Thermodynamics Practice Questions

Given, $\text{pT}^{3}= \text{constant}$ ...(i) From ideal gas equation, we have $\text{pV}=\text{nRT}$ $\Rightarrow \text{p}=\frac{\text{nRT}}{\text{V}}$ From Eq. (i), we get $\therefore (\frac{\text{nRT}}{\text{V}}) \text{T}^{3}= \text{constant}$ $\Rightarrow \text{nRT}^{4} \text{V}^{-1}= \text{constant}$ Differentiate above expression, we get \text{nR}\[4 \text{T}^{3} \text{V}^{-1} \text{dT}-\frac{\text{T}^{4}}{\text{V}^{2}} \text{dV}]$=0$$\Rightarrow 4 \text{dT}=\frac{\text{T}}{\text{V}} \text{dV}$...(ii) We know that,$\text{dV}=\gamma \text{VdT}$Here, γ = coefficient of volume expansion.$\Rightarrow \gamma =\frac{\text{dV}}{\text{VdT}}$...(iii) From Eq. (ii),$\frac{4}{\text{T}}=\frac{\text{dV}}{\text{VdT}}$...(iv) On comparing Eqs. (iii) and (iv), we get$\gamma =\frac{4}{\text{T}}$Thus, the coefficient of volume expansion is$4 /\text{T}$.

Given, Change in pressure with volume $\frac{\text{dp}}{\text{dV}}=-\text{ap}$ At $\text{V}=0, \text{p}=\text{p}_{0}$ Number of mole n = 1 Gas constant, $\text{R}=8.314 {\text{J}} {\text{K}}^{-1}$ From given equation, $\frac{\text{dp}}{\text{p}}=-\text{adV}$ $\int ^{\text{p}}_{\text{p}_0} \frac{\text{dp}}{\text{p}}=-a \int ^{v}_{0} \text{dV}$ $\Rightarrow [\ln \text{p}]_{\text{p}_{0}}^{\text{p}}=-a[\text{V}]_{0}^{\text{V}}$ $\Rightarrow \ln (\text{p} / \text{p}_{0})=-a(\text{V}-0)$ Taking anti log on both side, $\Rightarrow \text{p} / \text{p}_{0}={e}^{-{\text{aV}}}$ $\Rightarrow \text{p}=\text{p}_{0} {e}^{-{\text{aV}}}$ By using ideal gas law, pV = nRT $\Rightarrow (\text{p}_{0} {e}^{-{\text{aV}}}) \text{V}=\text{nRT}$ $\Rightarrow \frac{1}{\text{nR}}(\text{p}_{0} {e}^{-{\text{aV}}}) \text{V}=\text{T}$ $\Rightarrow \frac{1}{\text{R}}(\text{p}_{0} {e}^{-{\text{aV}}}) \text{V}=\text{T}$ [\because n = 1]...(i) For maximum temperature $(\text{T}_{\max })$, On differentiating both side with respect to V, we get \Rightarrow \frac{1}{n \text{R}}\[\text{p}{0} .{e}^{-a \text{V}} .(-a \text{V}) \text{V}+\text{p}{0} {e}^{-a \text{V}}]$=\frac{d \text{T}}{d \text{V}}=0$$\Rightarrow \text{p}{0} e^{-a \text{V}}(1-a \text{V}^{2})=0$$\Rightarrow 1=a \text{V}^{2}$$\Rightarrow \text{V}^{2}=1 / a$$\Rightarrow \text{V}=1 / a$Substituting the value in Eq. (i), we get$\text{T}=\frac{1}{\text{R}}(\text{p}{0} {e}^{-a .1 / a}) 1 / {a}=\frac{\text{p}{0} {e}^{-1}}{\text{R}}(1 / {a})=\frac{\text{p}{0}}{{e} \text{R} {a}}$

Given, thermal resistance of rod CD, $\text{R}_{\text{CD}}=10.0 {\text{kW}}^{-1}$ Temperature of end $\text{A}, \text{T}_{\text{A}}=200° {\text{C}}$ Temperature of end $\text{B}, \text{T}_{\text{B}}=100° {\text{C}}$ Temperature of end $\text{D}, \text{T}_{\text{D}}=125° {\text{C}}$ As rods are given as identical, then $\text{R}_{\text{CD}}=\text{R}_{\text{CB}}=10 {\text{kW}}^{-1}$ As, C is mid-point of AB, and as we know R ∝ length $\text{R}_{\text{AC}}=\text{R}_{\text{BC}}=5 {\text{kW}}^{-1}$ Now, we know that heat current is given as $\frac{\text{Q}}{t}=\frac{\Delta\text{T}}{\text{R}}$ Here, ∆T is temperature difference and R is thermal resistance. Now, for point C, consider temperature of point C as T. Consider at point C heat is coming in from A and going out at point B and D. Therefore, we can write, ${\text{T}_{\text{A}}-\text{T}}/{\text{R}_{\text{AC}}}={\text{T}-\text{T}_{\text{D}}}/{\text{R}_{\text{C} \text{D}}}={\text{T}-\text{T}_{\text{B}}}/{\text{R}_{\text{BC}}}$ $\Rightarrow {200-\text{T}}/{\text{R}_{\text{AC}}}={\text{T}-125}/{\text{R}_{\text{CD}}}+{\text{T}-100}/{\text{R}_{\text{BC}}}$ $\Rightarrow \frac{200-\text{T}}{5}=\frac{\text{T}-125}{10}, \frac{\text{T}-100}{5}$ $\Rightarrow 400-2 \text{T}=\text{T}-125+2 \text{T}-200$ $5 \text{T}=725° {\text{C}}$ $\text{T}=145° {\text{C}}$ The heat current in CD is $\text{I}_{\text{CD}}={\text{T}-\text{T}_{\text{D}}}/{\text{R}_{\text{CD}}}$ $=\frac{145-125}{10}=\frac{20}{10}=2 {\text{W}}$ Thus, the value of P is 2.

The rms speed of the molecule of gas is given by $v_{{\text{rms}}}=\\sqrt{\frac{3 \text{RT}}{\text{M}}}$ According to the, question T is constant. Moreover R is also a constant. So, we get $v_{{\text{rms}}} \propto{1}/{\\sqrt{\text{M}}}$ $(\text{M}_{{\text{H}}}) \text{M}_{{\text{hydrogen} }}=1.00784 {\text{u}} \approx 1 {\text{u}}$ $(\text{M}_{{\text{O}}}) \text{M}_{{\text{oxygen} }}=15.999 {\text{u}} \approx 16 {\text{u}}$ $(\text{M}_{{\text{CO}}_{2}}) \text{M}_{{\text{carbon} \text{dioxide} }}=44 {\text{u}}$ $(v_{{\text{rms}}})_{{\text{hydrogen} }} > (v_{{\text{rms}}})_{{\text{oxygen} }} > (v_{{\text{rms}}})_{{\text{carbon} \text{dioxide} }}$ $\Rightarrow v_{{\text{H}}} > v_{{\text{O}}} > v_{{\text{CO}}_{2}}$

Given masses of three liquids are equal, i.e. $m_{x}=m_{y}=m_{z}=m$ Temperature of mixture of liquids x and $y, \text{T}_{1}=16° {\text{C}}$ Temperature of mixture of liquids y and $z, \text{T}_{2}=26° {\text{C}}$ While mixing the liquids x and y, the heat energy lost by liquid y will be equal to the heat energy absorbed by liquid x. Consider the specific heat capacity of liquid x is $s_{x}$, of liquid y is $s_{y}$ and of liquid z is $s_{z}$. Now, for liquid x and y, using principle of calorimetry, Heat gained by liquid x = Heat lost by liquid y $m s_{x}(\text{T}_{1}-10)=m s_{y}(20-\text{T}_{1})$ $\Rightarrow s_{x}(16-10)=s_{y}(20-16)$ $\Rightarrow \frac{s_{x}}{s_{y}}=\frac{2}{3}$ ...(i) Similarly, for liquid y and z, using principle of calorimetry, Heat gained by liquid y = Heat lost by liquid z $m s_{y}(\text{T}_{2}-20)=m s_{z}(30-\text{T}_{2})$ $\Rightarrow s_{y}(26-20)=s_{z}(30-26)$ ...(ii) $\Rightarrow \frac{s_{y}}{s_{z}}=\frac{2}{3}$...(i) Multiply Eq. (i) by Eq. (ii), we get $\frac{s_{x}}{s_{z}}=\frac{4}{9}$ Consider the mixing of liquids x and z, let the temperature of mixture be $\text{T}_{3}$. Using principle of calorimetry for liquids x and z. Heat gained by liquid x = Heat lost by liquid z $m s_{x}(\text{T}_{3}-10)=m s_{z}(30-\text{T}_{3})$ $\Rightarrow s_{x}(\text{T}_{3}-10)=s_{z}(30-\text{T}_{3})$ $\Rightarrow \frac{s_{x}}{s_{z}}=\frac{30-\text{T}_{3}}{\text{T}_{3}-10}$ $\Rightarrow \frac{4}{9}=\frac{30-\text{T}_{3}}{\text{T}_{3}-10}$ $\Rightarrow 4 \text{T}_{3}-40=270-9 \text{T}_{3}$ $\Rightarrow \text{T}_{3}=23.84° {\text{C}}$

Given, rms speed of oxygen( $v_{\;\text{rms }\;})_{ {O}_{2}}=160$ ${m} / {s}$Let rms speed of hydrogen $=(v_{r m s})_{H_{2}}$Since, $\;\; v_{r m s}=\sqrt{\;\frac{3 R T}{M}}$where, $R=$ universal gas constant,$T=$ temperatureand $\;\; M=$ molecular mass.$\therefore \;\; \;{(v_{ {rms}})_{ {O}_{2}}}/{(v_{ {rms}})_{ {H}_{2}}}=\sqrt{\;{M_{ {H}_{2}}}/{M_{ {O}_{2}}}}$ $\Rightarrow \;\; \;{160}/{(v_{ {rms}})_{ {H}_{2}}}=\sqrt{\;\frac{1}{16}}$ $\Rightarrow \;\; \;{160}/{(v_{r m s})_{H_{2}}}=\;\frac{1}{4}$ $\Rightarrow \;\; (v_{r m s})_{H_{2}}=160 \times 4$ $=640 {ms}^{-1}$

On the basis of kinetic theory of gases, the gas exerts pressure because its molecules contain uniform speed, random motion and perform elastic collision with each other, as well as with the walls of container. As a result of which gaseous molecules suffer change in momentum when impinge on the walls of container.

Two identical metal wires of thermal conductivities $K_{1}$ and $K_{2}$ respectively connected in series are represented as followsThe above figure can also be represented as Therefore, the effective thermal conductivity of the combination will be given by $R_{\;\text{eff }\;}=\;\frac{I}{K_{1} A}+\;\frac{I}{K_{2} A}=\;\frac{2 l}{K_{e q} A}$ $\Rightarrow \;\; \;{2 I}/{K_{ {eq}} A}=\;\frac{1}{A}(\;\frac{1}{K_{1}}+\;\frac{1}{K_{2}}) \Rightarrow \;{2 I}/{K_{ {eq}} A}=\;\frac{I}{A}(\;\frac{K_{1}+K_{2}}{K_{1} K_{2}})$ $\Rightarrow \;\; \;{2}/{K_{ {eq}}}=\;{K_{1}+K_{2}}/{K_{ {l}} K_{2}} \Rightarrow K_{ {eq}}=\;\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}$

Given, ideal gas equation, $U=3 p V+4$ Here, $U=$ internal energy, $p=$ pressure and $V=$ volume. By ideal gas equation, $p V=n R \Delta T$ $=R \Delta T$ $( \because n=1) . . .( {i})$ $\Delta U \;\; =n C_{v} \Delta T=3 p V+4$ $\;\; =n \;\frac{f}{2} R \Delta T=3 p V+4$ $\Rightarrow \;f R \Delta T \;\; =6 p V+8$ $\Rightarrow \;f \cdot p V \;\; =6 p V+8$ [From Eq. (i)] $\Rightarrow \;f \;\; =6+\;\frac{8}{p V}$ [From Eq. (i)] As, degree of freedom is more than 6 . $\therefore$ Gas is polyatomic.

${C}_{{P}}-{C}_{{V}}={R}$ for ideal gas and gas behaves as ideal gas at high temperature so ${T}_{{P}}>{T}_{{Q}}$

We know that, in isothermal process, $\Delta T=0$ In isochoric process, $\Delta V=0$ In adiabatic process, $\Delta Q=0$ In isobaric process, $\Delta p=0$ So, the correct match is, ${A} \to 2, {B} \to 3, {C} \to 4, {D} \to 1$.

Given, pressure $(p) \propto k V^{3}$ $T_{1} \;\; =100^{\circ} {C}, T_{2}=300^{\circ} {C}$ $\Delta T \;\; =T_{2}-T_{1}=300-100=200^{\circ} {C}$ $\Rightarrow$ By using ideal gas equation, $p V=n R T$ $k V^{3} \cdot V=n R T \Rightarrow k V^{4}=n R T$ On differentiating both sides w.r.t temperature, we get $4 k V^{3} \;\frac{d V}{d T} \;\; =n R$ $\Rightarrow \;\; 4 k V^{3} d V \;\; =n R d T \Rightarrow k V^{3} d V=n R d T / 4$ $\Rightarrow \;\; p d V \;\; =n R d T / 4$ $\;\text{ As, work done }\;(W) \;\; =p d V=n R d T / 4$ $\;\; =\;\frac{n R}{4} \Delta T=\;\frac{n R}{4} \times 200=50 n R$

Process of isothermal ${W}={nRT} l{n}(\frac{{V}_{2}}{{V}_{1}})$ $=1 \\times 8.3 \\times 300 \\times \ln 2$ $=17258 \\times 10^{-1} {J}$

Given, Pressure of the gas, $p=1.01 \times 10^{5} {Pa}$ Absolute temperature of gas, $T=27^{\circ} {C}$ $=27+273=300 {K}$ Molecular diameter, $d=0.3 {nm}=0.3 \times 10^{-9} {m}$ Boltzmann constant, $k_{B}=1.38 \times 10^{-23} {J} / {K}$ Mean free path, $\lambda =\;\frac{k_{B} T}{\sqrt{2} \pi d^{2} p}$....(i) Substituting the given values in Eq. (i), we get $\lambda =\;\frac{1.38 \times 10^{-23} \times 300}{\sqrt{2} \times 3.14 \times (0.3 \times 10^{-9})^{2} \times 1.01 \times 10^{5}}$ $=102 {nm}$

Efficiency of Carnot heat engine is given by $η=1-\;\frac{T_{2}}{T_{1}}$ $=1-\;\frac{Q_{2}}{Q_{1}}=\;\frac{W}{Q_{1}}$.....(i) where. $W=$ net work done by the gas $Q_{1}=$ heat absorbed by the gas, $Q_{2}=$ heat released by the gas, $T_{1}=$ temperature of hot reservoir, and $T_{2}=$ temperature of cold reservoir. Using Eq. (i), we can write $1-\;\frac{T_{2}}{T_{1}}=\;\frac{W}{Q_{1}} \Rightarrow \;\frac{T_{2}}{T_{1}}=1-\;\frac{W}{Q_{1}}$ $\Rightarrow \;\; \;\frac{400}{800}=1-\;\frac{W}{Q_{1}} \Rightarrow \;\frac{W}{Q_{1}}=1-\;\frac{400}{800}$ $\Rightarrow \;\; \;\frac{W}{Q_{1}}=1-\;\frac{1}{2}=\;\frac{1}{2} \Rightarrow \;\frac{W}{Q_{1}}=\;\frac{1}{2}$ $\Rightarrow \;\; Q_{1}=2 \times W=2 \times 1200[ \because W_{1}=1200 J]$ $=2400 {J}$

${S}=\alpha ^{2} \beta \ln (\frac{\mu{KR}}{{J} \beta ^{2}}+3)$ ${S}=\frac{{Q}}{{T}}=$ joulek $/ {k}$ $[\alpha ^{2} \beta ]=$ Joule $/ {k}$ {PV}={nRT} \;\;\;\[\frac{\mu{KR}}{{J} \beta ^{2}}]$=1$${R}=\frac{\text{Joule }}{{K}}$$\Rightarrow{R}=\frac{\text{Joule }}{{K}}, {K}=\frac{\text{Joule }}{{R}}$$\Rightarrow\beta =(\frac{{Joule}}{{K}})$$\alpha ^{2} \beta =(\frac{{J}_{0} {c} l {e}}{{K}})$$\Rightarrow\alpha =$ dimensionless

${KE}=\frac{3}{2} {k} {T}$ ${PV}={{N}}/{{N}_{{A}}}{RT}$ ${N}=\frac{{PV}}{{kT}}$ $={N}=1.5 \\times 10^{11}$

Since we know that, the heat capacity ratio $\gamma$ for an ideal gas can be related to the degrees of freedom of gas molecules $(f)$ by formula $\gamma =1+\;\frac{2}{f}$.....(i) As each vibrational mode has 2 degrees of freedom, hence total vibrational degrees of freedom $=2 \times 24=48$ $\Rightarrow f=3$ (rotational) $+3$ (translational) $+48$ (vibrational) $\Rightarrow \;\; f=3+3+48 \Rightarrow f=54$ Now, put the value of $f$ in Eq. (i), we get $\gamma =1+\;\frac{2}{54} \Rightarrow \gamma =1+\;\frac{1}{27}$ $\Rightarrow \;\; \gamma =\;\frac{28}{27}$ or $\gamma =1.03$