 (vib.) = 14 & $\gamma =1+\frac{2}{{f}}$ $\gamma =1+\frac{2}{14}=\frac{8}{7}$ ${W}=\frac{{nR} \Delta{T}}{\gamma -1}=-582$ As ${W} < 0$. work is done on the gas.

Given, Volume, $\text{V}=500 {\text{cm}}^{3}=5 \times 10^{2} \times 10^{-6} {\text{m}}^{3}$ $=5 \times 10^{-4} {\text{m}}^{3}$ Pressure, $p=400 {\text{kPa}}=4 \times 10^{5} {\text{Pa}}$ Temperature, T = 300 K Total mass, m = 0.76 g We know that, Molar mass of hydrogen, $\text{M}_1$= 2 g Molar mass of oxygen, $\text{M}_2$ = 32 g Let $n_1$ and $n_2$ be the number of moles of hydrogen and oxygen, respectively. Total number of moles of mixture, $n=n_{1}+n_{2}$ By using ideal gas equation, pV = nRT Substituting the values, we get $4 \times 10^{5} \times 5 \times 10^{-4}=n \times 8.314 \times 300$ n= 0.08 mol Then,$n_{1}+n_{2}=0.08$ Let mass of hydrogen in mixture = $m_1$ and mass of oxygen in oxygen = $m_2$ $n_{1} \text{M}_{1}+n_{2} \text{M}_{2}=m$ $[\because n=\frac{m}{\text{M}} \Rightarrow m=n \text{M}]$ Substituting the values, we get $2 n_{1}+32 n_{2}=0.76$ $n_{1}+16 n_{2}=0.38$ ...(ii) Solving Eqs. (i) and (ii), we get $n_{1}=0.06$ and $n_{2}=0.02$ ∴ Ratio of masses, $\frac{m_{2}}{m_{1}}=\frac{n_{2} \text{M}_{2}}{n_{1} \text{M}_{1}}={0.02 \times 32}{0.06 \times 2}=\frac{16}{3}$ i.e. $m_{2}: m_{1}=16: 3$

As per the question, coefficient of expansion of metal $A$ is higher than that of $B$. It means, $\alpha _{A}>\alpha _{B}$ If the bimetallic strip is placed in a cold bath, then the length of both strips will decrease. i.e., $I_{A}>I_{B}$ $\therefore$ The bimetallic strip will bend towards left. It can be shown as below.

Let, $Q=$ heat current, $k=$ thermal conductivity, A = area, $I=$ length of capacitor and $\Delta \theta =$ change in temperature . $\therefore \;\; Q=\;\frac{k A \Delta \theta }{l}=\;\frac{\Delta \theta }{R} \;\; ( \because R=\;\frac{l}{k A})$ $\Rightarrow \;\; \;\frac{\theta _{2}-\theta }{R_{2}}=\;\frac{\theta -\theta _{1}}{R_{1}}$ $\Rightarrow \;R_{1} \theta _{2}-R_{1} \theta \;\; =R_{2} \theta -R_{2} \theta _{1}$ $\Rightarrow \;\theta \;\; =\;\frac{R_{1} \theta _{2}+R_{2} \theta _{1}}{R_{1}+R_{2}}$

Thermal stress is defined as the stress, experienced by any rod on heating between two fixed rigid supports. On heating, the size of the rod increases but, if the two ends are free, rod will not experience any stress. i.e, there is no thermal stress will be produced in it. Hence, option (a) is the correct.