Thermodynamics Practice Questions

$\;\sqrt{\;\frac{3 R T}{2}}=\sqrt{\;\frac{3 R(320)}{32}}$ $\;T=\;\frac{320}{16}=20 {\text{K}}$

$K E=\;\frac{f}{2} k T$Conceptual

For adiabatic process$\; {\text{TV}}^{\gamma -1}=\;\text{ constant }\;$ $\; {\text{T}}_{ {\text{a}}} \cdot {\text{V}}_{ {\text{a}}}^{\gamma -1}= {\text{T}}_{ {\text{d}}} \cdot {\text{V}}_{ {\text{d}}}^{\gamma -1}$ $\; (\;{ {\text{V}}_{ {\text{a}}}}/{ {\text{V}}_{ {\text{d}}}}) ^{\gamma -1}=\;{ {\text{T}}_{ {\text{d}}}}/{ {\text{T}}_{ {\text{a}}}}$ $\; {\text{T}}_{ {\text{b}}} \cdot {\text{V}}_{ {\text{b}}}^{\gamma -1}= {\text{T}}_{ {\text{c}}} \cdot {\text{V}}_{ {\text{c}}}^{\gamma -1}$ $\; (\;{ {\text{V}}_{ {\text{b}}}}/{ {\text{V}}_{ {\text{c}}}}) ^{\gamma -1}=\;{ {\text{T}}_{ {\text{c}}}}/{ {\text{T}}_{ {\text{b}}}}$ $\;\;{ {\text{V}}_{ {\text{a}}}}/{ {\text{V}}_{ {\text{d}}}}=\;{ {\text{V}}_{ {\text{b}}}}/{ {\text{V}}_{ {\text{c}}}} \;\; \binom{ \because {\text{T}}_{ {\text{d}}}= {\text{T}}_{ {\text{c}}}}{ {\text{T}}_{ {\text{a}}}= {\text{T}}_{ {\text{b}}}}$ $\;$

Steeper curve (B) is adiabaticAdiabatic $\Rightarrow {\text{PV}}^v=$ const.Or ${\text{P}} (\;{ {\text{T}}}/{ {\text{P}}}) ^v=$ const.$\;{ {\text{T}}^v}/{ {\text{P}}^{v-1}}=\;\text{ const. }\;$Curve (A) is isothermal${\text{T}}=$ const.${\text{PV}}=$ const.

Work done $= {\text{P}} \Delta {\text{V}}$ $=3 \times 10^5 \times 1600 \times 10^{-6}$ $=480 {\text{J}}$Only $10 \%$ of heat is used in work done.Hence $\Delta {\text{Q}}=4800 {\text{J}}$The rest goes in internal energy, which is $90 \%$ of heat.Change in internal energy $=0.9 \times 4800=4320 {\text{J}}$

Basic theoryTranslational $K . E$ on average of a molecule is $\;\frac{3}{2}$ KT which is independent of nature, pressure and volume.

Internal energy of ${\text{O}}_2=\;\frac{5}{2} {\text{nRT}}=\;\frac{5}{2} \times 2 {\text{RT}}=5 {\text{RT}}$Internal energy of ${\text{Ne}}= {\text{nRT}}=\;\frac{3}{2} \times {\text{nRT}}=\;\frac{3}{2} \times 4 {\text{RT}}=6 {\text{RT}}$ Total energy of mixture $($ system $)=5 R T+6 R T=11 R T$

For isochoric process$\;{ {\text{P}}}/{ {\text{T}}}= {\text{n}} \;{ {\text{R}}}/{ {\text{V}}}=\;\text{ constan }\; {\text{t}}$ ${\text{P}}=\;{ {\text{nR}}}/{ {\text{V}}}( {\text{t}}+273)$If ${\text{P}}=0 \Rightarrow {\text{t}}=-273^{\circ} {\text{C}}$

Work done $=$ Area of graph${\text{W}}\;=\;\frac{1}{2}(400-100)(4-2)$ ${\text{W}}\;=300 {\text{J}}$

$\;\;\frac{41^{\circ}-5^{\circ}}{95^{\circ}-5^{\circ}}=\;{ {\text{C}}-0^{\circ}}/{100^{\circ}-0^{\circ}}$ $\;\Rightarrow \;\; {\text{C}}=\;\frac{36}{90} \times 100=40^{\circ} {\text{C}}=313 {\text{K}}$

$\; {\text{V}}_{ {\text{mms}}}=\sqrt{\;{3 {\text{RT}}}/{ {\text{M}}}}$ $\; {\text{V}}_{ {\text{avg}}}=v=\sqrt{\;{8 {\text{RT}}}/{\pi {\text{M}}}}$ $\; {\text{V}}_{ {\text{mms}}}=\sqrt{\;\frac{3 \pi }{8} v}$ $\; {\text{V}}_{ {\text{mms}}}=\sqrt{\;\frac{3}{8} \times \;\frac{22}{7}} {\text{v}}= (\;\frac{33}{28}) ^{1 / 2} v$ $\; {\text{V}}_{ {\text{mms}}}= (1+\;\frac{5}{28}) ^{1 / 2} v$ $\; {\text{x}}=28$

$\Delta {\text{U}}= {\text{nC}}_{ {\text{V}}} \Delta {\text{T}}$, for all processFor isothermal process, $\Delta {\text{T}}=0$So, $\Delta {\text{U}}=0$That means internal energy of system remains constant.

$\;\;\text{ Kinetic energy per molecule }\;=\;\frac{3}{2} {\text{k}}_{ {\text{B}}} {\text{T}}$ $\;\propto \;{ {\text{k}}_{ {\text{Ar}}}}/{ {\text{k}}_{ {\text{H}}_2}}=\;{ {\text{T}}}/{ {\text{T}}}=1$

$\; {\text{VRMS}}=\sqrt{\;{\gamma {\text{RT}}}/{ {\text{m}}}}$ $\;\gamma =1+\;\frac{2}{f} \;\; \;\text{ so }\; \text{bo} {r}_{\;\text{monochromic }\;} > \text{bo} {r}_{\;\text{diatomic }\;} > \text{bo} {r}_{\;\text{poly }\;} \;\text{. }\;$ $\;V_{\;\text{mono }\;} > V_{\;\text{diatomic }\;} > V_{\;\text{poly. }\;}$ $\;\;\text{ Ans.(1) }\;$ $\;$

Statement-I$\; {\text{T}}_1=-73^{\circ} {\text{C}}=200 {\text{K}}$ $\; {\text{T}}_2=527^{\circ} {\text{C}}=800 {\text{K}}$ $\;\;{ {\text{V}}_1}/{ {\text{V}}_2}=\;{\sqrt{{3 {\text{RT}}_1}/{ {\text{M}}}}}/{\sqrt{\;{3 {\text{RT}}_2}/{ {\text{M}}}}}=\sqrt{\;{ {\text{T}}_1}/{ {\text{T}}_2}}$ $\;=\sqrt{\;\frac{200}{800}}=\;\frac{1}{2}$ $\; {\text{V}}_2=2 {\text{V}}_1 \;\text{ (True) }\;$Statement-II${\text{PV}}= {\text{nRT}}$Translational ${\text{KE}}=\;\frac{3}{2} {\text{nRT}}$ (False)

The rms speed of a gas molecule is$\; {\text{V}}_{ {\text{RMS}}}=\sqrt{\;{3 {\text{RT}}}/{ {\text{M}}}}$ $\; {\text{V}}_{ {\text{RMS}}} \propto \sqrt{ {\text{T}}}$

$\;\;\frac{X_T-X_L}{X_H-X_L}=\;\frac{T_F-32}{212-32}$ $\;\;\frac{-95^{\circ}- (-15^{\circ}) }{65^{\circ}- (-15^{\circ}) }=\;\frac{T_F-32}{180}$ $\;\;\frac{-80^{\circ}}{80^{\circ}}=\;\frac{T_F-32}{180^{\circ}}$ $\;-180= {\text{T}}_{ {\text{F}}}-32$ $\; {\text{T}}_{ {\text{F}}}=-180+32=-148^{\circ} {\text{F}}$ Ans. option $\to$ (2)

$\Delta {\text{U}}= {\text{nC}}_{ {\text{v}}} \Delta {\text{T}}$For isothermal process ${\text{T}}$ is constantSo $\Delta {\text{U}}=0$ ${\text{A}} \longrightarrow {\text{II}}$Adiabatic process$\;\Delta {\text{Q}}= 0$ $\;\Delta {\text{Q}}= \Delta {\text{U}}+\Delta {\text{W}}$ $\;\Delta {\text{U}}=-\Delta {\text{W}}$Work done by gas is positiveSo $\Delta {\text{U}}$ is negative$\;\text{ B }\; \longrightarrow \;\text{ I }\;$For Isochoric process $\Delta {\text{W}}=0$ ${\text{C}} \longrightarrow {\text{IV}}$For Isobaric process$\;\Delta {\text{W}}= {\text{P}} \Delta {\text{V}} \ne 0$ $\;\Delta {\text{U}}= {\text{nC}}_{ {\text{V}}} \Delta {\text{T}} \ne 0$Heat absorbed goes partly to increase internal energy and partly do work.