Q21JEE Main 2024MCQ

P-T diagram of an ideal gas having three different densities $\rho _1, \rho _2, \rho _3$ (in three different cases) is shown in the figure. Which of the following is correct :

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📖 Explanation

For ideal gas $\;P V=n R T$ $\;P V=\;\frac{m}{M} R T$ $\;P= (\;\frac{M}{V}) \;\frac{R T}{M}$ $\;P=\;\frac{\rho R T}{M}$ (Where ${\text{m}}$ is mass of gas and ${\text{M}}$ is molecular mass of gas)for same temperature ${\text{P}}_1 > {\text{P}}_2 > {\text{P}}_3$ So $\rho _1 > \rho _2 > \rho _3$ So correct answer is (2)

Q23JEE Main 2024MCQ

The speed of sound in oxygen at S.T.P. will be approximately:(Given, ${\text{R}}=8.3 {\text{JK}}^{-1}, \gamma =1.4$ ) [31-Jan-2024 Shift 2]

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📖 Explanation

$\; {\text{v}}=\sqrt{\;{\gamma {\text{RT}}}/{ {\text{M}}}}=\sqrt{\;\frac{1.4 \times 8.3 \times 273}{32 \times 10^{-3}}}$ $\;=314.8541 ≃ 315 {\text{m}} / {\ s}$

Q24JEE Main 2024MCQ

A thermodynamic system is taken from an original state $A$ to an intermediate state $B$ by a linear process as shown in the figure. It's volume is then reduced to the original value from ${\text{B}}$ to ${\text{C}}$ by an isobaric process. The total work done by the gas from A to B and B to C would be : [29-Jan-2024 Shift 1]

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📖 Explanation

Work done $\;\;\text{ AB }\;=\;\frac{1}{2}(8000+6000) \;\text{ Dyne }\; / {\text{cm}}^2 \times$ $\;4 {\text{m}}^3= (6000 {\text{Dyne}} / {\text{cm}}^2) \times 4 {\text{m}}^3$ Work done $\; {\text{BC}}=- (4000 {\text{Dyne}} / {\text{cm}}^2) \times 4 {\text{m}}^3$ Total work done $=2000 \;\text{ Dyne }\; / {\text{cm}}^2 \times 4 {\text{m}}^3$ $\;\; =2 \times 10^3 \times \;\frac{1}{10^5} \;{ {\text{N}}}/{ {\text{cm}}^2} \times 4 {\text{m}}^3$ $\;\; = 2 \times 10^{-2} \times \;{ {\text{N}}}/{10^{-4} {\text{m}}^2} \times 4 {\text{m}}^3$ $\;\; = 2 \times 10^2 \times 4 {\text{Nm}}=800 {\text{J}}$

Q25JEE Main 2024MCQ

Two vessels A and B are of the same size and are at same temperature. A contains $1 {\text{g}}$ of hydrogen and $B$ contains $1 {\text{g}}$ of oxygen. ${\text{P}}_{ {\text{A}}}$ and ${\text{P}}_{ {\text{B}}}$ are the pressures of the gases in A and ${\text{B}}$ respectively, then $\;{ {\text{P}}_{ {\text{A}}}}/{ {\text{P}}_{ {\text{B}}}}$ is : [29-Jan-2024 Shift 1]

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📖 Explanation

$\;\;\frac{P_A V_A}{P_B V_B}=\;\frac{n_A R T_A}{n_B R T_B}$ $\;\;\text{ Given }\; V_A=V_B$ $\;\;\text{ And }\; T_A=T_B$ $\;\;\frac{P_A}{P_B}=\;\frac{n_A}{n_B}$ $\;\;\frac{P_A}{P_B}=\;\frac{1 / 2}{1 / 32}=16$

Q26JEE Main 2024MCQ

${\text{N}}$ moles of a polyatomic gas $(f=6)$ must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of ${\text{N}}$ is : [29-Jan-2024 Shift 2]

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📖 Explanation

${\text{f}}_{ {\text{eq}}}=\;{ {\text{n}}_1 {\text{f}}_1+ {\text{n}}_2 {\text{f}}_2}/{ {\text{n}}_1+ {\text{n}}_2}$ For diatomic gas ${\text{f}}_{ {\text{eq}}}=5$ $\;5=\;{( {\text{N}})(6)+(2)(3)}/{ {\text{N}}+2}$ $\;5 {\text{N}}+10=6 {\text{N}}+6$ $\; {\text{N}}=4$

Q27JEE Main 2024MCQ

The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are $8 Ω$ and $10 Ω$ respectively. After inserting in a hot bath of temperature $400^{\circ} {\text{C}}$ , the resistance of platinum wire is :

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📖 Explanation

Given ${\text{R}}_0=8 Ω, {\text{R}}_{100}=10 Ω$ $\therefore {\text{R}}_{100}= {\text{R}}_0(1+\alpha \Delta {\text{T}})$ Also, ${\text{R}}_{400}= {\text{R}}_0 (1+\alpha \Delta {\text{T}}^1)$ $\therefore 10=8(1+\alpha \times 100) \Rightarrow 100 \alpha =\;\frac{1}{4}$ $\therefore {\text{R}}_{400}=8(1+400 \alpha )=8(1+1)=16 Ω$ Hence option (2)

Q28JEE Main 2024MCQ

The temperature of a gas is $-78^{\circ} {\text{C}}$ and the average translational kinetic energy of its molecules is ${\text{K}}$ . The temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2 {\text{K}}$ is :

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📖 Explanation

$\; {\text{K}} \cdot {\text{E}}=\;{ {\text{nf}}_1 {\text{RT}}}/{2}$ $\; {\text{T}}_{ {\text{i}}}=-78^{\circ} {\text{C}} \longrightarrow 273+ \$ [-78^{\circ} {\text{C}}]$ =195 {\text{K}} $K.E$ \alpha {\text{T}} $To double the K.E energy temp also become double$ ; {\text{T}}{ {\text{f}}}=390 {\text{K}}; {\text{T}}{ {\text{f}}}=117^{\circ} {\text{C}}$

Q29JEE Main 2024MCQ

Energy of 10 non rigid diatomic molecules at temperature ${\text{T}}$ is :

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📖 Explanation

$\;\;\text{ Degree of freedom }\;(f)=5+2(3 {\text{N}}-5)$ $\; \;\;\;\; {\text{f}}=5+2(3 \times 2-1)=7$ $\;\;\text{ energy of one molecule }\;=\;{ {\text{f}}}/{2} {\text{K}}_{ {\text{B}}} {\text{T}}$ $\;\;\text{ energy of }\; 10 \;\text{ molecules }\;$ $\;=10 (\;{ {\text{f}}}/{2} {\text{K}}_{ {\text{B}}} {\text{T}}) =10 (\;\frac{7}{2} {\text{K}}_{ {\text{B}}} {\text{T}}) =35 {\text{K}}_{ {\text{B}}} {\text{T}}$

Q30JEE Main 2024MCQ

A block of ice at $-10^{\circ} {\text{C}}$ is slowly heated and converted to steam at $100^{\circ} {\text{C}}$ . Which of the following curves represent the phenomenon qualitatively: [30-Jan-2024 Shift 2]

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Q31JEE Main 2024MCQ

If three moles of monoatomic gas $(\gamma =\;\frac{5}{3})$ is mixed with two moles of a diatomic gas $(\gamma =\;\frac{7}{5})$ , the value of adiabatic exponent $\gamma$ for the mixture is: [30-Jan-2024 Shift 2]

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📖 Explanation

$\; {\text{f}}_1=3, \;\; {\text{f}}_2=5$ $\; {\text{n}}_1=3, \;\; {\text{n}}_2=2$ $\; {\text{f}}_{\;\text{mixture }\;}=\;{ {\text{n}}_1 {\text{f}}_1+ {\text{n}}_2 {\text{f}}_2}/{ {\text{n}}_1+ {\text{n}}_2}$ $=\;{9+10}/{ {\text{f}}}=\;\frac{19}{5}$ $\;\gamma _{\;\text{mixture }\;}=1+\;\frac{2 \times 5}{19}$ $=\;\frac{29}{19}=1.52$

Q32JEE Main 2024MCQ

The temperature of a gas having $2.0 \times 10^{25}$ molecules per cubic meter at $1.38 {\text{atm}}$ (Given, ${\text{k}}=$ $1.38 \times 10^{-23} {\text{JK}}^{-1})$ is : [29-Jan-2024 Shift 2]

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📖 Explanation

$\; {\text{PV}}= {\text{nRT}}$ $\; {\text{PV}}=\;{ {\text{N}}}/{ {\text{N}}_{ {\text{A}}}} {\text{RT}}$ $\; {\text{N}}=\;\text{ Total no. of molecules }\;$ $\; {\text{P}}=\;{ {\text{N}}}/{ {\text{V}}} {\text{kT}}$ $\;1.38 \times 1.01 \times 10^5=2 \times 10^{25}$ $\times 1.38 \times 10^{-23} \times {\text{T}}$ $\;1.01 \times 10^5=2 \times 10^2 \times {\text{T}}$ $\; {\text{T}}=\;\frac{1.01 \times 10^3}{2} \approx 500 {\text{K}}$

Q33JEE Main 2024MCQ

A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is [31-Jan-2024 Shift 2]

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📖 Explanation

$\; {\text{U}}= {\text{nC}}_{ {\text{V}}} {\text{T}}$ $\;\Rightarrow {\text{U}}= {\text{n}}_1 {\text{C}}_{ {\text{V}}_1} {\text{T}}+ {\text{n}}_2 {\text{C}}_{ {\text{V}}_2} {\text{T}}$ $\;\Rightarrow 8 \times \;{3 {\text{R}}}/{2} \times {\text{T}}+6 \times \;{5 {\text{R}}}/{2} \times {\text{T}}$ $\;=27 {\text{RT}}$

Q34JEE Main 2024MCQ

The total kinetic energy of 1 mole of oxyge $27^{\circ} {\text{C}}$ is :[Use universal gas constant $( {\text{R}})=8.31 {\text{J}} / {\text{mole}} {\text{K}}$ [27-Jan-2024 Shift 2]

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📖 Explanation

$\;\;\text{ Kinetic energy }\;=\;{ {\text{f}}}/{2} {\text{nRT}}$ $\;=\;\frac{5}{2} \times 1 \times 8.31 \times 300 {\text{J}}$ $\;=6232.5 {\text{J}}$

Q35JEE Main 2024MCQ

The given figure represents two isobaric processes for the same mass of an ideal gas, then [31-Jan-2024 Shift 1]

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📖 Explanation

$\; {\text{PV}}= {\text{nRT}}$ $\; {\text{V}}= (\;{ {\text{nR}}}/{ {\text{P}}}) {\text{T}}$ $\;\;\text{ Slope }\;=\;{ {\text{nR}}}/{ {\text{P}}}$ $\;\;\text{ Slope }\; \propto \;{1}/{ {\text{P}}}$ $\;(\;\text{ Slope }\;)_2>(\;\text{ Slope }\;)_1$ $\; {\text{P}}_2< {\text{P}}_1$

Q36JEE Main 2024MCQ

The volume of an ideal gas $(\gamma =1.5)$ is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is:

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📖 Explanation

For Adiabatic process $\;P_i V_i=P_f V_f^\gamma$ $\;P_i(5)^{1.5}=P_f(4)^{1.5}$ $\;\;\frac{P_i}{P_f}= (\;\frac{4}{5}) ^{\;\frac{3}{2}}=\;\frac{4}{5} \cdot (\;\frac{4}{5}) ^{\;\frac{1}{2}} \Rightarrow \;{8}/{5 {\sqrt{5}}}$

Q37JEE Main 2024MCQ

If ${\text{n}}$ is the number density and ${\text{d}}$ is the diameter of the molecule, then the average distance covered by a molecule between two successive collisions (i.e. mean free path) is represented by :

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📖 Explanation

${\text{n}}=$ number of molecule per unit volume ${\text{d}}=$ diameter of the molecule $\lambda =\;{1}/{{\sqrt{2}} \pi {\text{d}}^2 {\text{n}}}$ (By Theory)

Q38JEE Main 2024MCQ

During an adiabatic process, if the pressure of a gas is found to be proportional to the cube of its absolute temperature, then the ratio of $\;\frac{C_P}{C_V}$ for the gas is :

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📖 Explanation

${\text{P}} \propto {\text{T}}^3$ ${\text{PT}}^{-3}=$ constant $\because \;{ {\text{PV}}}/{ {\text{T}}}= {\text{nR}}=$ constant from ideal gas equation(P) $( {\text{PV}})^{-3}=$ constant ${\text{P}}^{-2} {\text{V}}^{-3}=$ cosntant $\because$ Process equation for adiabatic process is ${\text{PV}}^{ {\text{y}}}=$ constantComparing equation (1) and (2) $\;{ {\text{C}}_{ {\text{P}}}}/{ {\text{C}}_{ {\text{V}}}}= {\text{y}}=\;\frac{3}{2}$

Q39JEE Main 2024MCQ

The translational degrees of freedom ( $f_t)$ and rotational degrees of freedom $(f_r)$ of ${\text{CH}}_4$ molecule are :

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📖 Explanation

Since ${\text{CH}}_4$ is polyatomic Non-LinearD.O.F of ${\text{CH}}_4$ T. ${\text{DOF}}=3$ ${\text{R}} D O F=3$

Q40JEE Main 2024MCQ

Given below are two statements :Statement (I) : Dimensions of specific heat is $[ {\text{L}}^2 {\text{T}}^{-2} {\text{K}}^{-1}]$ Statement (II) : Dimensions of gas constant is $[ {\text{M}} {\text{L}}^2 {\text{T}}^{-1} {\text{K}}^{-1}]$

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📖 Explanation

$\;\Delta {\text{Q}}= {\text{mS}} \Delta {\text{T}}$ $\; {\ s}=\;{\Delta {\text{Q}}}/{ {\text{m}} \Delta {\text{T}}}$ $\;{[ {\ s}]= \$ [;{ {\text{ML}}^2 {\text{T}}^{-2}}/{ {\text{MK}}}]$ };{[ {\ s}]= $[ {\text{L}}^2 {\text{T}}^{-2} {\text{K}}^{-1}]$ } $Statement-(I) is correct$ ; {\text{PV}}= {\text{nRT}} \Rightarrow {\text{R}}=;{ {\text{PV}}}/{ {\text{nT}}};{[ {\text{R}}]=;{[ {\text{ML}}^{-1} {\text{T}}^{-2}]$ $[ {\text{L}}^3]$ }/{[ {\text{mol}}][ {\text{K}}]}};{[ {\text{R}}]= $[ {\text{ML}}^2 {\text{T}}^{-2} {\text{mol}}^{-1} {\text{K}}^{-1}]$ }$Statement-II is incorrect