Thermodynamics Practice Questions

Extensive variables depend on size and amount of system.Extensive : Volume, mass, internal energyIntensive : Pressure, temperature, density

$C_V\;=\;\frac{n_1 C_{v_1}+n_2 C_{v_2}}{n_1+n_2}$ $\;=\;\frac{2 \times \frac{3}{2} R+6 \times \frac{5}{2} R}{2+6}$ $\;=\;\frac{9}{4} R$

$\; {\text{W}}=\;{ {\text{nR}} \Delta {\text{T}}}/{1-\gamma }$ $\; {\text{TV}}^{\gamma -1}=\;\text{ cons tan }\; {\text{t}}= {\text{T}}_{ {\text{f}}}(2 {\text{V}})^{\gamma -1}$ $\; {\text{T}}_{ {\text{f}}}= {\text{T}} (\;\frac{1}{2}) ^{1 / 2}=\;{ {\text{T}}}/{{\sqrt{2}}}$ $\; {\text{W}}=\;{ {\text{R}} ({ {\text{T}}}/{{\sqrt{2}}}- {\text{T}}) }/{1-\;\frac{3}{2}}=2 {\text{RT}} \;{({\sqrt{2}}-1)}/{{\sqrt{2}}}$ $\;= {\text{RT}}(2-{\sqrt{2}})$

$\;\lambda =\;{ {\text{RT}}}/{{\sqrt{2}} \pi {\text{d}}^2 {\text{N}}_{ {\text{A}}} {\text{P}}}$ $\; {\text{KE}}=\;{ {\text{f}}}/{2} {\text{nRT}}$

For monoatomic molecule degree of freedom $=3$.$\; \therefore {\text{K}}_{\;\text{avg }\;}=\;\frac{3}{2} {\text{K}}_{ {\text{B}}} {\text{T}}$ $\; {\text{T}}=\;\frac{0.414 \times 1.6 \times 10^{-19} \times 2}{3 \times 1.38 \times 10^{-23}}$ $\;=3200 {\text{K}}$

$\;\gamma =1+\;{2}/{ {\text{f}}}=1.4 \Rightarrow \;{2}/{ {\text{f}}}=0.4$ $\;\Rightarrow {\text{f}}=5$ $\; {\text{W}}= {\text{n}} R \Delta {\text{T}}=200 {\text{J}}$ $\; {\text{Q}}= (\;{ {\text{f}}+2}/{2}) {\text{nR}} \Delta {\text{T}}$ $\;=\;\frac{7}{2} \times 200=700 {\text{J}}$

$\; {\text{V}}_{ {\text{rms}}}=\sqrt{\;{3 R T}/{ {\text{M}}}}$ $\;\;{ {\text{V}}_1}/{ {\text{V}}_2}=\sqrt{\;{ {\text{M}}_2}/{ {\text{M}}_1}}$ $\Rightarrow \;{2}/{ {\text{V}}_2}=\sqrt{\;\frac{32}{2}}$ $\; {\text{V}}_2=0.5 {\text{km}} / {\ s}$

For ${\text{PV}}^{ {\text{x}}}=$ constantIf work done by gas is asked then$W\;=\;\frac{n R \Delta T}{1-x}$ $\;\text{ Here }\; x\;=\;\frac{3}{2}$ $\therefore W\;=\;\frac{P_2 V_2-P_1 V_1}{-\;\frac{1}{2}}$ $\;=2 (P_1 V_1-P_2 V_2) . . . . . \;\text{ Option (1) is correct }\;$If work done by external is asked then ${\text{W}}=-2 ( {\text{P}}_1 {\text{V}}_1- {\text{P}}_2 {\text{V}}_2) . . . .$. Option (2) is correct

[ {\text{P}}]= \[;{ {\text{a}}}/{ {\text{V}}^2}]$ \Rightarrow [ {\text{a}}]= [ {\text{PV}}^2]$And$[ {\text{V}}]=[ {\text{b}}]$$;{[ {\text{a}}]}/{[ {\text{b}}^2]$ }=;{[ {\text{PV}}^2]$ }/{[ {\text{V}}^2]$ }=[ {\text{P}}]$

Assuming mean free path constant. $\; {\text{f}} \propto {\text{V}} \propto \sqrt{ {\text{T}}}$ $\;\;{ {\text{f}}_1}/{ {\text{f}}_2}=\sqrt{\;{ {\text{T}}_1}/{ {\text{T}}_2}}=\sqrt{\;\frac{300}{400}}$ $\; {\text{f}}_2=\sqrt{\;\frac{4}{3}}= {\text{f}}_1=\;{2}/{{\sqrt{3}}} {\text{Z}}$

${\text{V}}_{ {\text{rms}}}=\;\sqrt{\;{3 R T}/{ {\text{M}}_{ {\text{w}}}}}$ $\Rightarrow \;{ {\text{V}}_{ {\text{O}}_2}}/{ {\text{V}}_{ {\text{He}}}}\;=\sqrt{\;{ {\text{M}}_{ {\text{w}}, {\text{He}}}}/{ {\text{M}}_{ {\text{w}}, {\text{O}}_2}}}$ $\;=\sqrt{\;\frac{4}{32}}=\;{1}/{2 {\sqrt{2}}}$ $\;{ {\text{V}}_{ {\text{He}}}}/{ {\text{V}}_{ {\text{O}}_2}}\;=\;{2 {\sqrt{2}}}/{1}$

We know that per ${ }^{\circ} {\text{C}}$ change is equivalent to $1.8^{\circ}$ change in ${ }^{\circ} {\text{F}}$.$\therefore 40^{\circ}$ change on celcius scale will corresponds to $72^{\circ}$ change on Fahrenhejt scale.Hence option (3)

$1^{\;\text{st }\;}$ law of thermodynamics$\;\Delta {\text{Q}}=\Delta {\text{U}}+ {\text{W}}$ $\;\Rightarrow +48= {\text{nC}}_{ {\text{v}}} \Delta {\text{T}}+ {\text{W}}$ $\;\Rightarrow 48=(1) (\;{3 {\text{R}}}/{2}) (2)+ {\text{W}}$ $\;\Rightarrow {\text{W}}=48-3 \times {\text{R}}$ $\;\Rightarrow {\text{W}}=48-3 \times (8.3)$ $\;\Rightarrow {\text{W}}=23.1 \;\text{ Joule }\;$

$\; {\text{P}} \propto {\text{T}}^3 \Rightarrow {\text{PT}}^{-3}=\;\text{ constant }\;$ $\; {\text{PV}}^\gamma =\;\text{ const }\;$ $\; {\text{P}} (\;{ {\text{nRT}}}/{ {\text{P}}}) ^\gamma =\;\text{ const }\;$ $\; {\text{P}}^{1-\gamma } {\text{T}}^\gamma =\;\text{ const }\;$ $\; {\text{PT}}^{\;\frac{\gamma }{1-\gamma }}=\;\text{ const }\;$ $\;\;\frac{\gamma }{1-\gamma }=-3$ $\;\gamma =-3+3 \gamma$ $\;3=2 \gamma$ $\;\gamma =\;\frac{3}{2}$

For process ${\text{A}}$ $\;\log P=\gamma \log {\text{V}} \Rightarrow {\text{P}}= {\text{V}}^\gamma ,(\gamma >1)$ $\;P V^{-\gamma }=\;\text{ Constant }\;$ $\;C_A=C_V+\;\frac{R}{1+\gamma } . . . . \;\text{ (i) }\;$ Likewise for process ${\text{B}} \longrightarrow P V^{-1}=$ Constant $\;C_B=C_v+\;\frac{R}{1+1}$ $\;C_B=C_v+\;\frac{R}{2} . . . ....$(ii) $\;C_P=C_v+R . . .....$ (iii) By (i), (ii) & (iii) $C_P>C_B>C_A>C_v$ [No answer matching]

$\;\Delta {\text{U}}=0(\;\text{ Cyclic process }\;)$ $\;\Delta {\text{Q}}= {\text{W}}=\;\text{ area of }\; {\text{P}}- {\text{V}} \;\text{ curve. }\;$ $\;=\pi \times (140 \times 10^3 {\text{Pa}}) \times (140 \times 10^{-6} {\text{m}}^3)$ $\;\Delta {\text{Q}}=61.6 {\text{J}}$

For Isobaric process$\; {\text{w}}= {\text{P}} \Delta {\text{v}}= {\text{nR}} \Delta {\text{T}}=100 {\text{J}}$ $\; {\text{Q}}=\Delta {\text{u}}+ {\text{w}}$ $\;\Delta {\text{Q}}=\;{ {\text{F}}}/{2} {\text{nR}} \Delta {\text{T}}+ {\text{nR}} \Delta {\text{T}}$ $\; (\;{ {\text{f}}}/{2}+1) {\text{nR}} \Delta {\text{T}}$ $\; (\;\frac{5}{2}+1) 100=350 {\text{J}}$

$\; {\text{TV}}^{\gamma -1}=\;\text{ constant }\;$ $\;\Rightarrow {\text{T}}( {\text{V}})^{\;\frac{3}{2}-1}= {\text{T}}_{ {\text{f}}}(2 {\text{V}})^{\;\frac{3}{2}-1}$ $\;\Rightarrow {\text{TV}}^{\;\frac{1}{2}}= {\text{T}}_{ {\text{f}}}(2)^{\;\frac{1}{2}}( {\text{V}})^{\;\frac{1}{2}}$ $\;\Rightarrow {\text{T}}_{ {\text{f}}}= (\;{ {\text{T}}}/{{\sqrt{2}}})$ $\;\;\text{ Now, W.D. }\;=\;{ {\text{nR}} \Delta {\text{T}}}/{1-\gamma }=\;{1 \cdot {\text{R}} [{ {\text{T}}}/{{\sqrt{2}}}- {\text{T}}] }/{1-\;\frac{3}{2}}$ $\;\Rightarrow \;\text{ W.D. }\;=2 {\text{RT}} [1-\;{1}/{{\sqrt{2}}}]$ $\;\Rightarrow \;\text{ W.D. F }\; {\text{RT}}[2-{\sqrt{2}}]$

$\;Q=\Delta U \;\text{ as work done is zero [constant volume] }\;$ $\;\Delta U=m s \Delta T$ $\;=0.08 \times (170 \times 4.18) \times 5$ $\;≃ 284 {\text{J}}$