datavorous/entrance-exam-dataset huggingface dataset:
The energy required to charge a parallel plate condenser of plate separation $d$ and plate area of cross-section A such that the uniform electric field between the plates is E, is

Options:
A $\frac{1}{2} \varepsilon_0 E^2 A d$
B $\frac{1}{2} \varepsilon_0 E^2 / A . d$
C $\varepsilon_0 E^2 / A d$
D $\varepsilon_0 \mathrm{E}^2 A d$

The correct answer is:
$\frac{1}{2} \varepsilon_0 E^2 A d$

Solution:
$$
\text { Energy required }=\frac{1}{2} C V^2=\frac{1}{2} \varepsilon_0 E^2 \mathrm{Ad}
$$

Topic: Physics, Capacitance, NEET, NEET 2008 (Mains)
The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section $A$ such that the uniform electric field between the plates is $E$, is

Options:
A $\frac{1}{2} \varepsilon_0 E^2 / A d$
B $\varepsilon_0 E^2 / A d$
C $\varepsilon_0 E^2 A d$
D $\frac{1}{2} \varepsilon_0 E^2 A d$

The correct answer is:
$\varepsilon_0 E^2 A d$

Solution:
Energy given by the cell
$E=C V^2$
Here, $C=$ capacitance of condenser $=\frac{A \varepsilon_0}{d}$ $V=$ potential difference across the plates $=E d$ Therefore,
$\begin{aligned}
E & =\left(\frac{A \varepsilon_0}{d}\right)(E d)^2 \\
& =A \varepsilon_0 E^2 d
\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited stale to the ground state?

Options:
A 24.2nm
B 11.4nm
C 35.8nm
D 8.6nm

The correct answer is:
11.4nm

Solution:
hcλ=13.6eVz21n12-1n22
n1=1
n2=3
hcλ=13.6eV32112-132
⇒hcλ=13.6eV989
Wavelength =12408×13.6 nm
λ=11.39 nm

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2020 (09 Jan Shift 2)
The energy required to remove the electron from a singly ionized Helium atom is 2.2 times the energy required to remove an electron from helium atom. The total energy required to ionize the Helium atom completely is close to

Options:
A 34 eV
B 20 eV
C 79 eV
D 109 eV

The correct answer is:
79 eV

Solution:
The energy is required to remove e − from singly ionized helium atom= 54.4 eV. The energy required to remove e − from a helium atom =x eV, given 54.4 eV=2.2 x⇒x=24.73 eV. The energy required to ionize helium atom = 79.13 eV≈79 eV.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2018 (15 Apr)
The energy required to remove the electron from a singly ionized Helium atom is 2.2 times the energy required to remove an electron from helium atom. The total energy required to ionize the Helium atom completely is close to

Options:
A 34 eV
B 20 eV
C 79 eV
D 109 eV

The correct answer is:
79 eV

Solution:
The energy required to remove e − from singly ionized helium atom = 54.4 eV
The energy required to remove e − from helium atom =x eV
given 54.4 eV=2.2 x⇒x=24.73 eV
The energy required to ionize helium atom = 79.13 eV≈79 eV

Topic: Physics, Atomic Physics, JEE Main
The energy required to remove the electron from a singly ionized Helium atom is $2.2$ times the energy required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is:

Options:
A
$20 \mathrm{eV}$

B
$79 \mathrm{eV}$

C
$109 \mathrm{eV}$

D
$34 \mathrm{eV}$

The correct answer is:
$79 \mathrm{eV}$

Solution:
Energy required to remove $\mathrm{e}^{-}$from singly ionized helium atom $=\frac{(13.6) Z^2}{1^2}=54.4 \mathrm{eV}$ $(\because \mathrm{Z}=2)$ Energy required to remove $\mathrm{e}^{-}$from helium atom $=x \mathrm{eV}$
According to question, $54.4 \mathrm{eV}=2.2 \mathrm{x} \Rightarrow \mathrm{x}$ $=24.73 \mathrm{eV}$
Therefore, energy required to ionize helium atom $=(54.4+24.73) \mathrm{eV}=79.12 \mathrm{eV}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The energy required to take a satellite to a height h above the Earth surface (radius of Earth =6.4×103 km ) is E1, and the kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is

Options:
A 1.28×104km
B 6.4Ã103km
C 3.2Ã103km
D 1.6Ã103km

The correct answer is:
3.2×103km

Solution:
Usurface +E1=Uh

KE of satellite is zero at earth surface and at height h

-GMemRe+E1=-GMemRe+h

E1=GMem1Re-1Re+h

E1=GMemRe+h×hRe

Gravitational attraction FG=maC=mv2Re+h

⇒mv2Re+h=GMemRe+h2

mv2=GMemRe+h

E2=mv22=GMem2Re+h

E1=E2

hRe=12⇒h=Re2=3200 km

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The energy spectrum of Î²-particles [number N E as a function of Î²-energy E] emitted from a radioactive source is

Options:
A
B
C
D

The correct answer is:

Solution:
In β - decay, energy emission is continuous and energy is shared between β - particles and daughter nuclei.
Thus certain β -particles have zero energy and certain have maximum energy.
Option (1) is most appropriate

Topic: Physics, Nuclear Physics, JEE Main
The energy spectrum of $\beta$-particles [number $N(E)$ as a function of $\beta$-energy $E$ ] emitted from a radioactive source is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2006
The energy stored in a condenser of capacity $C$ which has been raised to a potential $V$ is given by

Options:
A $\frac{1}{2} \mathrm{CV}$
B $\frac{1}{2} C v^2$
C $\mathrm{CV}$
D $\frac{1}{2 V C}$

The correct answer is:
$\frac{1}{2} C v^2$

Solution:
$U=\int_0^V C V d V=\frac{1}{2} C V^2$

Topic: Physics, Capacitance, JEE Main
The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4 μC charge, its radius will be: Take:14π∈0=9×109 N m2 C-2

Options:
A 32 mm
B 16 mm
C 28 mm
D 20 mm

The correct answer is:
16 mm

Solution:
Energy of sphere =Q22C
4.5=16×10-122C
C=16×10-129=4πε0R
R=16×10-129×14πε0
R=16×10-129×14πε0
=9×109×169×10-12
=16×10-3=16 mm

Topic: Physics, Electrostatics, JEE Main, JEE Main 2017 (08 Apr Online)
The energy that should be added to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is

Options:
A four times the initial energy.
B equal to the initial energy.
C twice the initial energy.
D thrice the initial energy.

The correct answer is:
thrice the initial energy.

Solution:
de-Broglie wavelength
Î»=h2mE
â´ Î»1Î»2=E2E1 â1Ã10-90.5Ã10-9=E2E1
â 2= E2E1 âE2E1=4
â´ E2= 4E1
â´ Energy to be added = E2-E1
=4E1-E1=3E1

Topic: Physics, Atomic Physics, JEE Main
The energy that should be added to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is

Options:
A four times the initial energy
B equal to the initial energy
C twice the initial energy
D thrice the initial energy

The correct answer is:
thrice the initial energy

Solution:
De-Broglie Wavelength
Î» = h 2 mE
â´ Î» 1 Î» 2 = E 2 E 1
â 1 Ã 1 0 - 9 0.5 Ã 1 0 - 9 = E 2 E 1
â 2 = E 2 E 1
â E 2 E 1 = 4
â´ E 2 = 4 E 1
â´ Energy to be added = E 2 - E 1
= 4 E 1 - E 1 = 3 E 1

Topic: Physics, Dual Nature of Matter, JEE Main
The engine of a train moving with speed 10 m s-1 towards a platform sounds a whistle at frequency 400 Hz. The frequency heard by a passenger inside the train is: (Neglect air speed. Speed of sound in air =330 m s-1)

Options:
A 400 Hz
B 200 Hz
C 412 Hz
D 388 Hz

The correct answer is:
400 Hz

Solution:
The formula to calculate the frequency of the sound as heard by the passenger inside the train can be written as

f'=f0v-vOv-vs ...1

Since the passenger is inside the train, the speeds of the source and the observer are the same.

Substitute the values of the known parameters into equation (1) to get the required frequency.

f'=400330-10330-10 Hz=400 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The engine of a truck moving along a straight road delivers constant power. The distance travelled by the truck in time $t$ is proportional to

Options:
A $t$
B $t^{2}$
C $\sqrt{\mathrm{t}}$
D $t^{3 / 2}$

The correct answer is:
$t^{3 / 2}$

Solution:
$\frac{\mathrm{mv}^{2}}{\mathrm{t}}=\frac{\mathrm{ms}^{2}}{\mathrm{t}^{3}}$

since both $\mathrm{P}$ and $\mathrm{m}$ are constants

$\therefore \frac{\mathrm{s}^{2}}{\mathrm{t}^{3}}=$ constant

Topic: Physics, Units and Dimensions, JEE Main
The enolic form of given compound contains
CH3-CH2-C||O-CH3
x, σ bonds, y, bonds, z lone pairs. The sum value of (x + y + z) is

Options:

The correct answer is:
15

Solution:
σ=12,π=1,lp=2
So x = 12, y = 1, z = 2
So x + y + z
= 12 + 1 + 2 = 15

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The enthalpies of combustion of carbon and carbon monoxide are $-393.5$ and $-283 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The enthalpy of formation of carbon monoxide per mole is

Options:
A
$110.5 \mathrm{~kJ}$

B
$-110.5 \mathrm{~kJ}$

C
$-676.5$ kJ

D
$676.5 \mathrm{~kJ}$

The correct answer is:
$-110.5 \mathrm{~kJ}$

Solution:
$\begin{array}{ll}\mathrm{C}+\mathrm{O}_2 \longrightarrow \mathrm{CO}_2 & \Delta \mathrm{H}=-393.5 \mathrm{~kJ} \\ 2 \mathrm{CO}+1 / 2 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2 & \Delta \mathrm{H}=-283 \mathrm{~kJ} \\ 2 \mathrm{C}+\mathrm{O}_2 \longrightarrow 2 \mathrm{CO} & \Delta \mathrm{H}=-110 \mathrm{~kJ}\end{array}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2004
The enthalpy change for a reaction does not depend upon

Options:
A
use of different reactants for the same product

B
the nature of intermediate reaction steps

C
the differences in initial or final temperatures of involved substances

D
the physical states of reactants and products

The correct answer is:
the nature of intermediate reaction steps

Solution:
No Solution Available

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2003
The enthalpy change for the adsorption process and micelle formation respectively are

Options:
A ΔHads>0 and ΔHmic<0
B ΔHads>0 and ΔHmic>0
C ΔHads<0 and ΔHmic>0
D ΔHads<0 and ΔHmic<0

The correct answer is:
ΔHads<0 and ΔHmic>0

Solution:
Adsorption is exothermic process due to decrease in surface energy. Micelle formation is endothermic, it is due to For any spontaneous process entropy should always increase, hence, ΔS>0 for micelle formation and micelle formation decreases the stability of the colloidal solution so, energy of the mixture should increase which means ΔH>0.

The adsorption process ∆H=–ve. For micelle formation ∆H is +ve and ∆S is +ve.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (10 Apr Shift 1)
The enthalpy change for the conversion of 12Cl2 (g) to Cl-(aq) is (-)
kJmol-1 (Nearest integer)
Given : Δdis HCl2( g)0=240kJmol-1.
ΔegHCl(g)o=-350kJmol-1,
ΔhydHCl(g)-o=-380kJmol-1

Options:

The correct answer is:
610

Solution:
12Cl2( g) → Cl(g) → Cl(g)- → Cl(aq.)-

Apply Hess's law,

∆H = 12 ∆dis HCl2 (g)0 + ∆egHCl(g)o + ∆hydHCl(g)-o

⇒ ∆H=12× 240 + (-350) + (-380)

⇒ ∆H = -610

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (31 Jan Shift 1)
The enthalpy change for the transition of liquid water to steam, $\Delta H_{v a p}=37.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $373 \mathrm{~K}$. The entropy change for the process is

Options:
A $111.9 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
B $37.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
C $100 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$
D $74.6 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$

The correct answer is:
$100 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$

Solution:
$\Delta S_{v a p}=\frac{\Delta H_{v a p}}{T}=\frac{37.3 \mathrm{KJ} \mathrm{mol}^{-1}}{373 \mathrm{~K}}$
$=0.1 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}=100 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$.

Topic: Chemistry, Thermodynamics (C), JEE Main
The enthalpy change on freezing of 1 mol of water at 5°C to ice at -5°C is:
(Given Δfus H=6 kJ mol-1 at 0°C,
CpH2O, l=75.3 J mol-1K-1
CpH2O, s=36.8 J mol-1K-1 )

Options:
A 6.00 kJ mol-1
B 5.81 kJ mol-1
C 5.44 kJ mol-1
D 6.56 kJ mol-1

The correct answer is:
6.56 kJ mol-1

Solution:
In order to calculate the enthalpy change for H2O at 5°C to ice at -5°C, we need to calculate the enthalpy change of all the transformation involved in the process.

(a) Energy change of 1 mol, H2Ol, at 5°C→1 mol, H2O l, 0°C

(b) Energy change of 1 mol,H2O l at 0°C→1 mol, H2O s ice, 0°C

(c) Energy change of 1 mol, Ice s, at 0°C→1 mol, Ice (s), -5°C

Total ΔH

=CPH2O(l)ΔT+ΔH freezing +CPH2O(s)ΔT

=75.3 J mol-1 K-1(-5)K+-6×103 J mol-1+36.8 J mol-1 K-1(-5)K

ΔH=-6.56 kJ mol-1 (exothermic process)

So, ΔH=6.56 kJ mol-1.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2017 (08 Apr Online)
The enthalpy changes for the following processes are listed below:
$\begin{array}{ll}\mathrm{Cl}_2(\mathrm{~g})=2 \mathrm{Cl}(\mathrm{g}), & 242.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{I}_2(\mathrm{~g})=2 \mathrm{I}(\mathrm{g}), & 151.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{ICl}(\mathrm{g})=\mathrm{I}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}), & 211.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{I}_2(\mathrm{~s})=\mathrm{I}_2(\mathrm{~g}), & 62.76 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{array}$
Given that the standard states for iodine and chlorine are $\mathrm{I}_2(\mathrm{~s})$ and $\mathrm{Cl}_2(\mathrm{~g})$, the standard enthalpy of formation for $\mathrm{ICl}(\mathrm{g})$ is

Options:
A
$-14.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$

B
$-16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

C
$+16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

D
$+244.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$+16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$$
\begin{aligned}
& \frac{1}{2} \mathrm{I}_2(\mathrm{~s})+\frac{1}{2} \mathrm{Cl}_2 \longrightarrow \mathrm{ICl}(\mathrm{g}) \\
& \Delta \mathrm{H}=\left[\frac{1}{2} \Delta \mathrm{H}_{\mathrm{b}_2(s) \rightarrow k_2(g)}+\frac{1}{2} \mu_{1-1}+\frac{1}{2} \mu_{\mathrm{cl}-\mathrm{cl}}\right]-\left[\mu_{1-a}\right] \\
& =\left(\frac{1}{2} \times 62.76+\frac{1}{2} \times 151.0+\frac{1}{2} \times 242.3\right)-(211.3) \\
& =228.03-211.3 \\
& \Delta \mathrm{H}=16.73 \\
&
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2006
The enthalpy (H) of an elementary exothermic reaction $\mathrm{A} \rightleftharpoons \mathrm{B}$ is schematically plotted against the reaction coordinate. The plots in the presence and absence of a catalyst are shown in dashed and solid lines, respectively. Identify the correct plot for the reaction.

Options:
A
B
C
D

The correct answer is:

Solution:
For exothermic reaction
$\begin{aligned}
\Delta \mathrm{H}=& \mathrm{E}_{\mathrm{P}}-\mathrm{E}_{\mathrm{R}} < 0 \\
& \mathrm{E}_{\mathrm{B}}-\mathrm{E}_{\mathrm{A}} < 0 \\
& \mathbf{E}_{\mathrm{B}} < \mathbf{E}_{\mathrm{A}}
\end{aligned}$
Catalyst does not changes the initial and final position of the reaction so correct answer is (A)

Topic: Chemistry, Chemical Kinetics, JEE Main
The enthalpy of combustion at $25^{\circ} \mathrm{C}$ of $H_2$, cyclohexane $\left(C_6 H_{12}\right)$ and cyclohexene $\left(C_6 H_{10}\right)$ are $-241,-3920$ and $800 \mathrm{KJ} /$ mole respectively. The heat of hydrogenation of cyclohexene is

Options:
A $-121 \mathrm{KJ} / \mathrm{mole}$
B $+121 \mathrm{KJ} / \mathrm{mole}$
C $-242 \mathrm{KJ} / \mathrm{mole}$
D $+242 \mathrm{KJ} /$ mole

The correct answer is:
$-121 \mathrm{KJ} / \mathrm{mole}$

Solution:
(i) $\mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O}, \Delta \mathrm{H}=-241 \mathrm{~kJ}$
(ii) $\mathrm{C}_6 \mathrm{H}_{10}+\frac{17}{2} \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O}, \Delta \mathrm{H}=-3800 \mathrm{~kJ}$
(iii) $\mathrm{C}_6 \mathrm{H}_{12}+9 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O}, \Delta \mathrm{H}=-3920 \mathrm{~kJ}$
$\mathrm{C}_6 \mathrm{H}_{10}+\mathrm{H}_2 \rightarrow \mathrm{C}_6 \mathrm{H}_{12}$
Eq. (i) + Eq. (ii) - Eq. (iii)
$\Delta H=-241-3800-(-3920)$
$=-4041+3920=-121 \mathrm{~kJ}$

Topic: Chemistry, Thermodynamics (C), JEE Main
The enthalpy of neutralisation of NH4OH and CH3COOH is -10.5 kcal mol-1​ and enthalpy of neutralisation of CH3COOH with strong base is -12.5 kcal mol-1.The enthalpy of ionisation of NH4OH will be

Options:
A 4.0 kcal mol-1
B 3.0 kcal mol-1
C 2.0 kcal mol-1
D 3.2 kcal mol-1

The correct answer is:
2.0 kcal mol-1

Solution:
Î Hâ¡ neutr. for strong acid with strong base = - 1 3 Â· 7 kcal equiv -1
Enthalpy of ionisation of acetic acid & NH4OH together
=â10.5â( â13.7 )=+3 .2 kcal mol -1
Enthalpy of ionisation of only acetic acid
=â12.5â( â13.7 )=+1 .2 kcal mol -1
Enthalpy of ionisation of only NH4OH
=3.2â1.2= 2 kcal mol -1

Topic: Chemistry, Thermodynamics (C), JEE Main
The enthalpy of neutralisation of $\mathrm{NH}_4 \mathrm{OH}$ with $\mathrm{HCl}$ is $-51.46 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and the enthalpy of neutralisation of $\mathrm{NaOH}$ with $\mathrm{HCl}$ is $-55.90 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The enthalpy of ionisation of $\mathrm{NH}_4 \mathrm{OH}$ is

Options:
A
$-107.36 \mathrm{~kJ} \mathrm{~mol}^{-1}$

B
$-4.44 \mathrm{~kJ} \mathrm{~mol}^{-1}$

C
$+107.36 \mathrm{~kJ} \mathrm{~mol}^{-1}$

D
$+4.44 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$+4.44 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$\mathrm{HCl} \longrightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}$
Strong acid (Complete ionisation)
$$
\mathrm{NH}_4 \mathrm{OH} \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-}
$$
Weak base $\Delta \mathrm{H}=x \mathrm{~kJ} \mathrm{~mol}^{-1}$
$$
\begin{gathered}
\mathrm{H}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_2 \mathrm{O} \\
\Delta \mathrm{H}=-55.90 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{gathered}
$$
(from neutralisation of strong acid and strong base)
From equation (i), (ii) and (iii)
$$
\begin{aligned}
& \mathrm{NH}_4 \mathrm{OH}+\mathrm{HCl} \longrightarrow \mathrm{NH}_4^{+}+\mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O} \\
& \Delta \mathrm{H}=-51.46 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
\therefore \quad & x+(-55.90)=-51.46 \\
& x=-51.46+55.90 \\
& =4.44 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
$\therefore$ Enthalpy of ionisation of
$$
\mathrm{NH}_4 \mathrm{OH}=4.44 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2012 (19 May Online)
The enthalpy of neutralization is about $57.3 \mathrm{~kJ}$ for the pair

Options:
A $\mathrm{HCl}$ and $\mathrm{NH}_4 \mathrm{OH}$
B $\mathrm{NH}_4 \mathrm{OH}$ and $\mathrm{HNO}_3$
C $\mathrm{HCl}$ and $\mathrm{NaOH}$
D $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{NaOH}$

The correct answer is:
$\mathrm{HCl}$ and $\mathrm{NaOH}$

Solution:
Heat of neutralisation between strong acid and a strong base is about $-13.7 \mathrm{Kcal}$

Topic: Chemistry, Thermodynamics (C), JEE Main
The enthalpy of neutralization of which of the following acids and bases is nearly $-13.6 \mathrm{Kcal}$

Options:
A $\mathrm{HCN}$ and $\mathrm{NaOH}$
B $\mathrm{HCl}$ and $\mathrm{KOH}$
C $\mathrm{HCl}$ and $\mathrm{NaOH}$
D $\mathrm{HCl}$ and $\mathrm{NH}_4 \mathrm{OH}$

The correct answers are:
$\mathrm{HCl}$ and $\mathrm{KOH}$, $\mathrm{HCl}$ and $\mathrm{NaOH}$

Solution:
Correct option is $\mathrm{3} . \mathrm{HCl}$ and $\mathrm{NaOH}$
Heat of neutralisation of a strong acid and strong base is equal to $-13.7 \mathrm{kcal}$, which is always constant. (1 mole)
Options 2 and 3 are correct.

Topic: Chemistry, Thermodynamics (C), JEE Main
The enthalpy of vaporization of liquid water using the data:

(i). H2(g)+12O2gâH2O(l)

âH=-285.77 kJ/mol

(ii). H2g+12O2gâH2O(g) ,

âH=-241.84 kJ/mol in kJ/mol is

Options:
A +43.93
B -43.93
C +527.61
D -527.61

The correct answer is:
+43.93

Solution:
Enthalpy of vapourisation of H 2 O is
H 2 O l â H 2 O g
i.e. (ii)-(i)
Enthalpy of vaporization of liquid water =285.77-241.84=+43.93 kJ mol-1

Topic: Chemistry, Thermodynamics (C), JEE Main
The entire graph of the expression \( y=x^{2}+k x-x+9 \) is strictly above the \( x \)-axis if and only if

Options:
A \( k < 7 \)
B \( -5 < k < 7 \)
C \( k>-5 \)
D None of these

The correct answer is:
\( -5 < k < 7 \)

Solution:
y=x2+kx-x+9

Entire graph is above x-axis only if it does not have real roots and a>0.

⇒a>0, D<0.

⇒b2-4ac<0

⇒k-12-419<0

⇒k-12<36

⇒-6<k-1<6

⇒-5<k<7

Topic: Mathematics, Quadratic Equation, JEE Main
The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is:
(Specific heat of water liquid and water vapour are 4.2 kJ K- and 2.0 kJ K-1kg-1 ; heat of liquid fusion and vaporization of water are 334 kJ kg-1 and 2491 kJ kg-1 , respectively). ( log 273=2.436,log⁡373=2.572,log⁡383=2.583 )

Options:
A 9.26 kJ kg-1K-1
B 2.64Â kJÂ kg-1K-1
C 8.49Â kJÂ kg-1K-1
D 7.90Â kJÂ kg-1K-1

The correct answer is:
9.26 kJ kg-1K-1

Solution:
(i ) Phase change H 2 O( l )→ H 2 O( s )
So
(ii) Temperature change phase same (fusion)
H 2 O( l ) → Δ S 2 Δ H 2 O( l )
ΔS2= mS ln T2T1=1×4.2 ln 373273=1.31
(iii) Phase change (vapourisation)
ΔS3=ΔHvap373=2491373=6.67
(iv) Temperature change phase same
H2O(g)373 K→ΔS4H2O(g)383 K
ΔS4=2 ln383373=0.0529
ΔS  total=ΔS1+ΔS2+ΔS3+ΔS4=1.223+1.31+6.67+0.0529=9.26  kJ/kg K

Topic: Chemistry, Thermodynamics (C), JEE Main
The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is:
(Specific heat of water liquid and water vapour are 4.2 kJ K- and 2.0 kJ K-1kg-1 ; heat of liquid fusion and vaporization of water are 334 kJ kg-1 and 2491 kJ kg-1 , respectively). ( log 273=2.436,log⁡373=2.572,log⁡383=2.583 )

Options:
A 9.26 kJ kg-1K-1
B 2.64Â kJÂ kg-1K-1
C 8.49Â kJÂ kg-1K-1
D 7.90Â kJÂ kg-1K-1

The correct answer is:
9.26 kJ kg-1K-1

Solution:
(i ) Phase change H 2 O( l )→ H 2 O( s )
So
(ii) Temperature change phase same (fusion)
H 2 O( l ) → Δ S 2 Δ H 2 O( l )
ΔS2= mS ln T2T1=1×4.2 ln 373273=1.31
(iii) Phase change (vapourisation)
ΔS3=ΔHvap373=2491373=6.67
(iv) Temperature change phase same
H2O(g)373 K→ΔS4H2O(g)383 K
ΔS4=2 ln383373=0.0529
ΔS  total=ΔS1+ΔS2+ΔS3+ΔS4=1.223+1.31+6.67+0.0529=9.26  kJ/kg K

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (09 Jan Shift 2)
The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of $10 \mathrm{dm}^3$ to a volume of $100 \mathrm{dm}^3$ at $27^{\circ} \mathrm{C}$ is :

Options:
A
$35.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

B
$32.3 \mathrm{~J} \mathrm{~mol}{ }^{-1} \mathrm{~K}^{-1}$

C
$42.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

D
$38.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

The correct answer is:
$38.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

Solution:
For an ideal gas, for isothermal reversible process,
$$
\begin{aligned}
& \Delta S=2.303 \mathrm{nR} \log \left(\frac{\mathrm{v}_2}{\mathrm{v}_1}\right) \\
& =2.303 \times 2 \times 8.314 \times \log \left(\frac{100}{10}\right) \\
& =38.3 \mathrm{~J} \mathrm{~mol}^{-1} \cdot \mathrm{k}^{-1}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2011
The entropy change when an ideal gas under atmospheric condition at room temperature is allowed to expand from 0.5âL to 1.0âL and also is simultaneously heated to 373âK will be
(Given: C v,âm â=â12 .50 J K â1 mol â1 and logâ1.25â=â0.1 )

Options:
A 0.18 J K-1
B 0.36Â JÂ K-1
C 0.90Â JÂ K-1
D 0.72Â JÂ K-1

The correct answer is:
0.18 J K-1

Solution:
n=PVRT=0.02 moles
ΔS = nRln V 2 V 1 + nC v,m ln T 2 T 1
ΔS = 0.12 + 0.06
ΔS = 0.18  JK −1

Topic: Chemistry, Thermodynamics (C), JEE Main
The entropy of a sample of a certain substance increases by $0.836 \mathrm{~J} \mathrm{~K}^{-1}$ on adding reversibly $0.3344 \mathrm{~J}$ of heat at constant temperature. The temperature of the sample is:

Options:
A
$2.5 \mathrm{~K}$

B
$0.3 \mathrm{~K}$

C
$0.016 \mathrm{~K}$

D
$0.4 \mathrm{~K}$

The correct answer is:
$0.4 \mathrm{~K}$

Solution:
$\Delta S=\frac{\Delta H}{T} ; T=\frac{\Delta H}{\Delta S}=\frac{0.3344}{0.836}=0.4 \mathrm{~K}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2012 (07 May Online)