Thermodynamics Practice Questions

$\gamma$ is independent of temperatureOption 2

$\;\Delta {\text{Q}}=184 \times 10^3$ $\; {\text{m}}=0.600 {\text{kg}} \;\text{ at }\;-12^{\circ} {\text{C}}$ $\; {\text{S}}=222.3 {\text{J}} / {\text{kg}} /{ }^{\circ} {\text{C}}$ $\; {\text{L}}=336 \times 10^3 {\text{J}} / {\text{kg}}$ $\; {\text{Q}}_1=0.600 \times 2222.3 \times 12=16000.56 {\text{J}}$ $\;\;\text{ Remaining heat }\; \Delta {\text{Q}}_1=184000-16000.56$ $\;=167999.44 {\text{J}}$ $\;\;\text{ For meeting at }\; 0{ }^{\circ} {\text{C}}$ $\;\Delta {\text{Q}}_2=0.600 \times 336000=201600 {\text{J}} \;\text{ needed }\;$ $\; \therefore 100 \% \;\text{ ice is not melted }\;$ $\;\;\text{ Amount of ice melted }\;$ $\;167999.44= {\text{m}} \times 336000=0.4999 {\text{kg}}$ $\; \therefore \;\text{ mass of water }\;=0.4999 {\text{kg}}$ $\;\;\text{ Mass of ice }\;=0.1001$ $\; \therefore \;\text{ Ratio }\;=\;\frac{0.1001}{0.4999} \approx 1: 5$

Engine is operating between FP and BP of water.We asked to find efficiency :$\; {\text{T}}_1=100^{\circ} {\text{C}} \;\; \Rightarrow 100+273=373 {\text{K}}$ $\; {\text{T}}_2=0^{\circ} {\text{C}} \;\; \Rightarrow 0+273=273 {\text{K}}$We know$\;η=1-\;\frac{T_2}{T_1}$ $\;η=1-\;\frac{273}{373}$ $\;η=\;\frac{373-273}{373}=\;\frac{100}{373}$ $\;η=0.268$and$η \%\;=0.268 \times 100$ $η \%\;=26.8 \%$

mean free path $\lambda$ $\;\lambda =\;{ {\text{RT}}}/{{\sqrt{2}} {\text{Cd}}^2 {\text{N}}_{ {\text{A}}} {\text{P}}}$ $\;\lambda \propto {\text{T}}$ $\;\;{1500 {\text{d}}}/{\lambda }=\;\frac{273}{373}$ $\;\lambda =2049 {\text{d}}$

${\text{T}}_{ {\text{H}}}=99^{\circ} {\text{C}}=99+273$ $\;\;\;\; =372 {\text{K}}$ $1-\;{ {\text{T}}_{ {\text{C}}}}/{ {\text{T}}_{ {\text{H}}}}=\;\frac{1}{3}$ ${l}\;{ {\text{T}}_{ {\text{C}}}}/{ {\text{T}}_{ {\text{H}}}}=\;\frac{2}{3} \;\; (1) \Rightarrow {\text{T}}_{ {\text{C}}}=\;\frac{2}{3} \times 372$ $=2 \times 124=248 {\text{K}}$ $1-\;{ {\text{T}}_{ {\text{C}}}+ {\text{X}}}/{ {\text{T}}_{ {\text{H}}}}=\;\frac{1}{6}$ $\;\frac{5}{6}=\;{ {\text{T}}_{ {\text{C}}}+ {\text{X}}}/{ {\text{T}}_{ {\text{H}}}}$ $\;\frac{5}{6}=\;{248+ {\text{X}}}/{372}$ $248+ {\text{X}}=5 \times 62$ ${\text{X}}=310-248=62 {\text{K}}$

$\;\sqrt{\;\frac{3 R T}{M}}=\sqrt{\;\frac{\alpha +5}{\alpha }} \sqrt{\;\frac{8}{\pi } \;\frac{R T}{M}}$ $\;3=\;\frac{\alpha +5}{\alpha } \;\frac{8}{\pi }$ $\;\alpha =28$

$\;\;\text{ Using }\; {\text{v}}=\sqrt{\;{\gamma {\text{RT}}}/{ {\text{m}}}}$ $\;\;{ {\text{U}}_{ {\text{H}}_2}}/{ {\text{v}}_{ {\text{O}}_2}}=\sqrt{\;{ {\text{m}}_{ {\text{O}}_2}}/{ {\text{m}}_{ {\text{H}}_2}}}=\sqrt{\;\frac{32}{2}}=\sqrt{\;\frac{16}{1}}=4: 1$(since both hydrogen and oxygen are di-atomic, $\gamma$ will be same)

$\; {\text{V}}_{\;\text{rms }\;}=\sqrt{\;{3 {\text{RT}}}/{ {\text{M}}_{ {\text{w}}}}}=\sqrt{\;{3 {\text{RT}}}/{ {\text{mN}}_{ {\text{A}}}}}=\sqrt{\;{3 {\text{KT}}}/{ {\text{m}}}}$ $\;=\sqrt{\;\frac{3 \times 1.4 \times 10^{-23} \times 300}{4.6 \times 10^{-26}}}$ $\;=\sqrt{\;\frac{9 \times 1.4 \times 10^5}{4.6}}$ $\;={\sqrt{2.73 \times 10^5}}$ $\;={\sqrt{27.3 \times 10^4}}$ $\;=522.4 {\text{ms}}^{-1}$ $\;=523 {\text{ms}}^{-1}$

$\;{[b]=[V]}$ \; \;\;\;\; \[;\frac{a}{b^2}]$ =[P] ;; \therefore ;; $[;\frac{b^2}{a}]$ =;\frac{1}{[P]}=;\frac{1}{[B]}=[K]$

As, gas in suddenly compressed, the process is adiabatic So equation of gas for adiabatic process is:$\; {\text{PV}}^\gamma =\;\text{ constant }\;$ $\; {\text{P}}_0 {\text{V}}_0^\gamma = {\text{P}}_2 (\;{ {\text{V}}_0}/{4}) ^\gamma$ $\; {\text{P}}_2= {\text{P}}_0(4)^\gamma$

${\text{PT}}^2=$ constant, ${\text{Using}} {\text{PV}}= {\text{nRT}}$ ${\text{P}}=\;{ {\text{nRT}}}/{ {\text{V}}}$ ${\text{PT}}^2=\;{ {\text{nRT}}}/{ {\text{V}}} \times {\text{T}}^2=$ constant$\Rightarrow {\text{T}}^3= {\text{KV}}$So, $\;{ {\text{d}}}/{ {\text{dT}}}( {\text{KV}})=3 {\text{T}}^2$ $\Rightarrow \;{ {\text{KdV}}}/{ {\text{dT}}}=3 {\text{T}}^2$ $\Rightarrow {\text{dV}}=\;{3 {\text{T}}^2}/{ {\text{K}}} {\text{dT}}$ ${\text{dV}}= {\text{V}} \gamma {\text{dT}}$ $\Rightarrow \gamma {\text{V}}=\;{3 {\text{T}}^2}/{ {\text{K}}} \;\; \Rightarrow \gamma =\;{3 {\text{T}}^2}/{ {\text{KV}}}=\;{3 {\text{T}}^2}/{ {\text{T}}^3}=\;{3}/{ {\text{T}}}$

Method-1Using exact law of cooling$\; {\text{T}}- {\text{T}}_{ {\ s}}= ( {\text{T}}_0- {\text{T}}_{ {\ s}}) {\text{e}}^{- {\text{Kt}}}$ $\;\;\text{ Case-I: }\;(40-10)=(60-10) {\text{e}}^{-7 {\text{K}}}$ $\;30=50 {\text{e}}^{-7 {\text{K}}}$ $\;\;\text{ Case-II: }\;( {\text{T}}-10)=(40-10) {\text{e}}^{-7 {\text{K}}} \;\text{ or }\; {\text{T}}-10=30 {\text{e}}^{-7 {\text{K}}}$ $\;\;\text{ Dividing }\;(2) \;\text{ by }\;(1)$ $\;\;{ {\text{T}}-10}/{30}=\;\frac{30}{50}$ $\;\Rightarrow {\text{T}}-10=\;\frac{30 \times 30}{50}=18$ $\;\;\text{ or }\; {\text{T}}=28^{\circ} {\text{C}}$Methode-2Using newton's average law of cooling$\;\;{ {\text{T}}_{ {\text{i}}}- {\text{T}}_{ {\text{f}}}}/{ {\text{t}}}= {\text{k}} (\;{ {\text{T}}_{ {\text{i}}}+ {\text{T}}_{ {\text{f}}}}/{2}- {\text{T}}_{ {\ s}})$ $\;\;\text{ Case-I:- }\; \;\frac{60-40}{7}= {\text{R}} [\;\frac{60+40}{2}-10] \Rightarrow \;\frac{20}{7}= {\text{k}}[40]$ . . . (i)$\;\;\text{ Case-II:- }\; \;{40- {\text{T}}}/{7}= {\text{R}} [\;{20+ {\text{T}}}/{2}]$ . . . (2) $\;\;\text{ Dividing (2) by (1) }\;$ $\;\;{40- {\text{T}}}/{20}=\;{20+ {\text{T}}}/{80}$ $\;160-4 {\text{T}}=20+ {\text{T}}$ $\;5 {\text{T}}=140$ $\; {\text{T}}=28^{\circ} {\text{C}}$

Initially $η=\;\frac{1}{2}$ But $\;η=1-\;{ {\text{T}}_2}/{ {\text{T}}_1}$ $\; \therefore \;\frac{1}{2}=1-\;{ {\text{T}}_2}/{600}$ $\;\Rightarrow \;{ {\text{T}}_2}/{600}=\;\frac{1}{2}$ $\;\;\text{ Now efficiency is increased to }\; 70 \% \;\text{ and }\; {\text{T}}_2=300$ $\; {\text{K}}, \;\text{ Let temp of source }\; {\text{T}}_1= {\text{T}}$ $\;\Rightarrow \;\frac{7}{10}=1-\;{300}/{ {\text{T}}}$ $\;\Rightarrow \;{300}/{ {\text{T}}}=1-\;\frac{7}{10}$ $\;\Rightarrow \;{300}/{ {\text{T}}}=\;\frac{3}{10} \;\; \therefore {\text{T}}_2=300 {\text{K}}$

By Isothermal Process for (A)$\; {\text{P}}_1 {\text{V}}_1= {\text{P}}_2 {\text{V}}_2$ $\; {\text{PV}}= {\text{P}}_2 \;{ {\text{V}}}/{8}$ $\; {\text{P}}_2=8 {\text{P}}$For B adiabatically $\gamma _{\;\text{mono }\;}=\;\frac{5}{3}$ $\; {\text{P}}_1 {\text{V}}_1^\gamma = {\text{P}}_2 {\text{V}}_2^\gamma$ $\; {\text{PV}}^{5 / 3}= {\text{P}}_2 (\;{ {\text{V}}}/{8}) ^{5 / 3}$ $\; {\text{P}}_2=(8)^{5 / 3} {\text{P}}$ $\;\;{ {\text{P}}_2}/{ {\text{P}}_1}=\;{8^{5 / 3}}/{8 {\text{P}}}=(8)^{\;\frac{2}{3}}=4$

$\;\;\frac{\;\text{ reading on scale }\;-\;\text{ Lower fixed point }\;}{\;\text{ upper fixed point-lower fixed point }\;}=\;\text{ constant }\;$ $\;\;\frac{t_P-30}{180-30}=\;\frac{t_Q-0}{100-0}$ $\;\;\frac{t_P-30}{150}=\;\frac{t_Q}{100}$

${\text{K}}_{ {\text{av}}}\;=\;\frac{5}{2} {\text{kT}}$ $\;\text{ Ratio }\;\;=1: 1$

$\;\frac{2}{5} \times \;\frac{1}{2} m v^2=m L+m s \Delta T$ $\Rightarrow \;\frac{v^2}{5}=2.5 \times 10^{4+}125+200$ $\Rightarrow \;\frac{v^2}{5}=5 \times 10^4$ $\Rightarrow v=500 {\text{m}} / {\ s}$

$m L=\Delta Q=m s \Delta T$ $\Rightarrow m=\;\frac{5 \times 0.39 \times 10^3 \times 500}{335}$ $=2.9 {\text{kg}}$