Thermodynamics Practice Questions

Given, temperature of source, $T_{H}=127^{\circ} {C}$ $=273+127=400 {K}$ Efficience, $η=60 \%=0.6$ The efficiency of Carnot (ideal) heat engine is given by $η=(1-\;\frac{T_{L}}{T_{H}})$ where, $T_{L}=$ temperature of sink, $0.6=(1-\;\frac{T_{L}}{T_{H}}) \Rightarrow \;\frac{T_{L}}{T_{H}}=1-0.6$ $\;\frac{T_{L}}{T_{H}}=0.4 \Rightarrow T_{L}=0.4 \times T_{H}$ $=0.4 \times 400=160 {K}$ $=160-273=-113^{\circ} {C}$

Given, heat supplied to the system, $\frac{\Delta\text{Q}}{\Deltat}=6000 {\text{J}} / {\min}=\frac{6000}{60} {\text{J}} / {\text{s}}=100 {\text{J}} / {\text{s}}$ Power delivered, $\text{P}=\frac{\Delta\text{W}}{\text{t}}=90 {\text{W}}$ Increase in internal energy, $\Delta\text{U}=2.5 \times 10^{3} {\text{J}}$ From first law of thermodynamics, we have $\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$ or $\frac{\Delta\text{Q}}{\Deltat}=\frac{\Delta\text{U}}{\Deltat}+\frac{\Delta\text{W}}{\Deltat}$ ...(i) Substituting the given values in Eq. (i), we get $100=\frac{2.5 \times 10^{3}}{\Deltat}+90$ $\Rightarrow 10=\frac{2.5 \times 10^{3}}{\Deltat}$ $\Rightarrow \Deltat=\frac{2.5 \times 10^{3}}{10}$ ⇒ $\Deltat=2.5 \times 10^{2} {\text{s}}$

$\because$ mean free path $\lambda=\frac{1}{\\sqrt{2} \pi {d}^{2} {n}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{{d}_{2}^{2} {n}_{2}}{{d}_{1}^{2} {n}_{1}}$ $=(\frac{5}{10})^{2}=0.25=25 \\times 10^{-2}$

According to Statement $1 :$ In diatomic molecule the rotational energy at a given temperature obeys Maxwell's distribution is correct. But, according to Statement $2:$ In diatomic molecule, the rotational energy at a given temperature equals translational energy for each molecule is false. Because kinetic energy $( {KE})$ of gaseous molecule is $\;\frac{f}{2} K_{B} T$. where, $f$ is the degree of freedom, and $f($ diatomic atom $)=5=[3$ (translational) and 2 (rotational)] Therefore, translational ${KE}$ of gas $=\;\frac{3}{2} K_{B} T$ . . . (i) and rotational ${KE}$ of gas $=2 / 2 K_{B} T=K_{B} T$ . . . (ii) $\because$ Eq. (i) is not equal to Eq. (ii) Hence, option (c) is the correct.

We know that, $\gamma =3 \alpha$. ... (i) where, $\alpha$ is the coefficient of linear expansion and $\gamma$ is the coefficient of volume expansion. We know that, $\;\frac{\Delta V}{V}\;=\gamma \Delta T\;$ $\;\frac{\Delta V}{V}\;=3 \alpha \Delta T\; \;\; [\;\text{ from Eq. (i)] }\;$ \Rightarrow \;\; \Delta V\;=3 a^{3} \alpha \Delta T\;{[ \because \;\text{ volume of cube }\;=a^{3}]\}$

$\gamma =\alpha _x+\alpha _y+\alpha _z$ $=5\times 10^{-5}+5\times 10^{-6}+5\times 10^{-6}$ $=(50+5+5)\times 10^{-6}$ $\gamma =60\times 10^{-6}$ $C = 60$.

$\text{table} \frac{{Cal}}{20 {gm}} ,\frac{{H}_{2} \ {O}}{180 {gm}}, \frac{\text{ Sterm}}{\ {m}}; 25°{C} , 25°{C} , 100°{C}$ $200 \\times 1 \\times (31-25)$ $=\ {m} \\times 540+\ {m} \\times 1$ $\times (100-31)$

$P V=n R T$ $P \DeltaV+V \DeltaP=0$(for constant temp.) $P \DeltaV=n R \DeltaT$ (for constant pressure) $\DeltaT= \frac{P \DeltaV}{n R}$ $\Delta{P}=-\frac{{P} \Delta{V}}{ {V}} (\Delta{V}$ is same in both cases $)$ $\frac{\Delta{T}}{\Delta{P}}=\frac{{P} \Delta{V}}{ {nR}} \frac{{V}}{- {P} \Delta{V}}$ $= \frac{- {V}}{ {nR}} =-\frac{{T}}{ {P}}$ $( {PV}= {nRT})$ $$\frac{{V}}{ {nR}}=\frac{{T}}{ {P}} )$$|\ \frac{\Delta{T}}{\Delta{P}} |=|\frac{-300}{2} |=150$

$PV^\gamma =\text{constant}$ $TV^{\gamma -1}=C$ $300\times V^{7/5-1}=T_2(V/16)^{7/5-1}$ $300\times 2^{4\times 2/5}=T_2$Isobaric process$V={nRT}/P$ $V_2=kT_2$... (1) $2V_2=KT_f$..... (2)$1/2=T_2/T_f\Rightarrow T_f=2T_2$ $T_f=2\times 300\times 2^{8/5}=1818.85$ ∴ Correct answer 1819

$C_{\text{P}_{\text{eq}}}={n_1C_{\text{P}_{1}}+n_2C_{\text{P}_{2}}}/{n_1+n_2}$ $C_{\text{V}_{\text{eq}}}={n_1C_{\text{V}_{1}}+n_2C_{\text{V}_{2}}}/{n_1+n_2}$ $\gamma _{\text{eq}}=C_{\text{P}_{\text{eq}}}/C_{\text{V}_{\text{eq}}}=\frac{2\times {5R}/2+3\times {8R}/2}{2\times {3R}/2+3\times {6R}/2}$ $=\frac{5+12}{3+9}=17/12$ ≃ 1.42

In adiabatic process ${PV}^{\gamma }=$ constant ${P} (\frac{{m}}{\rho } )^{\gamma }= {\text{constant}}$ as mass is constant ${P} \propto\rho ^ \gamma$ $\frac{P_{f}}{P_{i}}=(\frac{\rho _{f}}{\rho _{i}} )^{\gamma }=(32)^{7 / 5}$ $=2^{7}=128$

${v}_{{\text{avg}}} \propto \sqrt{{T}}$ ${t}_{0}:$ mean time $\lambda:$ mean free path ${t}_{0}={\lambda}/{{v}_{{\text{avg} }}} \propto {1}/{\sqrt{{T}}}$

${u}=\frac{{f}_{1} {n}_{1} {RT}}{2}+\frac{{f}_{2} {n}_{2} {RT}}{2}$ ${u}= \frac{5}{2} \\times 3 {RT}+ \frac{3 \\times 5 {RT}}{2}$ $=15 {RT}$

${w}+ {Q}_{1}= {Q}_{2}$ ${w}= {Q}_{2}- {Q}_{1}$ C.O.P. $=\frac{{Q}_{1}}{ {w}}=\frac{{Q}_{1}}{ {Q}_{2}- {Q}_{1}}$ $= \frac{273}{300-273}=\frac{{Q}_{1}}{ {W}}$ ${w}= \frac{27}{273} \\times 80 \\times 100 \\times 4.2$ ${Q}_{2}= {w}+\theta _{1}$ ${Q}_{2}= \frac{27}{273} \\times 80 \\times 100$ $\times 4.2+80 \\times 100 \\times 4.2$ ${Q}_{2}=\ \frac{300}{273} \\times 80 \\times 100$ $=8791.2 \ {cal}$

Degree of freedom of a diatomic molecule if vibration is absent = 5Degree of freedom of a diatomic molecule if vibration is present = 7$\therefore C_V^A=f_A/2R=5/2R$ & $C_V^B=f_B/2R=7/2R$ $\therefore C_V^A/C_V^B=5/7$

Refrigerator cycle is :$η=W/Q_{+}=W/{W+Q_{-}}$ $1/10=1/{10+Q_{-}}$ $Q_{-}=90J$Heat absorbed from the reservoir at lower temperature is 90 J

According to table and applying law of calorimetry $1T _1+ 2T_2$ = (1 + 2)60° = 180.... (1) $1T_2 + 2T_3$ = (1 + 2)30° = 90....(2) $2T_1+1T_3$ = (1 + 2)60°= 180 ....(3) Adding (1) + (2) + (3) $3 (T_1+ T_2 + T_3) = 450$ $T_1+ T_2 + T_3 = 150°$Hence,$T_1 + T_2 + T_3 = (1 + 1 + 1)\theta$ $150 = 3\theta$ $\theta = 50°C$

M × 540 + M + 60 = 200 × 80 + 200 × 1× (40 - 0) ⇒ M = 40