Q241JEE Main 2020MCQ

The plot that depicts the behavior of the mean free time t (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature (T), qualitatively, is: (Graphs are schematic and not drawn to scale)

🚀 Solve in Practice Mode

📖 Explanation

Mean free time $=\text{ Mean free path}/\text{Average speed}$ $={1/{\sqrt2\pi D^2n}}/{\sqrt{\frac{8RT}{\pi M_w}}}$ $t\propto 1/\text{sqrtT}$

Q242JEE Main 2020NAT

Nitrogen gas is at $300°{C}$ temperature. The temperature (in ${K}$ ) at which the ${\text{rms}}$ speed of a ${H}_{2}$ molecule would be equal to the rms speed of a nitrogen molecule, is _____. (Molar mass of ${N}_{2}$ gas $28{g}$ )

🚀 Solve in Practice Mode

📖 Explanation

${V}_{ {\text{rms}}} =\\sqrt{ \frac{3 {RT}}{ {M}}}$ ${V}_{{N}_{2}} = {V}_{ {H}_{2}}$ $\sqrt{{3 {RT}_{ {N}_{2}}}/{ {M}_{ {N}_{2}}}}=\sqrt{{3 {RT}_{ {H}_{2}}}/{ {M}_{ {H}_{2}}}}$ $\frac{573}{28}={{T}_{ {H}_{2}}}/{2}$ $\Rightarrow {T}_{ {H}_{2}}=40.928$

Q243JEE Main 2020MCQ

A bullet of mass $5 \ { g}$ , travelling with a speed of 210 $\ {m} / {s},$ strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is $0.030 \ {cal} /({g}-{ }°{C} )$ (1 cal $=4.2 \\times 10^{7}$ ergs) close to :

🚀 Solve in Practice Mode

📖 Explanation

$\frac{1}{2} {mv}^{2}\times \frac{1}{2}= {ms} \Delta{T}$ $\Delta{T}=\frac{{v}^{2}}{4 \\times 5}$ $= \frac{210^{2}}{4 \\times 30 \\times 4.200}$ $=87.5°{C}$

Q244JEE Main 2020NAT

An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to ________.

🚀 Solve in Practice Mode

📖 Explanation

${\text{W}}_{{\text{ABCDA}}}=2 {\text{P}}_{0} {\text{V}}_{0}$ ${\text{Q}}_{{\in}}={\text{Q}}_{{\text{AB}}}+{\text{Q}}_{{\text{BC}}}$ ${\text{Q}}_{{\text{AB}}}={\text{nC}}({\text{T}}_{{\text{B}}}-{\text{T}}_{{\text{A}}})$ $={{\text{n}} 3 {\text{R}}}/{2}({\text{T}}_{{\text{B}}}-{\text{T}}_{{\text{A}}})$ $=\frac{3}{2}({\text{P}}_{{\text{B}}} {\text{V}}_{{\text{B}}}-{\text{P}}_{{\text{A}}} {\text{V}}_{{\text{A}}})$ $=\frac{3}{2}(3 {\text{P}}_{{\text{B}}} {\text{V}}_{0}={\text{P}}_{0} {\text{V}}_{0})$ $=3 {\text{P}}_{0} {\text{V}}_{0}$ ${\text{Q}}_{{\text{BC}}}={\text{nC}}_{{\text{P}}}({\text{T}}_{{\text{C}}}-{\text{T}}_{{\text{B}}})$ $={{\text{n}} 5 {\text{R}}}/{2}({\text{T}}_{{\text{C}}}-{\text{T}}_{{\text{B}}})$ $=\frac{5}{2}({\text{P}}_{{\text{C}}} {\text{V}}_{{\text{C}}}-{\text{P}}_{{\text{B}}} {\text{V}}_{{\text{B}}})$ $=\frac{5}{2}(6 {{\text{P}}}_{0} {\text{V}}_{0}-{3} {{\text{P}}}_{0} {\text{V}}_{0})$ $=\frac{15}{2} {\text{P}}_{0} {\text{V}}_{0}$ $\eta ={{\text{W}}}/{{\text{Q}}_{{\in}}} \times 100$ $={2 {\text{P}}_{0} {\text{V}}_{0}}/{3 {\text{P}}_{0} {\text{V}}_{0}+{15}{2} {\text{P}}_{0} {\text{V}}_{0}} \times 100$ $\eta =\frac{400}{21}=19.04 \approx 19$ $\eta =19$

Q245JEE Main 2020MCQ

Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom. The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of $\gamma (={{C}_{{P}}}/{{C}_{{v}}})$ given, respectively, by:

🚀 Solve in Practice Mode

📖 Explanation

Total degree of freedom $=3+2=5$ ${U}={{\text{nfRT}}}/{2} \Rightarrow {5 {\text{RT}}}/{2}$ $\gamma \Rightarrow {\text{C}_{\text{P}}}/{\text{C}_{\text{V}}} \Rightarrow 1+\frac{2}{f}$ $\Rightarrow 1+\frac{2}{5} \Rightarrow \frac{7}{5}$

Q246JEE Main 2020MCQ

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, $T _1$ and $T_2$ . The temperature of the hot reservoir of the first engine is $T _1$ and the temperature of the cold reservoir of the second engine is $T_2$ . T is temperature of the sink of first engine which is also the source for the second engine. How is T related to $T _1$ and $T _2$ , if both the engines perform equal amount of work ?

🚀 Solve in Practice Mode

📖 Explanation

$Q_H/Q_L=T_1/T$ and $W=Q_H-Q_L$ ....(1) $Q_L/Q'_L=T /T_2$ and $W=Q_L-Q'_L$ ....(2)

Q247JEE Main 2020MCQ

A thermodynamic cycle xyzx is shown on a V-T diagram.The P-V diagram that best describes this cycle is : (Diagrams are schematic and not to scale)

🚀 Solve in Practice Mode

Q248JEE Main 2020MCQ

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from $\tau _1$ to $\tau _2$ . If $C_p/C_v=\gamma$ for this gas then a good estimate for $\tau _2/ \tau _1$ is given by :

🚀 Solve in Practice Mode

📖 Explanation

$\;\text{ Mean free path, }\; \lambda =(\;\frac{ R T }{\sqrt{ 2 } \pi d ^{2} N _{ A } P })$ $V _{ rms }=\sqrt{\;\frac{ 3 R T }{ M }}$ $\therefore \tau =\;\frac{\lambda }{ V _{ rms }}=\;\frac{ R T }{\sqrt{ 2 }\pi d^{2}{N_A} \;P} \times \sqrt{\;\frac{ M }{ 3 R T }}$ $\tau \propto \;\frac{\sqrt{ T }}{ P }$ $\;\frac{\tau _{1}}{\tau _{2}}=\;{\sqrt{ T _{1}}}/{ P _{1}} \times \;{ P _{2}}/{\sqrt{ T _{2}}}=(\;\frac{ P _{2}}{ P _{1}}) \sqrt{\;\frac{ T _{1}}{ T _{2}}}$ $PV = nRT$ $\;\frac{ P _{1}}{ P _{2}} \times \;\frac{ V _{1}}{ V _{2}}=\;\frac{ T _{1}}{ T _{2}}$ $\Rightarrow 2^{\gamma } \times \;\frac{ V }{2 V }=\;\frac{ T _{1}}{ T _{2}}$ $\Rightarrow \;\frac{ T _{1}}{ T _{2}}=(2^{\gamma -1})$ $\therefore \;\frac{\tau _{1}}{\tau _{2}}=(\;\frac{ P _{2}}{ P _{1}}) \times \sqrt{\;\frac{ T _{1}}{ T _{2}}}=\;\frac{1}{2^{\gamma }} \times \sqrt{2^{\gamma -1}}=\;{2^{\frac{\gamma -1}{2}}}/{2^{\gamma }}$ $\;\frac{\tau _{2}}{\tau _{1}}=\;{2^{\gamma }}/{2^{(\;\frac{\gamma -1}{2})}}=2^{\gamma -(\;\frac{\gamma -1}{2})}=2^{\;\frac{\gamma +1}{2}}$ $\Rightarrow \;\frac{\tau _{1}}{\tau _{2}}=(\;\frac{1}{2})^{\;\frac{\gamma +1}{2}}$

Q249JEE Main 2020MCQ

A litre of dry air at STP expands adiabatically to a volume of 3 litres. If $\gamma = 1.40$ , the work done by air is : $(3^1.4 = 4.6555)$ [Take air to be an ideal gas] [7-Jan-2020 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$w=\frac{nR(T_1-T_2)}{\gamma -1}$ $={P_1V_1-P_2V_2}/0.4$ $={100-100/4.6555\times 3}/0.4=88.90$

Q250JEE Main 2020MCQ

Match the thermodynamic processes taking place in a system with the correct conditions. In the table : $\DeltaQ$ is the heat supplied, $\DeltaW$ is the work done and $\Delta{U}$ is change in internal energy of the system : Process Condition (I) Adiabatic (A) $\Delta{W}=0$ (II) Isothermal (B) $\DeltaQ=0$ (III) Isochoric (C) $\Delta{U} \ne 0, \Delta{W} \ne 0$ , $\Delta Q \ne 0$ (IV) Isobaric (D) $\Delta{U}=0$

🚀 Solve in Practice Mode

📖 Explanation

(I) Adiabatic process $\Rightarrow\DeltaQ=0$ No exchange of heat takes place with surroundings (II) Isothermal proess $\Rightarrow$ Temperature remains constant $(\Delta{T}=0)$ $\Delta{u}=\frac{{F}}{2} {nR} \Delta{T} \Rightarrow \Delta{u}=0$ No change in internal energy $[\Delta{u}=0]$ (III) Isochoric process Volume remains constant $\Delta{V}=0$ ${W}=∫{P.dV}=0$ Hence work done is zero. (IV) Isobaric process $\Rightarrow$ Pressure remains constant ${W}= {P.} \Delta{V} \ne 0$ $\Delta{u}=\frac{{F}}{2} {nR} \Delta{T}$ $=\frac{{F}}{2}[ {P} \Delta{V}] \ne 0$ $\Delta{Q}= {nC}_{ {p}} \Delta{T} \ne 0$

Q251JEE Main 2020MCQ

The specific heat of water $=4200 {J} {kg}^{-1} {K}^{-1}$ and the latent heat of ice $=3.4\times 10^{5} \ {J} \ {kg}^{-1}$ 100 grams of ice at $0° {C}$ is placed in $200 \ {g}$ of water at $25°{C}$ . The amount of ice that will melt as the temperature of water reaches $0°{C}$ is close to (in grams):

🚀 Solve in Practice Mode

📖 Explanation

Here the water will provide heat for ice to melt therefore ${m}_{ {w}} {s}_{ {w}} \Delta\theta = {m}_{ {\text{ice}}} {L}_{ {\text{ice}}}$ $\ {m}_{ {\text{ice}}}=\ \frac{0.2 \\times 4200 \\times 25}{3.4 \\times 10^{5}}$ $=0.0617 \ {kg}$ $=61.7 \ {gm}$ Remaining ice will remain un-melted so correct answer is 1

Q252JEE Main 2020MCQ

Three different processes that can occur in an ideal monoatomic gas are shown in the $\ {P}$ vs $\ {V}$ diagram. The paths are labelled as $\ {A} \to {B}, {A}\to {C}$ and $\ {A}\to {D} .$ The change ininternal energies during these process are taken as $\ {E}_{ {AB}}, {E}_{ {AC}}$ and ${E}_{ {AD}}$ and the workdone as ${W}_{ {AB}}$ ${W}_{ {AC}}$ and ${W}_{ {AD}}$ . The correct relation between these parametersare :

🚀 Solve in Practice Mode

📖 Explanation

(b) Temperature change $\Delta T$ is same for all three processes $A \to B ; A \to C$ and $A \to D$ $\Delta U=n C_{v} \Delta T=$ same $E_{A B}=E_{A C}=E_{A D}$ Work done, $W=P \times \Delta V$ $A B \to$ volume is increasing $\Rightarrow W_{A B}>0$ $A D \to$ volume is decreasing $\Rightarrow W_{A D}<0$ $A C \to$ volume is constant $\Rightarrow W_{A C}=0$

Q253JEE Main 2020MCQ

Two gases-argon (atomic radius 0.07 nm, atomic weight 40) and xenon (atomic radius 0.1 nm, atomic weight 140) have the same number density and are at the same temperature. The raito of their respective mean free times is closest to :

🚀 Solve in Practice Mode

📖 Explanation

$\lambda =1/{\sqrt2\pi n_vd^2}$ $\tau =\lambda /v=1/{\sqrt2\pi n_vd^2v}$ $=1/{\sqrt2\pi n_vd^2}\sqrt{m/{3RT}}$ $\tau _1/\tau _2=\sqrt{M_1/M_2}d_2^2/d_1^2$ $=\sqrt{40/140} (0.1)^2/(0.07)^2=1.09$ ∴ Nearest possible answer (3)

Q254JEE Main 2020MCQ

A heat engine is involved with exchange ofheat of $1915 {J},-40 {J},+125 {J}$ and ${QJ},$ duringone cycle achieving an efficiency of $50.0 \%$ . The value of ${Q}$ is:

🚀 Solve in Practice Mode

📖 Explanation

$\eta = {\text{Work done }}{\text{ Heat supplied}}$ $\frac{1}{2}=\eta = \frac{1915-40+125-{Q}}{1915+125}$ $\frac{1}{2}= \frac{2000- {Q}}{2040}$ $2040=4000-2 {Q}$ $2 \ {Q}=1960$ ${Q}=980 {J}$

Q255JEE Main 2020MCQ

Two different wires having lengths $L_{1}$ and $L_{2}$ , and respective temperature coefficient of linear expansion $\alpha _{1}$ and $\alpha _{2},$ are joined end-to-end. Then the effective temperature coefficient of linear expansion is :

🚀 Solve in Practice Mode

📖 Explanation

At ${T}°{C} \;\;\;{L}= {L}_{1}+ { L}_{2}$ At ${T}+\Delta{T} \;\;\;{L}'_{ {\text{eq}}}= {L}'_{1}+ { L}'+_{2}$ where $L_{1}'=L_{1} (1+\alpha _{1} \DeltaT )$ ${L}_{2}'= {L}_{2} (1+\alpha _{2} {\DeltaT} )$ ${L}_{{\text{eq}}}'=({L}_{1}+{L}_{2} ) (1+\alpha _{ {\text{avg}}} \Delta{T} )$ $\Rightarrow ({L}_{1}+{L}_{2}$ (1+\alpha { {\text{avg}}} {ST} )= {L}{1}+{L}{2}+{L}{1} \alpha {1} \Delta{T}+ {L}{2} \alpha {2} \Delta{T}\Rightarrow ({L}{1}+{L}_{2} ) \alpha { {\text{avg}}}= {L}{1} \alpha {1}+{L}{2} \alpha {2}\Rightarrow \alpha { {\text{avg}}}=\frac{{L}{1} \alpha {1}+{L}{2} \alpha {2}}{{L}{1}+{L}{2}}$

Q256JEE Main 2020NAT

A closed vessel contains $0.1 \ {\text{mole}}$ of a monoatomic ideal gas at 200 K. If 0.05 mole of the same gas at $400 \ {K}$ is added to it, the final equilibrium temperature (in $\ {K}$ ) of the gas in the vessel will be closed to ______.

🚀 Solve in Practice Mode

📖 Explanation

As work done on gas and heat supplied to the gas are zero, total internal energy of gases remain same $\ {u}_{1}+\ {u}_{2}=\ {u}_{1}'+\ {u}_{2}'$ $(0.1) {C}_{ {v}}(200)+(0.05) {C}_{ {v}}(400)$ $=(0.15) {C}_{ {v}} {T}$ ${T}= \frac{800}{3} {k}=266.67 \ {k}$

Q257JEE Main 2020MCQ

Consider a gas of triatomic molecules. Themolecules are assumed to the triangular andmade of massless rigid rods whose vertices areoccupied by atoms. The internal energy of a mole of the gas at temperature $\ {T}$ is :

🚀 Solve in Practice Mode

📖 Explanation

$\ {DOF}=3+3=6$ $\ {U}=\frac{{f}}{2} {nRT}=3 \ {RT}$

Q258JEE Main 2020MCQ

To raise the temperature of a certain mass of gas by $50°{C}$ at a constant pressure, 160 calories of heat is required. When the same mass of gas is cooled by $100°{C}$ at constant volume, 240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?

🚀 Solve in Practice Mode

📖 Explanation

${nC}_{ {P}}(50)=160$ ${nC}_{ {v}}(100)=240$ $\Rightarrow \frac{C_{p}}{2 C_{v}}= \frac{160}{240}= \frac{\gamma }{2}$ $\therefore \gamma = \frac{4}{3}$ and $f= \frac{2}{\gamma -1}=6$

Q259JEE Main 2020MCQ

Match the ${C}_{ {p}} / {C}_{ {V}}$ ratio for ideal gases with different type of molecules: Molecular type ${C}_{ {p}} / {C}_{ {V}}$ (A) Monoatomic (I) $7 / 5$ (B) Diatomic rigidmolecules (II) $9 / 7$ (C) Diatomic non-rigidmolecules (III) $4 / 3$ (D) Triatomic rigidmolecules (IV) $5 / 3$

🚀 Solve in Practice Mode

📖 Explanation

$\gamma = \frac{C_{p}}{C_{v}}=1+ \frac{2}{f}$ where 'f' is degree of freedom (A) Monoatomic $\ {f}=3, \gamma =1+ \frac{2}{3}= \frac{5}{3}$ (B) Diatomic rigid molecules, ${f}=5, \gamma =1+\ \frac{2}{3}= \frac{7}{5}$ (C) Diatomic non-rigid molecules ${f}=7, \gamma =1+ \frac{2}{7}= {9}/ {7}$ (D) Triatomic rigid molecules ${f}=6, \gamma =1+ \frac{2}{6}= \frac{4}{3}$

Q260JEE Main 2020NAT

A bakelite beaker has volume capacity of $500 \ {cc}$ at $30°{C}$ . When it is partially filled with $\ {V}_{ {m}}$ volume(at $30°$ ) of mercury, it is found that the unfilledvolume of the beaker remains constant astemperature is varied. If $\gamma _{(\text{beaker })}=6 \\times 10^{-6}{ }°{C}^{-1}$ and $\gamma _{(\text{mercury })}=1.5 \\times 10^{-4}{ }°{C}^{-1},$ where $\gamma$ is thecoefficient of volume expansion, then $\ {V}_{ {m}}$ (in cc) is close to _____.

🚀 Solve in Practice Mode

📖 Explanation

$[\Delta{V}=({V}_{0}- {V}_{ {m}} )$ After increasing temperature $\Delta{V}'=({{V}_{0}'- {V}_{ {m}}'})$ $\Delta{V}'=\Delta{V}$ ${V}_{0}- {V}_{ {m}}= {V}_{0} (1+\gamma _{ {b}} \Delta{T} )$ $- {V}_{ {m}}(1+\gamma _{ {M}} \Delta{T} )$ $\ {V}_{0} \gamma _{ {b}}=\ {V}_{ {m}} \gamma _{ {m}}$ ${V}_{ {m}}={{V}_{0} \gamma _{ {b}}}/{\gamma _{ {m}}}$ $= \frac{(500) (6 \\times 10^{-6} )}{ (1.5 \\times 10^{-4} )}$ $=20 \ {CC}$