datavorous/entrance-exam-dataset huggingface dataset:
An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M. The piston and the cylinder have an equal cross-sectional area A. Atmospheric pressure is p0 and when the piston is in equilibrium, the volume of the gas is V0. The piston is now displaced slightly from its equilibrium position. Assuming that the system, is completely isolated from, its surroundings, what is the frequency of oscillation. An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M. The piston and the cylinder have an equal cross-sectional area A. Atmospheric pressure is p0 and when the piston is in equilibrium, the volume of the gas is V0. The piston is now displaced slightly from its equilibrium position. Assuming that the system, is completely isolated from, its surroundings, what is the frequency of oscillation.

Options:
A f=12πγ p0A2+MgAV0M
B fâ¡ = 1 2 π 1 γ p 0 A 2 + Mg A V 0 M
C fâ¡ = 1 2 π p 0 A 2 + Mg A V 0 M
D fâ¡ = 1 2 π A p 0 A + Mg A V 0 M

The correct answer is:
f=12πγ p0A2+MgAV0M

Solution:
In equilibrium, pressure inside the cylinder = pressure just outside it
or p = p 0 + Mg A
When piston is displaced slightly by an amount x, change in volume,
dV = - Ax
Since, the cylinder is isolated from the surroundings, nature of the process is adiabatic. In adiabatic process,
dp dV = - Î³ p V
or increase in pressure inside the cylinder,

dp = - Î³ p V dV = Î³ p 0 + Mg A V 0 Ax â
This increase in pressure when multiplied with area of cross-section A will give a net upward force (or the restoring force). Hence,
F = - dp A = - Î³ p 0 A 2 + MgA V 0 x
Or a = F M = - Î³ p 0 A 2 + MgA V 0 M x
Since, a â - x , motion of the piston is simple harmonic in nature. Frequency of this oscillation is given by
f = 1 2 Ï a x = 1 2 Ï Î³ p 0 A 2 + MgA V 0 M

Topic: Physics, Oscillations, JEE Main
An ideal gas is expanded from p1,V1,T1 to (p2,V2,T2) under different conditions. The incorrect statement among the following is?

Options:
A The word done by the gas is less when it is expanded reversibly from V1 to V2 under adiabatic conditions as compared to that when expanded reversibly from V1 to V2 under isothermal conditions
B The change in internal energy of the gas is (i) zero, if it is expanded reversibly with T1=T2 and (ii) positive, if it is expanded reversibly under adiabatic conditions with T1≠T2
C If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic
D The work done on the gas is maximum when it is compressed irreversibly from p2,V2 to p1,V1 against constant pressure p1

The correct answer is:
The change in internal energy of the gas is (i) zero, if it is expanded reversibly with T1=T2 and (ii) positive, if it is expanded reversibly under adiabatic conditions with T1≠T2

Solution:
Area under curve in reversible isothermal is more. So, more work will be done by gas.
T1=T2
ΔU=nCVΔT=0
In reversible adiabatic expansion, T2<T1
So ΔT=-ve , ΔU=-ve
In free expansion, Pext=0
So W=0
If carried out isothermally ΔU=0
q=0 (adiabatic); from I law
If carried out adiabatically q=0
ΔU=0 (isothermal); From I law

During irreversible compression, maximum work is done on the gas (corresponding to shaded area).

Topic: Chemistry, Thermodynamics (C), JEE Main
An ideal gas is expanding such as PT2=constant. The coefficient of volume expansion of the gas is

Options:
A 1T
B 2T
C 3T
D 4T

The correct answer is:
3T

Solution:
PT2=constant

nRTVT2=nRT3V-1=constant

Differentiating the equation, we get
3nRVT2dT-nRT3dVV2=0

⇒γ=1VdVdT=3T

Topic: Physics, Thermal Properties of Matter, JEE Main
An ideal gas is expanding such that pT2= constant. The coefficient of volume expansion of the gas is

Options:
A 1T
B 2T
C 3T
D 4T

The correct answer is:
3T

Solution:
Given, pT2 = constant
∴ nRTVT2=constant
or T3V-1= constant
Differentiating the equation, we get
3T2V.dT-T3V2.dV=0
or 3.dT=TV.dV
From the equation, dV=VγdT
γ= coefficient of volume expansion of gas
= dVV.dT
∴ γ=dVV.dT=3T

Topic: Physics, Thermal Properties of Matter, JEE Main
An ideal gas is expanding such that PT3= constant. The coefficient of volume expansion of the gas is:

Options:
A 2T
B 3T
C 1T
D 4T

The correct answer is:
4T

Solution:
γ=1VdVdt

PT3= constant

TVT3= Constant

T4 V-1=C

V-14 T3 dT+T4-1 V2 d V=0

dVdT=4T3/VT4/V2=4VT ∴γ=4T

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2021 (27 Aug Shift 1)
An ideal gas is initially at temperature T and volume V . It's volume increases by ΔV due to an increase in temperature of ΔT , pressure remaining constant. The quantity μ=ΔVVΔT varies with temperature as -

Options:
A

B

C

D

The correct answer is:

Solution:
V=KTâÎV=KÎT
â´ ÎV ÎT =K . . . . . .( i )
âÎ¼=ÎVVÎTâÎ¼Î±1VâÎ¼Î±1T
So graph bet. Î¼ & T will be rectangular hyperbola.

Topic: Chemistry, States of Matter, NEET
An ideal gas is initially at temperature T and Volume V. Its volume increases by ΔV due to an increase in temperature ΔT, (P remains constant). The quantity δδ=ΔVVΔT varies with temperature as

Options:
A
B
C
D

The correct answer is:

Solution:
As PV=RT
So PδV=R·δT
Or δVδT=RP=R·VRT=VT
δ=δVV·δT=1T
Hence Plot in option C is Correct.

Topic: Chemistry, States of Matter, JEE Main
An ideal gas is initially at temperature T and volume V. Its volume is increased by ÎV due to an increase in temperature ÎT, pressure remaining constant. The quantity Î´=ÎV/VÎT varies with temperature as

Options:
A
B
C
D

The correct answer is:

Solution:
For an ideal gas: pV = nRT
For, p = constant
pÎV=nRÎT

â´ ÎVÎT=nRp=nRnRTV=VT
â´ ÎVVÎT=1T or Î´=1T

Topic: Physics, Thermodynamics, NEET
An ideal gas is initially at temperature T and volume V. Its volume is increased by ÎV due to an increase in temperature ÎT, pressure remaining constant. The quantity Î´=ÎV/VÎT varies with temperature as

Options:
A
B
C
D

The correct answer is:

Solution:
For an ideal gas: pV = nRT
For, p = constant
pΔV=nRΔT

∴ ΔVΔT=nRp=nRnRTV=VT
∴ ΔVVΔT=1T or δ=1T

Topic: Physics, Thermodynamics, NEET
An ideal gas is taken around $A B C A$ as shown in the above $P-V$ diagram. The work done during a cycle is

Options:
A $2 P V$
B PV
C $1 / 2 P V$
D Zero

The correct answer is:
$2 P V$

Solution:
Work done $=$ Area enclosed by triangle $A B C$
$=\frac{1}{2} A C \times B C=\frac{1}{2} \times(3 V-V) \times(3 P-P)=2 P V$

Topic: Physics, Thermodynamics, JEE Main
An ideal gas is taken around the cycle, ABCA as shown in P-V diagram. The net work done during the cycle is equal to

Options:
A 12P1V1
B 6P1V1
C 5P1V1
D P1V1

The correct answer is:
5P1V1

Solution:
Work done = Area bounded by the triangle ABC
=12×2V1×5P1
=5P1V1

Topic: Chemistry, Thermodynamics (C), JEE Main
An ideal gas is taken around the cycle ABCA as

Options:
A 12P1V1
B 6P1V1
C 3P1V1
D P1V1

The correct answer is:
3P1V1

Solution:
Work done in the cyclic process = Area bounded (ABCA)
=12×AC×AB
=12×2V1×3P1=3P1V1 .

Topic: Chemistry, Thermodynamics (C), JEE Main
An ideal gas is taken around the cycle \( \mathrm{ABCA} \) as shown in \( \mathrm{P}-\mathrm{V} \) diagram. The net work done during the cycle is equal to -

Options:
A \( 12 \mathrm{P}_{1} \mathrm{~V}_{1} \)
B \( 6 \mathrm{P}_{1} \mathrm{~V}_{1} \)
C \( 5 \mathrm{P}_{1} \mathrm{~V}_{1} \)
D \( \mathrm{P}_{1} \mathrm{~V}_{1} \)

The correct answer is:
\( 5 \mathrm{P}_{1} \mathrm{~V}_{1} \)

Solution:
From the graph, it is clear that the work done in the cyclic process is equal to the area of a triangle.
Area of triangle=12×base×height =12×(3V1-V1)×(6P1-P1) =5P1V1

Topic: Chemistry, Thermodynamics (C), JEE Main
An ideal gas is taken path as shown in figure. The net work done in the whole cycle is

Options:
A $3 p_{1} V_{1}$
B $-3 \mathrm{p}_{1} \mathrm{~V}_{1}$
C $6 p_{1} V_{1}$
D Zero

The correct answer is:
$-3 \mathrm{p}_{1} \mathrm{~V}_{1}$

Solution:
The net work done in the whole cycle is
$$
=\frac{1}{2} \times\left(3 \mathrm{p}_{1}\right) \times\left(2 \mathrm{~V}_{1}\right)=3 \mathrm{p}_{1} \mathrm{~V}_{1}
$$
If the cycle is anticlockwise work done is negative $\therefore \quad \mathrm{W}_{\text {cycle }}=-3 \mathrm{p}_{1} \mathrm{~V}_{1}$

Topic: Physics, Thermodynamics, JEE Main
An ideal gas is taken reversibly around the cycle a-b-c-d-a as shown on the $\mathrm{T}$ (temperature) - S (entrophy) diagram
The most appropriate representation of above cycle on a U (internal energy) - V (volume) diagrame is

Options:
A
B
C
D

The correct answer is:

Solution:
$\mathrm{bc} \Rightarrow$ Isothermal process so $\mathrm{U}$ remain constant
$\mathrm{cd} \Rightarrow$ Isentropic process so $\mathrm{S}$ remain constant
$\therefore$
bc should be straight line parallel to $V \&$ cd graph should be

Topic: Physics, Thermodynamics, KVPY, KVPY 2016 (SB/SX)
An ideal gas is taken through the cycle AâBâCâA, as shown in the figure. If the net heat supplied to the gas in the cycle is 5âJ, the work done by the gas in the process CâA is

Options:
A â5 J
B −10 J
C −15 J
D −20 J

The correct answer is:
â5 J

Solution:
Δ W AB = p Δ V = 1 0 2 - 1 = 10 J
ΔWBC=0 as V = constant
From first law of thermodynamics,
Δ Q = Δ W + Δ U
Δ U = 0
Process ABCA is cyclic,
∴ Δ Q = Δ W AB + Δ W BC + Δ W CA
∴ Δ W CA = Δ Q - Δ W AB - Δ W BC
= 5 - 1 0 - 0 = - 5 J

Topic: Physics, Thermodynamics, JEE Main
An ideal gas obeying kinetic theory of gases can be liquefied if

Options:
A Its temperature is more than critical temperature $T_c$
B Its pressure is more than critical pressure $P_C$
C Its pressure is more than $P_c$ at a temperature less than $T_c$
D It cannot be liquefied at any value of $P$ and $T$

The correct answer is:
It cannot be liquefied at any value of $P$ and $T$

Solution:
Absence of inter molecular attraction ideol gas cannot be liquefied at any volume of $P$ and $T$.

Topic: Chemistry, States of Matter, JEE Main
An ideal gas of molar mass $\mathrm{M}$ is contained in a very tall vertical cylindrical column in the uniform gravitational field. Assuming the gas temperature to be $\mathrm{T}$, the height at which the centre of gravity of the gas is located is (R: universal gas constant)

Options:
A $\frac{\mathrm{RT}}{\mathrm{g}}$
B $\frac{\mathrm{RT}}{\mathrm{Mg}}$
C $\mathrm{MgR}$
D RTg

The correct answer is:
$\frac{\mathrm{RT}}{\mathrm{Mg}}$

Solution:
$\because$ field is uniform COM and COG coincides.
$$
\text { So } \mathrm{y}_{\mathrm{cem}}=\frac{\int_{0}^{\infty} \mathrm{ydm}}{\int_{0}^{\infty} \mathrm{dm}}=\frac{\int_{0}^{\infty} \mathrm{y} \mathrm{Ady} \rho}{\int_{0}^{\infty} \mathrm{Ady} \rho}=\frac{\int_{0}^{\infty} \mathrm{y} \rho \mathrm{dy}}{\int_{0}^{\infty} \rho \mathrm{dy}}
$$
From Barometric formula :- $\rho=\rho_{0} \mathrm{e}^{-\mathrm{Mgy} / \mathrm{RT}}$
So $y_{\text {cen }}=\frac{\int_{0}^{\infty} y e^{-M g y / R T}}{\int_{0}^{\infty} e^{-M g y / R T}}=\frac{R T}{M g}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An ideal gas undergoes a cyclic process as shown in Figure.
$\begin{aligned}
&\Delta \mathrm{U}_{\mathrm{BC}}=-5 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{q}_{\mathrm{AB}}=2 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
&\mathrm{~W}_{\mathrm{AB}}=-5 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{~W}_{\mathrm{CA}}=3 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$
Heat absorbed by the system during process CA is:

Options:
A
$-5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

B
$+5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

C
$18 \mathrm{~kJ} \mathrm{~mol}^{-1}$

D
$-18 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$+5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$\Delta \mathrm{U}_{\mathrm{AB}}=\mathrm{q}_{\mathrm{AB}}+\mathrm{W}_{\mathrm{AB}}=2+(-5)=-3 \mathrm{~kJ} / \mathrm{mol}$
$\Delta \mathrm{U}_{\mathrm{BC}}=-5 \mathrm{~kJ} / \mathrm{mol}$
For cyclic process, $\Delta \mathrm{U}=0$
$\begin{aligned}
&\Delta \mathrm{U}_{\mathrm{AB}}+\Delta \mathrm{U}_{\mathrm{BC}}+\Delta \mathrm{U}_{\mathrm{CA}}=0 \\
&\Delta \mathrm{U}_{\mathrm{CA}}=-\Delta \mathrm{U}_{\mathrm{AB}}-\Delta \mathrm{U}_{\mathrm{BC}} \\
&\Delta \mathrm{U}_{\mathrm{CA}}=-(-3)-(-5)=8 \mathrm{~kJ} / \mathrm{mol} \\
&\Delta \mathrm{U}_{\mathrm{CA}}=\mathrm{q}_{\mathrm{CA}}+\mathrm{W}_{\mathrm{CA}} \\
&8=\mathrm{q}_{\mathrm{CA}}+3 \\
&\mathrm{q}_{\mathrm{CA}}=+5 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}$
Heat absorbed has positive sign.

Topic: Chemistry, Thermodynamics (C), JEE Main
An ideal gas undergoes a cyclic process as shown in Figure.
$\begin{aligned}
&\Delta \mathrm{U}_{\mathrm{BC}}=-5 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{q}_{\mathrm{AB}}=2 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
&\mathrm{~W}_{\mathrm{AB}}=-5 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{~W}_{\mathrm{CA}}=3 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$
Heat absorbed by the system during process CA is:

Options:
A
$-5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

B
$+5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

C
$18 \mathrm{~kJ} \mathrm{~mol}^{-1}$

D
$-18 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$+5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$\Delta \mathrm{U}_{\mathrm{AB}}=\mathrm{q}_{\mathrm{AB}}+\mathrm{W}_{\mathrm{AB}}=2+(-5)=-3 \mathrm{~kJ} / \mathrm{mol}$
$\Delta \mathrm{U}_{\mathrm{BC}}=-5 \mathrm{~kJ} / \mathrm{mol}$
For cyclic process, $\Delta \mathrm{U}=0$
$\begin{aligned}
&\Delta \mathrm{U}_{\mathrm{AB}}+\Delta \mathrm{U}_{\mathrm{BC}}+\Delta \mathrm{U}_{\mathrm{CA}}=0 \\
&\Delta \mathrm{U}_{\mathrm{CA}}=-\Delta \mathrm{U}_{\mathrm{AB}}-\Delta \mathrm{U}_{\mathrm{BC}} \\
&\Delta \mathrm{U}_{\mathrm{CA}}=-(-3)-(-5)=8 \mathrm{~kJ} / \mathrm{mol} \\
&\Delta \mathrm{U}_{\mathrm{CA}}=\mathrm{q}_{\mathrm{CA}}+\mathrm{W}_{\mathrm{CA}} \\
&8=\mathrm{q}_{\mathrm{CA}}+3 \\
&\mathrm{q}_{\mathrm{CA}}=+5 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}$
Heat absorbed has positive sign.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
An ideal gas undergoes a cyclic process as shown in the figure
∆UBC=-5KJmol-1,qAB=2KJmol-1
WAB=-5KJmol-1,WCA=3KJmol-1
Heat absorbed by the system during process CA is

Options:
A -5KJmol-1
B +5KJmol-1
C -18KJmol-1
D 18KJmol-1

The correct answer is:
+5KJmol-1

Solution:
AB→ isobaric
BC→ isochoric
CA→ not defined

For cyclic process, ΔU=0
∆UAB=q+W
=2-5=-3
∆UABC=∆UAB+∆UBC
=-3-5=-8 KJ
∆UCA=+8
=Q+ W
8=Q+3
Q=+5 kJ

Therefore, the heat absorbed by the system during process CA is +5 KJmol−1.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2018 (15 Apr)
An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A→B→C→A as shown in the diagram. The total work done in the process is _____J.

Options:

The correct answer is:
200

Solution:
Work done is given by area enclosed in the P vs V cyclic graph or V vs P cyclic graph

Sign of work is positive for clockwise cyclic process for V vs P graph.

Work done in cyclic process = area inside the figure

W=12×(30-10)×(30-10)=200 kPa-dm3

=200×1000 Pa-L=2 L-bar=200 J

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (30 Jan Shift 1)
An ideal gas undergoes a quasi-static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn= constant, then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively) :

Options:
A n=CP-CC-CV
B n=C-CVC-CP
C n=CPCV
D n=C-CPC-CV

The correct answer is:
n=C-CPC-CV

Solution:
PVn=k
C=Cv+R1-n ; C-Cv=R1-n
1-n=RC-Cv ; n=1-RC-Cv
n=C-Cv-RC-Cv ; n=C-Cv-Cp-CvC-Cv
n=C-Cv-Cp+CvC-Cv ; n=C-CpC-Cv

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2016 (03 Apr)
An ideal gas undergoes isothermal compression from 5 m3 to 1 m3 against a constant external pressure of 4 N m-2. The heat released in this process is 24 J mol-1K-1 and is used to increase the pressure of 1 mole of Al. The temperature of Al increases by:

Options:
A 23 K
B 1 K
C 32 K
D 2 K

The correct answer is:
23 K

Solution:
Work done on isothermal irreversible compression for an ideal gas:
W=-4 1-5=16 J
For an isothermal process,∆U=0
q=-16J; heat released is 16 J.
q=nCmΔT
16=1×24×∆T
∆T=23 K

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (10 Jan Shift 2)
An ideal gas undergoes isothermal expansion at constant pressure. During the process:

Options:
A Enthalpy remains constant but entropy increases
B Enthalpy decreases but entropy increases
C Enthalpy increases but entropy decreases
D Both enthalpy and entropy remain constant

The correct answer is:
Enthalpy remains constant but entropy increases

Solution:
ΔH =  nC p  ΔT = 0,    ∴ ΔT = 0
â´ Pext = defined
so process is irreversible.
ΔS = nRln V f /V i ≥0

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2017 (09 Apr Online)
An ideal gas undergoes isothermal expansion at constant pressure. During the process:

Options:
A Enthalpy remains constant but entropy increases
B Enthalpy decreases but entropy increases
C Enthalpy increases but entropy decreases
D Both enthalpy and entropy remain constant

The correct answer is:
Enthalpy remains constant but entropy increases

Solution:
ΔH =  nC p  ΔT = 0,    ∴ ΔT = 0
â´ Pext = defined
so process is irreversible.
ΔS = nRln V f /V i ≥0

Topic: Chemistry, Thermodynamics (C), NEET
An ideal gas undergoes the cyclic process abca as shown in the figure. The internal energy change of the gas along the path ca is â180 J . The gas absorb 250 J of heat along the path ab and also absorbs 60 J of heat along the path bc . Find the work done by the gas along the path abc

Options:
A 130 J
B 140 J
C 120 J
D 100 J

The correct answer is:
130 J

Solution:
Change in internal energy for a cyclic process is zero. Thus,

∆Uabc+∆Uca=0∆Uabc=-∆Uca=180 J
ΔQ=ΔW+ΔU
250+60=ΔW+180
ΔW=130 J

Topic: Physics, Thermodynamics, JEE Main
An ideal gas with pressure $\mathrm{P}$, volume $\mathrm{V}$ and temperature $\mathrm{T}$ is expanded isothermally to a volume $2 \mathrm{~V}$ and a final pressure $\mathrm{P}_{\mathrm{i}}$. The same gas is expanded adiabatically to a volume $2 \mathrm{~V}$, the final pressure is $\mathrm{P}_{\mathrm{a}}$. In terms of the ratio of the two specific heats for the gas ' $\gamma$ ', the ratio $\frac{\mathrm{P}_i}{\mathrm{P}_{\mathrm{a}}}$ is

Options:
A $2^{\gamma+1}$
B $2^{\gamma-1}$
C $2^{1-\gamma}$
D $2^\gamma$

The correct answer is:
$2^{\gamma-1}$

Solution:
For isothermal expansion we have
$$
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\
& \therefore \mathrm{P}_2=\mathrm{P}_1 \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\mathrm{P}_1 \times \frac{1}{2}=\frac{\mathrm{P}}{2} \\
& \therefore \mathrm{P}_{\mathrm{i}}=\frac{\mathrm{P}}{2}
\end{aligned}
$$
For adiabatic process:
$$
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\
& \therefore \mathrm{P}_2=\mathrm{P}_1\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=\mathrm{P}_1\left(\frac{1}{2}\right)^\gamma=\frac{\mathrm{P}}{2^\gamma} \\
& \therefore \mathrm{P}_{\mathrm{a}}=\frac{\mathrm{P}}{2^\gamma} \\
& \therefore \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{a}}}=\frac{2^\gamma}{2}=2^{\gamma-1}
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main
An ideal gaseous mixture of ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ and ethene $\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)$ occupies 28 litre at 1 atm and $273 \mathrm{~K}$. The mixture reacts completely with $128 \mathrm{~g}$ $\mathrm{O}_{2}$ to produce $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O} .$ Mole fraction at $\mathrm{C}_{2} \mathrm{H}_{6}$ in the mixture is:

Options:
A 0.6
B 0.4
C 0.5
D 0.8

The correct answer is:
0.4

Solution:
$\mathrm{C}_{2} \mathrm{H}_{6}+3.5 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O} ;$

$\mathrm{C}_{2} \mathrm{H}_{4}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$

Let volume of ethane is $\mathrm{x}$ litre, $22.4 \times 4=3.5 \mathrm{x}+3(28-\mathrm{x})$

$\Rightarrow \mathrm{x}=11.2$ litre

at constant $\mathrm{T}$ and $\mathrm{P}, \mathrm{V} \propto \mathrm{n} ;$

$\Rightarrow$ Mole fraction of $\mathrm{C}_{2} \mathrm{H}_{6}$ in mixture $=\frac{11.2}{28}=0.4$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
An ideal liquid flows through a horizontal tube of variable diameter. The pressure is lowest where the _____.

Options:
A Velocity is highest
B Velocity is lowest
C

Diameter is largest

D Velocity is intermediate

The correct answer is:
Velocity is highest

Solution:
Using Bernoulli's theorem, which states the total mechanical energy, potential energy and kinetic energy of an ideal and steady fluid flow remain conserved at a section during flow process.

⇒Pρ+v22+gh=constant

P is the pressure of fluid at the section, v is the velocity of fluid at that section and h is the height of fluid flow.

Pρ→pressure head

v22→velocity head

gh→elevation head

According to Bernoulli's theorem, total head remain conserved between the two section of continuous fluid flow in the pipe.

Therefore, if the pressure is the lowest then the velocity of the fluid will be maximum there considering the fact that height of pipe is same.

Topic: Physics, Mechanical Properties of Fluids, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 1)
An ideal mixture of two liquids A and B is put in a cylinder containing piston. Piston is pulled out isothermally so that the volume of liquid decreases but that of vapours increases. Negligibly small amount of liquid was left and mole-fraction of A in vapour is 0.4. If P A â = 0.4 and P B â = 1.2 atm at the experimental temperature, which of the following is the total pressure at which the liquid is almost evaporated?

Options:
A 0.22 atm
B 0.431 atm
C 0.667 atm
D 1 atm

The correct answer is:
0.667 atm

Solution:
Mole-fraction in vapour phase, Y for A
Y A = 0.4 , Y B = 0.6
P A â = 0.4 P B â = 1.2 } given
P T = P A â x A + P B â x B ...(i)
Y A = 0.4 = P A â x A P T â 0.4 x A P T = 0.4
or P T = x A ...(ii)
Y B = 0.6 = P B â x B P T â 1.2 x B P T = 0.6
or P T = 2 x B ...(iii)
Comparing (ii) and (iii),
x A = 2 x B
we know, x A + x B = 1
Hence, x B = 1 3 and x A = 2 3
Putting in Equation (i)
P T = 0.4 Ã 2 3 + 1.2 Ã 1 3 = 0.667 atm

Topic: Chemistry, Solutions, JEE Main
An ideal monatomic gas is confined in a cylinder by a spring-loaded massless piston of cross-section 8Ã10â3 m2. The piston can slide on the walls of the cylinder without any friction. Initially, the gas is at 300 K and occupies a volume of 2.4Ã10â3 m3 and the spring is in its relaxed position. Now, gas is slowly heated by a small electric heater and the piston moves out slowly by 0.1 m. Calculate the final temperature of the gas and the heat supplied by the heater. The force constant of the spring is 8000 N/m and atmospheric pressure is 1Ã105 N/m2.

Options:
A T2=600 K, Q=680 J
B T2=800 K, Q=600 J
C T2=600 K, Q=720 J
D T2=800 K, Q=720 J

The correct answer is:
T2=800 K, Q=720 J

Solution:
In the initial condition

P1=105 Pa, V1=2.4×10-3 m3, T1=300 K

Finally,

P2=P1+kxA=2×105 Pa

V2=V1+Ax=3.2×10-3 m3
P1V1T1=P2V2T2, on solving we get

T2=800 K

Applying work-energy theorem for the piston

Wgas+Watmos+Wspring=0

Wgas=P0Ax+12kx2=120 J

ΔU=nCVΔT=nRΔTγ-1=P2V2-P1V1γ-1=600 J

Q=ΔU+Wgas=600+120=720 J

Topic: Physics, Thermodynamics, JEE Main
An ideal monatomic gas with pressure $P$, volume $V$ and temperature $T$ is expanded isothermally to a volume $2 V$ and a final pressure $P_i$. If the same gas is expanded adiabatically to a volume $2 \mathrm{~V}$, the final pressure is $P_a$. The ratio $\frac{P_a}{P_i}$ is

Options:
A
$2^{-1 / 3}$

B
$2^{1 / 3^6}$

C
$2^{2 / 3}$

D
$2^{-2 / 3}$

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

Solution:
For isothermal process :
$$
\begin{aligned}
& P V=P_i 2 V \\
& P=2 P_i
\end{aligned}
$$
For adiabatic process
$$
\mathrm{PV}^\gamma=\mathrm{P}_{\mathrm{a}}(2 \mathrm{~V})^\gamma
$$
$(\because$ for monatomic gas $\gamma=5 / 3)$
[From (i) ]
or, $\quad 2 \mathrm{P}_{\mathrm{i}} \mathrm{V}^{\frac{5}{3}}=P_a(2 \mathrm{~V})^{\frac{5}{3}}$
$$
\begin{aligned}
& \Rightarrow \frac{P_a}{P_i}=\frac{2}{2^{\frac{5}{3}}} \\
& \Rightarrow \frac{P_a}{P_i}=2^{\frac{-2}{3}}
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2012 (26 May Online)