Thermodynamics Practice Questions

$V_{{RMS}}=\\sqrt{\frac{3 {RT}}{{M}}}$ ${m}_{{A}} < {m}_{{B}} < {m}_{{C}}$ $\Rightarrow {v}_{{A}}>{v}_{{B}}>{v}_{{C}}$ $\Rightarrow {1}/{{v}_{{A}}} < {1}/{{v}_{{B}}} < {1}/{{v}_{{C}}}$

We know that, The average value of energy of an ideal gas in thermal equilibrium temperature $T$, $E=\;\frac{f}{2} k_{B} T$ Given, degree of freedom, $f=1$ Energy per degree of freedom, $E=\;\frac{1}{2} k_{B} T$

Given, initial efficiency of engine, $\eta _{1}=1 / 4$ Let, initial temperature of $\sin k=\text{T}_{2}$ Final temperature of $\sin k=\text{T}_{2}-58° {\text{C}}$ Final efficiency of engine, $\eta _{2}=2 \eta _{1}=2 / 4=1 / 2$ Temperature of source $=\text{T}_{1}$ Since, $\eta =1-\frac{\text{T}_{2}}{\text{T}_{1}}$ $\therefore \eta _{1}=\frac{1}{4}=1-\frac{\text{T}_{2}}{\text{T}_{1}}$ $\Rightarrow \frac{\text{T}_{2}}{\text{T}_{1}}=1-\frac{1}{4}$ $\Rightarrow \text{T}_{2} / \text{T}_{1}=3 / 4$ $\Rightarrow \text{T}_{1}=\frac{4}{3} \text{T}_{2}$ ...(i) and$\eta _{2}=\frac{1}{2}=1-\frac{\text{T}_{2}-58}{\text{T}_{1}}$ $\Rightarrow \frac{\text{T}_{2}-58}{\text{T}_{1}}=1-\frac{1}{2}$ $\Rightarrow \frac{\text{T}_{2}-58}{4 / 3 \text{T}_{2}}=\frac{1}{2}$[From Eq. (i)] $\Rightarrow 2 \text{T}_{2}-116=\frac{4}{3} \text{T}_{2}$ $\Rightarrow \frac{2 \text{T}_{2}}{3}=116$ $\Rightarrow \text{T}_{2}=\frac{116 \times 3}{2}=58 \times 3=174° {\text{C}}$

Here, $p_{1}$ and $p_{2}, T_{1}$ and $T_{2}, V_{1}$ and $V_{2}$ are initial and final pressures, temperatures and volumes, respectively. Given, $\;\; V_{2}=2 V_{1}$ $p V^{1 / 2}=\;\text{ constant }\;$ From graph, $\gamma =$ adiabatic constant $=1 / 2$ $\Rightarrow \;\; p_{1} V_{1}^{1 / 2}=p_{2} V_{2}^{1 / 2} \Rightarrow \;\frac{p_{1}}{p_{2}}=(\;\frac{2 V_{1}}{V_{1}})^{1 / 2}=2^{1 / 2}$ Also, $\;\; p^{1-\gamma } T^{\gamma }=$ constant (for adiabatic process) $\Rightarrow \;\; p_{1}^{1-\gamma } T_{1}^{\gamma }=p_{2}^{1-\gamma } T_{2}^{\gamma }$ $\Rightarrow \;\; (\;\frac{p_{1}}{p_{2}})^{1-\gamma }=(\;\frac{T_{2}}{T_{1}})^{\gamma } \Rightarrow (\;\frac{p_{1}}{p_{2}})^{\;\frac{1-\gamma }{\gamma }}=\;\frac{T_{2}}{T_{1}}$ $\therefore \;\; T_{2} / T_{1}=(2^{1 / 2})^{\;\frac{1-0.5}{0.5}}=\sqrt{2}$

Given, height of fall, $H=63 {m}$Acceleration due to gravity, $g=9.8 {ms}^{-2}$Specific heat of water, $s=1 {cal} {g}^{-1} {C}^{-1}$ $=4.2 \times 10^{3} {J} {kg}^{-1} {C}^{-1}$Since, energy, $E=m g H=m s \Delta T$where, $m=$ mass and $\Delta T=$ change in temperature.$\therefore \Delta T=\;\frac{g H}{s}=\;\frac{9.8 \times 63}{4.2 \times 10^{3}}=0.147^{\circ} {C}$

According to question, two ideal atomic gases at temperatures $T_{1}$ and $T_{2}$ are mixed. Let the final temperature of this mixture be $T$. As per question there is no loss of energy, it means $\Delta U=0$...(i) As, we know, $\Delta U=\;\frac{f_{1} n_{1} R \Delta T}{2}+\;\frac{f_{2} n_{2} R \Delta T}{2}$....(ii) From Eqs. (i) and (ii), we get $\;\frac{f_{1} n_{1} R \Delta T}{2}+\;\frac{f_{2} n_{2} R \Delta T}{2}=0$ $\Rightarrow \;\; f_{1} n_{1} R \Delta T+f_{2} n_{2} R \Delta T=0$ $\Rightarrow \;\; f_{1} n_{1} R(T_{1}-T)+f_{2} n_{2} R(T_{2}-T)=0$ $\Rightarrow \;\; f_{1} n_{1}(T_{1}-T)+f_{2} n_{2}(T_{2}-T)=0$ $\Rightarrow \;\; f_{1} n_{1} T_{1}-f_{1} n_{1} T+f_{2} n_{2} T_{2}-f_{2} n_{2} T=0$ $\Rightarrow \;\; f_{1} n_{1} T+f_{2} n_{2} T=f_{1} n_{1} T_{1}+f_{2} n_{2} T_{2}$ $\Rightarrow T(f_{1} n_{1}+f_{2} n_{2})=f_{1} n_{1} T_{1}+f_{2} n_{2} T_{2}$ $T=\;\frac{f_{1} n_{1} T_{1}+f_{2} n_{2} T_{2}}{f_{1} n_{1}+f_{2} n_{2}}$

Entropy present in one part of the system is $S_{1}$. Entropy present in other part of the system is $S_{2}$. After removing piston,We know that, $\Delta S_{\;\text{unvi }\;}=\Delta S_{\;\text{sys }\;}+\Delta S_{\;\text{surr }\;}$ Thus, after removing the piston, the total entropy present in system, $S_{\;\text{tot }\;}=S_{1}+S_{2}$

As per Equi-partition law : Each degree of freedom contributes $\frac{1}{2} {k}_{{B}} {T}$ Average Energy In monoatomic gas D.O.F. $=3$ $\Rightarrow$ Average energy $=3 \\times \frac{1}{2} {k}_{{B}} {T}=\frac{3}{2} {k}_{{B}} {T}$

Given The final pressure of the adiabatic process, $p_{f}=200 {N} / {m}^{2}$ The initial pressure of the adiabatic process, $p_{i}=100 {N} / {m}^{2}$ The initial volume of the adiabatic process, $V_{1}=4 {m}^{3}$ The final volume of the adiabatic process, $V_{f}=3 {m}^{3}$ We know that, for an adiabatic process $W=\;\frac{p_{f} V_{f}-p_{i} V_{i}}{1-\gamma }$ $\Rightarrow \;\; W=\;\frac{200(3)-(100)(4)}{1-1.4}$ $\Rightarrow \;\; W=-500 J$ So, the work done by an adiabatic process is $500 {J}$ and negative sign represents the work done on the system.

As, $A B$ is isothermal process, so work done by isothermal process is given by $\;W_{A B}=n R T \ln (\;\frac{V_{2}}{V_{1}})$ $\;W_{A B}=n R T \ln (\;\frac{2 V_{1}}{V_{1}})=n R T \ln 2$ As, $B C$ is isobaric process, so work done by isobaric process is given by $W_{B C} \;\; =p \Delta V=p_{2}(V_{1}-V_{2})$ $\;\; =p_{2}(\;\frac{V_{2}}{2}-V_{2})=-\;\frac{p_{2} V_{2}}{2}$ $\;\; =\;\frac{-n R T}{2}$ As, $C A$ is isochoric process, so work done by isochoric process is given by, $W_{C A}=0 \;\; [ \because \Delta V=0]$ Total work done in the cycle $A B C A$, $W_{A B C A} \;\; =W_{A B}+W_{B C}+W_{C A}$ $\;\; =n R T \ln 2+(\;\frac{-n R T}{2})+0$ $\;\; =n R T(\ln 2-\;\frac{1}{2})$

Given, number of mole, $n=1$ Heat supplied $=Q$ Work done $=Q / 5$ By using first law of thermodynamics, $\Delta Q \;\; =\Delta U+\Delta W$ $Q \;\; =\Delta U+Q / 5$ $\Delta U \;\; =4 Q / 5=n C_{v} \Delta T$. . . (i) where, $C_{V}$ is heat coefficient at constant volume and $\Delta T$ is change in temperature. Degree of freedom of diatomic gas, $f=5$ $\therefore \;\; Q=\;\frac{5}{4} \;\frac{f}{2} R \Delta T$ [from Eq. (i)] $\Rightarrow \;\; \;\frac{Q}{\Delta T}=C=\;\frac{5}{4} \times \;\frac{5}{2} R=\;\frac{25}{8} R$ Hence, $x=25$

For polyatomic gas molecule has 3 rotational degrees of freedom, 3 translational degrees of freedom, and 2 vibrational modes. So, number of vibrational degrees of freedom $=2(2)=4$ $\therefore$ Total number of degrees of freedom, $f=3+3+4=10$ Here, ratio of molar specific heats is given as $\beta =\;\frac{C_{p}}{C_{v}}$ In terms of degrees of freedom, $\beta =1+\;\frac{2}{f}$ Substituting the value of $f$ in above equation, we get $\beta =1+\;\frac{2}{10} \Rightarrow \beta =1.2$

Given, volume of 1 st chamber, $V_{1}=4.5 {L}$ Volume of 2 nd chamber, $V_{2}=5.5 {L}$ $\;n_{1}=3 {mol}$ $\;p_{1}=2 {atm}$ $\;n_{2}=4 {mol}$ $\;p_{2}=3 {atm}$ By using ideal gas equation, $p V=n R T$ $\; \therefore \;p_{1} V_{1}+p_{2} V_{2}\;=p(V_{1}+V_{2})$ $\Rightarrow \;2 \times 4.5+3 \times 5.5\;=p \times 10$ $\Rightarrow \;9+16.5=10 p \Rightarrow \;\frac{25.5}{10}=p$ $\therefore \;p=25.5 \times 10^{-1} {atm}$ Hence, $x=25.5 {atm}$

{W}_{{A}}=1-\frac{{Q}_{2}}{{Q}_{1}}=1-\frac{{T}}{{T}_{1}} \ $\Rightarrow \frac{{Q}_{2}}{{Q}_{1}}=\frac{{T}}{{T}_{1}}$ {W}_{{B}}=1-\frac{{Q}_{3}}{({Q}_{2} / 2)}=1-\frac{{T}_{3}}{{T}} \ $\Rightarrow \frac{2 {Q}_{3}}{{Q}_{2}}=\frac{{T}_{3}}{{T}}$ ${\text{Now}}, {W}_{{A}}={W}_{{B}}$ ${Q}_{1}-{Q}_{2}=\frac{{Q}_{2}}{2}-{Q}_{3}$ $\Rightarrow\frac{2 {Q}_{1}}{{Q}_{2}}+\frac{2 {Q}_{3}}{{Q}_{2}}=3$ $\Rightarrow\frac{2 {T}_{1}}{{T}}+\frac{{T}_{3}}{{T}}=3$ $\frac{2 {T}_{1}}{3}+\frac{{T}_{3}}{3}={T}$

Work done in adiabatic process $=\frac{-{nR}}{\gamma -1}({T}_{{f}}-{T}_{{i}})$ $\therefore {W}_{{AD}}=\frac{-{nR}}{\gamma -1}({T}_{2}-{T}_{1})$ and ${W}_{{BC}}=\frac{-{nR}}{\gamma -1}${T}{2}-{T}{1})$$\therefore {W}{{AD}}={W}{{BC}}$

$\eta =1-{{T}_{{L}}}/{{T}_{{H}}} \........$ (i) $2 \eta =1-{({T}_{{L}}-62)}/{{T}_{{H}}}=1-{{T}_{{L}}}/{{T}_{{H}}}+{\frac{62}{T}_{{H}}}$ \Rightarrow\eta ={62}/{{T}_{{H}}} \Rightarrow \frac{1}{6}={62}/{{T}_{{H}}} \ $\Rightarrow {T}_{{H}}=6 \\times 62=372 {K}$ ${\text{In}}°{C} \Rightarrow372-273=99°{C}$

Given, power consumed by refrigerator $\text{P}=35 {\text{W}}=35 {\text{Js}}^{-1}$ ∴ Energy consumed by refrigerator per second, $\text{W}=35 {\text{J}}$ Lower temperature limit, $\text{T}_{\text{L}}=-10° {\text{C}}=263 {\text{K}}$ Higher temperature limit, $\text{T}_{\text{H}}=25° {\text{C}}=298 {\text{K}}$ The coefficient of performance of refrigerator is given as ${\text{COP}}={\text{T}_{\text{L}}}/{\text{T}_{\text{H}}-\text{T}_{\text{L}}}=\frac{\text{Q}}{\text{W}}$ Here, Q is the heat transferred by the refrigerator. Substituting the values, we get $\frac{263}{298-263}=\frac{\text{Q}}{35}$ $\Rightarrow \text{Q}=\frac{263}{35} \times 35=263 {\text{J}}/ {\text{s}}$ Thus, the average heat transfer by refrigerator per second is $263 {\text{Js}}^{-1}$.

$(\frac{\DeltaQ}{\Deltat})_{A}=(\frac{\DeltaQ}{\Deltat})_{B}$ $m S_{A}(\frac{\DeltaT}{\Deltat})_{{A}}=m S_{B}(\frac{\DeltaT}{\Deltat})_{{B}}$ $\frac{S_{A}}{S_{B}}=\frac{(\frac{\DeltaT}{\Deltat})_{A}}{(\frac{\DeltaT}{\Deltat})_{B}}$ $=\frac{90 / 6}{120 / 3}=\frac{15}{40}=\frac{3}{8}$

Given, volume of container, $\text{V}=4.0 \times 10^{-3} {\text{m}}^{3}$ Gas constant, $\text{R}=8.3 {\text{J}} \;{\text{mol}}^{-1} {\text{K}}^{-1}$ Number of moles of hydrogen, $n_{1}=1$ Number of moles of carbon dioxide, $n_{2}=2$ Temperature of mixture, T = 400 K Using Dalton's law to calculate the pressure of mixture as follows $p=p_{1}+p_{2}$ $=\frac{n_{1} \text{R} \text{T}}{\text{V}}+\frac{n_{2} \text{R} \text{T}}{\text{V}}$ $=\frac{\text{R} \text{T}}{\text{V}}(n_{1}+n_{2})$ $=\frac{\text{R} \text{T}}{\text{V}}(1+2)=\frac{3 \text{R} \text{T}}{\text{V}}$ $=\frac{3 \times 8.3 \times 400}{4.0 \times 10^{-3}}=24.9 \times 10^{5} {\text{Pa}}$ Thus, the pressure of mixture of gases is $24.9 \times 10^{5} {\text{Pa}}$. Thus, the pressure of mixture of gases is $24.9 \times 10^{5} {\text{Pa}}$.