datavorous/entrance-exam-dataset huggingface dataset:
A gas is heated in such a way that its pressure and volume both becomes double. Then by decreasing temperature, some of the molecules of air has been added in a container to maintain the doubled volume and pressure. Assuming 14th of the initial number of moles has been added in for the purpose. By what fraction the temperature must have been raised finally of initial absolute temperature?

Options:
A 4Â times
B 165 times
C 45times
D 15Â times

The correct answer is:
165 times

Solution:
I initially : Pressure P volume V temp T moles n.
II on 2P 2V then temperature heating becomes 4T After decreasing temperature :
III Pressure volume Temperature moles 2P 2V Tâ² (assume) n+n4=5n4 for II case ideal
gas equation 2P Ã 2V=n Ã R Ã 4T
III case
2P Ã 2V=5n4 Ã R Ã Tâ² â Tâ²=165 T

Topic: Physics, Thermodynamics, JEE Main
A gas is undergoing an adiabatic process. At a certain stage, the volume and absolute temperature of the gas are V0, T0 and the magnitude of the slope of the V-T curve is m. molar specific heat of the gas at constant pressure is [Assume the volume of the gas is taken on the y-axis and absolute temperature of the gas taken on x-axis]

Options:
A mRT0V0
B mRT02V0
C V0+mT0RV0
D V0+mT0R2V0

The correct answer is:
V0+mT0RV0

Solution:
For adiabatic process TVÎ³-1= constant Differentiating with respect to T
V Î³â1 +T( Î³â1 ) V Î³â2 dV dt =0
â dVdT=-VTÎ³-1, dVdTT0, V0=m
âV0T0Î³-1=m
Î³-1=V0T0m
CV=RÎ³-1=RT0mV0
CP=CV+R=1+T0mV0R
=( V 0 +m T 0 V 0 )R

Topic: Physics, Thermodynamics, JEE Main
A gas mixture consists of 2 moles of oxygen and 4 moles argon at temperature $T$. Neglecting all vibrational modes, the total internal energy of the system is

Options:
A $4 R T$
B $15 R T$
C $9 R T$
D $11 R T$

The correct answer is:
$11 R T$

Solution:
For an ideal gas, the internal energy is given by,
$\mathrm{U}=\frac{\mathrm{f}}{2} \mathrm{nRT}$.
For a monoatomic gas, there are only 3 translational degrees of freedom and for a diatomic gas there are 3 translational +2 rotational degrees of freedom.
So,
$\mathrm{U}=2 \times \frac{5}{2} \times \mathrm{RT}+4 \times \frac{3}{2} \mathrm{RT}=11 \mathrm{RT}$

Topic: Physics, Thermodynamics, JEE Main
A gas mixture consists of 2 moles of oxygen and 4 moles argon at temperature $T$. Neglecting all vibrational modes, the total internal energy of the system is

Options:
A $4 R T$
B $15 R T$
C $9 R T$
D $11 R T$

The correct answer is:
$11 R T$

Solution:
Oxygen is diatomic gas, hence its energy of two moles $=2 \times \frac{5}{2} R T=5 R T$
Argon is a monoatomic gas, hence its internal energy of 4 moles $=4 \times \frac{3}{2} R T=6 R T$ monoatomic gas, hence its internal energy of 4 moles
Total Internal energy $=(6+5) R T=11 R T$

Topic: Physics, Thermodynamics, JEE Main
A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes we calculated the total internal energy of the system to be xRT. What is the value of x?

Options:

The correct answer is:
11

Solution:
The internal energy of n moles of a gas is U=12nFRT
where F = number of degrees of freedom.
The internal energy of 2 moles of oxygen at temperature T is
U1=12×2×5RT=5RT (F = 5 for oxygen molecule)
The total internal energy of 4 moles of argon at temperature T is
U2=12×4×3RT=6RT (F = 3 for argon molecule)
Total internal energy =U1+U2=11RT

⇒x=11

Topic: Physics, Thermodynamics, JEE Main
A gas mixture consists of 2 moles of oxygen and 4 moles of neon at temperature T. Neglecting all vibrational modes, the total internal energy of the system will be:

Options:
A 11RT
B 8RT
C 4RT
D 16RT

The correct answer is:
11RT

Solution:
The internal energy of a gas is given by U=nCVT.

A diatomic gas has 3 translational and 2 rotational degrees of freedom. A monoatomic gas has 3 translational degrees of freedom.

Hence, the internal energy of the mixture =n1CV1T+n2CV2T

=2×5R2T+4×3R2T=5RT+6RT

=11RT

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (10 Apr Shift 2)
A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Considering only translational and rotational modes, the total internal energy of the system is

Options:
A $15 \mathrm{RT}$
B $12 \mathrm{RT}$
C $4 \mathrm{RT}$
D $20 \mathrm{RT}$

The correct answer is:
$15 \mathrm{RT}$

Solution:
$U=\frac{f_{1}}{2} n_{1} R T+\frac{f_{2}}{2} n_{2} R T$

Considering translational and rotational modes, degrees of freedom $\mathrm{f}_{1}=5$ and $\mathrm{f}_{2}=3$ $\therefore \mathrm{u}=\frac{5}{2}(3 \mathrm{RT})+\frac{3}{2} \times 5 \mathrm{RT}$

$\mathrm{U}=15 \mathrm{RT}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2019 (11 Jan Shift 1)
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

Options:
A 29 RT
B 20 RT
C 27 RT
D 21 RT

The correct answer is:
27 RT

Solution:
As we know, internal energy is given by, U=nCvT. Also argon gas is monoatomic and oxygen gas is diatomic.

Therefore, total internal energy will beUnet=n1CV1T+n2CV2T

=8×3R2×T+6×5R2×T

=27RT

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (31 Jan Shift 2)
A gas mixture contains $n_1$ moles of a monoatomic gas and $n_2$ moles of gas of rigid diatomic molecules. Each molecule in monoatomic and diatomic gas has 3 and 5 degrees of freedom respectively. If the adiabatic exponent $\left(\frac{C_p}{C_V}\right)$ for this gas mixture is 1.5 , then the ratio $\frac{n_1}{n_2}$ will be

Options:
A 1
B 1.5
C 2
D 2.5

The correct answer is:
1

Solution:
For a gas mixture,
$$
C_V=\frac{n_1 C_{V_1}+n_2 C_{V_2}}{n_1+n_2} \text { and } C_p=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1+n_2}
$$
For monoatomic gas,
$$
n=n_1, C_{V_1}=\frac{3}{2} R, \quad C_{p_1}=\frac{5}{2} R
$$
For diatomic gas,
$$
n=n_2, C_{V_2}=\frac{5}{2} R \text { and } C_{p_2}=\frac{7}{2} R
$$
Now given for gas mixture,
$$
\begin{gathered}
\quad \frac{C_p}{C_V}=1.5 \\
\text { So, } 1.5=\frac{\frac{n_1}{n_2}\left(\frac{5}{2}\right)+\frac{7}{2}}{\frac{n_1}{n_2}\left(\frac{3}{2}\right)+\frac{5}{2}} \quad \frac{9}{2}\left(\frac{n_1}{n_2}\right)+\frac{15}{2}=\left(\frac{n_1}{n_2}\right) 5+7 \\
\Rightarrow \frac{n_1}{n_2}=1
\end{gathered}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A gas molecule of mass M at the surface of the earth has kinetic energy equivalent to 0 °C. If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than the radius of the earth. (kB is Boltzmann constant)

Options:
A 273kB2Mg
B 819kB2Mg
C 0
D 546kB3Mg

The correct answer is:
819kB2Mg

Solution:
Mgh=12Mvrms2

=12M3kTm

⇒vrms2=819kM

Now apply v2=u2−2gh undefined

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2014 (19 Apr Online)
A gas occupies 2 litres at STP. It is provided 58.63 joule heat so that its volume becomes 2.5 litre at 1 atm. Change in its internal energy in joules is

Options:
A 8.63
B 7.62
C 12.9
D 5.54

The correct answer is:
8.63

Solution:
Work is done by a system irreversibly (Wsys)=-PextâV â(Wsys)=-(1Ã105)N/m2Ã(2.5-2)Ã10-3m3=50
From 1st Law of Thermodynamics: âU=wsys+qsys âU=-50+58.63J=8.63 J

Topic: Chemistry, Thermodynamics (C), JEE Main
A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A.

Options:
A 6 J of the work will be done by the gas
B 6 J of the work will be done by the surroundings on gas
C 10 J of the work will be done by the surroundings on gas
D 10 J of the work will be done by the gas

The correct answer is:
6 J of the work will be done by the surroundings on gas

Solution:
for A to Bq=5W= −8 J,ΔUAB=dq+dw=5−8ΔUAB=−3
for B to AΔUBA=dq+dwΔUBA= +33=-3 +WBA WBA=6 J

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2017 (09 Apr Online)
A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A:

Options:
A 6 J of the work will be done by the gas
B 6 J of the work will be done by the surrounding on gas
C 10 J of the work will be done by the surrounding on gas
D 10 J of the work will be done by the gas

The correct answer is:
6 J of the work will be done by the surrounding on gas

Solution:
for A to B q=5 W= â8 J, Î U AB =dq+dw â 5â8 Î U AB =â3
for B to A q=5 W= â8 J, Î U BA =dq+dw Î U BA = +3 dw=+3+3 W BA =6 J

Topic: Chemistry, Thermodynamics (C), JEE Main
A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A.

Options:
A 6 J of the work will be done by the gas
B 6 J of the work will be done by the surroundings on gas
C 10 J of the work will be done by the surroundings on gas
D 10 J of the work will be done by the gas

The correct answer is:
6 J of the work will be done by the surroundings on gas

Solution:
for A to Bq=5W= −8 J,ΔUAB=dq+dw=5−8ΔUAB=−3
for B to AΔUBA=dq+dwΔUBA= +33=-3 +WBA WBA=6 J

Topic: Chemistry, Thermodynamics (C), JEE Main
A gas undergoes physical adsorption on a surface and follows the given Freundlich adsorption isotherm equation
xm=kp0.5
Adsorption of the gas increases with:

Options:
A Decrease in p and increase in T
B Increase in p and decrease in T
C Increase in p and increase in T
D Decrease in p and decrease in T

The correct answer is:
Increase in p and decrease in T

Solution:
xm=kp 0.5
Physical adsorption increases with decrease in Temperature.
According to relation xm↑ on P↑
So adsorption increases on increasing pressure.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2019 (10 Apr Shift 1)
A gaseous carbon compound is soluble in dilute $\mathrm{HCl}$. The solution on treating with $\mathrm{NaNO}_{2}$ gives off nitrogen leaving behind a solution, which smells of wood spirit. The carbon compound is

Options:
A $\mathrm{HCHO}$
B $\mathrm{CO}$
C $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}$
D $\mathrm{CH}_{3} \mathrm{NH}_{2}$

The correct answer is:
$\mathrm{CH}_{3} \mathrm{NH}_{2}$

Solution:
The gaseous carbon compound, methyl amine $\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)$ is soluble in dil. $\mathrm{HCl}$ and forms soluble hydrogen chloride salt solution.
This solution on treating with $\mathrm{NaNO}_{2}$ gives nitrogen leaving behind a methanol solution with wood spirit smell.

Topic: Chemistry, Amines, JEE Main
A gaseous compound of nitrogen and hydrogen contains $12.5 \%$ (by mass) of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is:

Options:
A
$\mathrm{NH}_2$

B
$\mathrm{N}_3 \mathrm{H}$

C
$\mathrm{NH}_3$

D
$\mathrm{N}_2 \mathrm{H}_4$

The correct answer is:
$\mathrm{N}_2 \mathrm{H}_4$

Solution:
In an unknown compounds containing $\mathrm{N}$ and $\mathrm{H}$
given $\%$ of $\mathrm{H}=12.5 \%$
$$
\therefore \% \text { of } N=100-12.5=87.5 \%
$$
$$
\begin{array}{|c|c|c|c|}
\hline \text { Element } & \text { Percentage } & \text { Atomic ratio } & \text { Simple ratio } \\
\hline \mathrm{H} & 12.5 \% & \frac{12.5}{1}=12.5 & \frac{12.5}{6.25}=2 \\
\hline \mathrm{N} & 87.5 & \frac{87.5}{14}=6.25 & \frac{6.25}{6.25}=1 \\
\hline
\end{array}
$$
$2 \times$ vapour density $=$ Mol. wt $=16 \times 2=32$.
Molecular formula $=n \times$ empirical formula mass
$$
n=\frac{32}{16}=2
$$
$\therefore$ Molecular formula of the compound will be $=\left(\mathrm{NH}_2\right)_2$ $=\mathrm{N}_2 \mathrm{H}_4$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2014 (11 Apr Online)
A gaseous compound of nitrogen and oxygen is paramagnetic in nature. When it is cooled below 0°C its molecular mass increases and paramagnetism is lost. The behaviour is reversed on heating. The compound is

Options:
A N2O3
B N2O
C NO2
D N2O4

The correct answer is:
NO2

Solution:
NO2 is paramagnetic because of unpaired electron. On cooling NO2 below 0°C , it dimerises to N2O4 which does not have any unpaired electrons. Hence, no paramagnetism is show by the dimer, N2O4 .

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A gaseous hydrocarbon on combustion gives 0.72 g of water and 3.08 g CO2. What is the empirical formula of the hydrocarbon?

Options:
A C6H5
B C7H8
C C2H4
D C3H4

The correct answer is:
C7H8

Solution:
C x H y + x + y 4 O 2 ⟶ xC O 2 + y 2 H 2 O
Moles of CO2 = 3 . 0 8 4 4 = 0 . 0 7 moles = x
Moles of H2O = 0 . 7 2 1 8 = 0 . 0 4 moles = y 2
x = 0.07 g atoms of carbon
y = 0.08 g atoms of hydrogen

x : y0.07 0.08 7 8

Empirical formula becomes C7H8.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2013 (07 Apr)
A gaseous mixture consists of $16 \mathrm{~g}$ of helium and $16 \mathrm{~g}$ of oxygen. The ratio $C_{p} / C_{V}$ of the mixture is

Options:
A $1.4$
B $1.54$
C $1.59$
D $1.62$

The correct answer is:
$1.62$

Solution:
Mass of helium, $m_{1}=16 \mathrm{~g}$
Mass of oxygen, $m_{2}=16 \mathrm{~g}$
$\therefore$ Number of moles of He gas,
$$
n_{1}=\frac{m_{1}}{4}=\frac{16}{4}=4
$$
Number of moles of $\mathrm{O}_{2} \mathrm{gas}$,
$$
n_{2}=\frac{m_{2}}{32}=\frac{16}{32}=\frac{1}{2}
$$
We know that for the mixture,
$\begin{aligned} C_{V} &=\frac{n_{1} C_{V}+n_{2} C_{V_{2}}}{n_{1}+n_{2}}, \text { where } C_{V}=\frac{f R}{2} \\ \text { and } C_{p} &=\frac{n_{1} C_{p_{1}}+n_{2} C_{p_{2}}}{n_{1}+n_{2}}, \text { where } C_{p}=\left(\frac{f}{2}+1\right) R \end{aligned}$
For He gas, $f=3, n_{1}=4$
For $\mathrm{O}_{2}$ gas, $f=5, n_{2}=\frac{1}{2}$
$$
\begin{aligned}
\therefore \quad \frac{C_{p}}{C_{V}} &=\frac{n_{1} C_{p_{1}}+n_{2} C_{p_{2}}}{n_{1} C_{V_{1}}+n_{2} C_{V_{2}}} \\
&=\frac{4 \times\left(\frac{3}{2}+1\right) R+\frac{1}{2}\left(\frac{5}{2}+1\right) R}{4 \times \frac{3}{2} R+\frac{1}{2} \times \frac{5}{2} R} \\
&=\frac{\left(4 \times \frac{5}{2} R\right)+\left(\frac{1}{2} \times \frac{7}{2} R\right)}{\left(4 \times \frac{3}{2} R\right)+\left(\frac{1}{2} \times \frac{5}{2} R\right)}=\frac{47}{29} \\
&=1.62
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A gaseous mixture consists of $16 \mathrm{~g}$ of helium and $16 \mathrm{~g}$ of oxygen. The ratio $\frac{C_p}{C_v}$ of the mixture is

Options:
A
$1.59$

B
$1.62$

C
$1.4$

D
$1.54$

The correct answer is:
$1.62$

Solution:
$c_v=\frac{n_1 c_{v 1}+n_2 c_{v 2}}{n_1+n_2}=\frac{29 R}{18}$
$c_P=\frac{47 R}{18}, \quad \frac{c_P}{c_v}=1.62$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A gaseous mixture consists of $16 \mathrm{~g}$ of helium and $16 \mathrm{~g}$ of oxygen. The ratio $\frac{C_p}{C_v}$ of the mixture is

Options:
A
$1.59$

B
$1.62$

C
$1.4$

D
$1.54$

The correct answer is:
$1.62$

Solution:
$c_v=\frac{n_1 c_{v 1}+n_2 c_{v 2}}{n_1+n_2}=\frac{29 R}{18}$
$c_P=\frac{47 R}{18}, \quad \frac{c_P}{c_v}=1.62$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2005
A gaseous mixture of two substances A and B, under a total pressure of 0.8 atm is in equilibrium with an ideal liquid solution. The mole fraction of substance A is 0.5 in the vapour phase and 0.2 in the liquid phase. The vapour pressure of pure liquid A is____atm.(Nearest integer)

Options:

The correct answer is:
2

Solution:
Raoult’s law states that a solvent’s partial vapour pressure in a solution (or mixture) is equal or identical to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.

Raoult’s law is valid only in the case of ideal solutions

PA=PoAχAAccording to Raoult's law

PA=PTYAAccording to Dalton's lawPT=Total pressurePA=vapour pressure of A in solution

PoA=Vapour pressure of pure liquid A

χA=Mole fraction of component A in liquid phase

YA=Mole fraction of component A in gaseous phase

So, PA=PoAχA=YAPT

PAo0.2=0.5×0.8

PAo=2 atm

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A gene encoding for a polypeptide of 60 amino acids get mutated at 30th codon UAU becoming UAA The result would be

Options:
A A polypeptide of 29 amino acids
B Two polypeptides one with 29 amino acids and second with 31 amino acids.
C A polypeptide with 59 amino acids
D A polypeptide of 25 amino acids.

The correct answer is:
A polypeptide of 29 amino acids

Solution:
In this example, 30th codon is mutated to form stop codon. Thus the first 29 codons will be read and 29 amino acids will form a polypeptide chain. But the stop codon will not allow further translation to occur and the remaining codons will not be translated

Topic: Biology, Molecular Basis of Inheritance, NEET
A generator has armature resistance of $0.1 \Omega$ and develops an induced emf of $120 \mathrm{~V}$ when driven at its rated speed. Its terminal voltage when a current of $50 \mathrm{~A}$ is being drawn is

Options:
A
$120 \mathrm{~V}$

B
$5 \mathrm{~V}$

C
$115 \mathrm{~V}$

D
$70 \mathrm{~V}$

The correct answer is:
$115 \mathrm{~V}$

Solution:
Armature resistance $R=0.1 \Omega$
Induced emf, $e=120 \mathrm{~V}$
Current drown, $I=50 \mathrm{~A}$
We know that,
$e=V+I R$
or, $V=e-I R=120-50 \times 0.1=115 \mathrm{~V}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2012 (19 May Online)
A geostationary satellite is orbiting around an arbitrary planet P at a height of 11R above the surface of P, R being the radius of P. The time period of another satellite in hours at a height of 2R from the surface of P is ________ has the time period of 24 hours.

Options:
A 62
B 62
C 3
D 5

The correct answer is:
3

Solution:
T∝R3/2

24 T=12R3R3/2⇒T=3 hr

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (17 Mar Shift 2)
A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a sky satellite orbiting a few 100 km above the earth's surface R=64000 km will approximately be

Options:
A 12 h
B 1 h
C 2 h
D 4 h

The correct answer is:
2 h

Solution:
By Kepler's law of planetary motion,

T2∝r3, hence T12∝r13

Similarly T22∝r23

∴ T12∝360003

T22∝6400+h3

Therefore,

T22=T126400+h360003

>T12640036003

>2428453

∴ T2>24×8×845×45

>1.8 h

So, T2≅2 h

Topic: Physics, Gravitation, JEE Main
A geyser heats water flowing at a rate of 2.0 kg per minute from 30°C to 70°C. If geyser operates on a gas burner, the rate of combustion of fuel will be _____ g min-1.

[Heat of combustion =8×103Jg-1, Specific heat of water =4.2Jg-1C∘-1]

Options:

The correct answer is:
42

Solution:
Given here, mass flow rate of water, dmdt=2 kg min-1

The geyser heats the water, raising the temperature from θ1=30°C to θ2=70°C.

Total heat utilised in raising the temperature, ΔQΔt=dmdt×sΔθ

ΔQΔt=260×4.2×40=mt×L

L= heat of combination=8×103 J g-1

Now, rate of consumption of fuel is mt=2×4.2×4060×8×103

=4.26×10-3=0.7×10-3 kg sec-1

=42 g min-1

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2022 (26 Jun Shift 2)
A girl swings on cradle in a sitting position. If she stands what happens to the time period of girl and cradle?

Options:
A Time period decreases
B Time period increases
C Remains constant
D First increases and then remains constant

The correct answer is:
Time period decreases

Solution:
As the girl stands up, the effective length of pendulum decreases due to the reason that the centre of gravity rises up. Hence, according to
T=2πlg
T will decrease, Since L is the effective distance of COM from the point of suspension of the simple pendulum.

Topic: Physics, Oscillations, NEET
A given ideal gas with $\gamma=\frac{C_p}{C_v}=1.5$ at a temperature $T$. If the gas is compressed adiabatically to one-fourth of its initial volume, the final temperature will be

Options:
A
$2 \sqrt{2} T$

B
$4 T$

C
$2 T$

D
$8 T$

The correct answer is:
$2 T$

Solution:
$T V^{\gamma-1}=$ constant
$$
\begin{aligned}
& T_1 V_1^{\gamma-1}=\mathrm{T}_2 V_2^{\gamma-1} \\
& \Rightarrow T(V)^{1 / 2}=\mathrm{T}_2\left(\frac{V}{4}\right)^{1 / 2} \\
& {\left[\because \gamma=1.5, T_1=T, V_1=V \text { and } V_2=\frac{V}{4}\right]} \\
& \therefore \quad T_2=\left(\frac{4 V}{V}\right)^{1 / 2} T=2 T
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2012 (12 May Online)
A given mass of a gas is compressed isothermally until its pressure is doubled. It is then allowed to expand adiabatically until its original volume is restored and its pressure is then found to be 0.75 of its initial pressure. The ratio of the specific heats of the gas is approximately :

Options:
A 1.20
B 1.41
C 1.67
D 1.83

The correct answer is:
1.41

Solution:
isothermal process, temperature of the gas remains constant,so the gas obeys Boyle's law. That is,
$P \propto \frac{1}{V}$
$\Rightarrow \quad \frac{P_2}{P_1}=\frac{V_1}{V_2}$
$\Rightarrow \quad \frac{2 P}{P}=\frac{V_1}{V_2}$
$\therefore \quad \frac{V_1}{V_2}=2$ $\ldots(1)$
Now, the gas is expanded adiabatically, so $P V^\gamma=$ constant
$\frac{P_1}{P_2}=\left(\frac{V_2}{V_1}\right)^\gamma$
$\Rightarrow \quad \frac{2 P}{0.75 P}=\left(\frac{2}{1}\right)^\gamma$
(since volume is restored)
$\Rightarrow \quad \log \left(\frac{8}{3}\right)=\gamma \log 2$
$\Rightarrow \quad \log 8-\log 3=\gamma \log 2$
$\therefore \quad \gamma=1.41$

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2006
A given mass of a gas is compressed isothermally until its pressure is doubled. It is then allowed to expand adiabatically until its original volume is restored and its pressure is then found to be 0.75 of its initial pressure. The ratio of the specific heats of the gas is approximately :

Options:
A 1.20
B 1.41
C 1.67
D 1.83

The correct answer is:
1.41

Solution:
isothermal process, temperature of the gas remains constant,so the gas obeys Boyle's law. That is,
$P \propto \frac{1}{V}$
$\Rightarrow \quad \frac{P_2}{P_1}=\frac{V_1}{V_2}$
$\Rightarrow \quad \frac{2 P}{P}=\frac{V_1}{V_2}$
$\therefore \quad \frac{V_1}{V_2}=2$ $\ldots(1)$
Now, the gas is expanded adiabatically, so $P V^\gamma=$ constant
$\frac{P_1}{P_2}=\left(\frac{V_2}{V_1}\right)^\gamma$
$\Rightarrow \quad \frac{2 P}{0.75 P}=\left(\frac{2}{1}\right)^\gamma$
(since volume is restored)
$\Rightarrow \quad \log \left(\frac{8}{3}\right)=\gamma \log 2$
$\Rightarrow \quad \log 8-\log 3=\gamma \log 2$
$\therefore \quad \gamma=1.41$

Topic: Physics, Thermodynamics, TS EAMCET, TS EAMCET 2006
A given mass of gas expands from state A to state B by three paths 1, 2 and 3 as shown in the figure

If w1,w2 and w3 respectively are be the works done by the gas along three paths, then

Options:
A w1>w2>w3
B w1<w2<w3
C w1=w2=w3
D w1<w2;w1<w3

The correct answer is:
w1<w2<w3

Solution:
Work is equal to area under P-V graph (when P is plotted along y-axis). As area under graph 3 is maximum and area under graph 1 is minimum, so w3 is maximum and w1 is minimum.
So w1<w2<w3

Topic: Chemistry, Thermodynamics (C), JEE Main
A given mass of gas expands from state A to state B by three paths \( 1,2 \) and \( 3 \) as shown in the figure
If \( \mathrm{w}_{1}, \mathrm{w}_{2} \) and \( \mathrm{w}_{3} \) respectively are be the works done by the gas along three paths, then

Options:
A \( \mathrm{w}_{1}>\mathrm{w}_{2}>\mathrm{w}_{3} \)
B \( \mathrm{w}_{1} < \mathrm{w}_{2} < \mathrm{w}_{3} \)
C \( \mathrm{w}_{1}=\mathrm{w}_{2}=\mathrm{w}_{3} \)
D \( \mathrm{w}_{1} < \mathrm{w}_{2} ; \mathrm{w}_{1}>\mathrm{w}_{3} \)

The correct answer is:
\( \mathrm{w}_{1} < \mathrm{w}_{2} < \mathrm{w}_{3} \)

Solution:
Work is equal to area under P-V graph (when P is plotted along y-axis). As area under graph 3 is maximum and area under graph 1 is minimum, so w3 is maximum and w1 is minimum.
So w1<w2<w3

Topic: Chemistry, Thermodynamics (C), JEE Main