An air bubble of volume 2.9 , cm3 rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17 , . The volume of the bubble when it reaches the surface, where the water temperature is 27 , , is ____ cm3 . (g = 10 , m/s2 , density of water = 103 , kg/m3 , and 1 atm pressure is 105 , Pa )

For an air bubble rising in water, the number of moles of gas remains constant. P1 = Pressure at bottom = Patm + gh T1 = Temperature at bottom = 17 = 290 , K V1 = 2.9 , cm3 P2 = Pressure at surface = Patm T2 = Temperature at surface = 27 = 300 , K Standard combined gas law application. Always remember to convert Celsius to Kelvin and carefully compute gauge pressure plus atmospheric pressure for the submerged state. Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Mechanical Properties of Fluids

An air bubble of volume 2.9 , cm3 rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17 , . The volume of the bubble when it reaches the surface, wher

For an air bubble rising in water, the number of moles of gas remains constant. P1 = Pressure at bottom = Patm + gh T1 = Temperature at bottom = 17 = 290 , K V1 = 2.9 , cm3 P2 = Pressure at surface = Patm T2 = Temperature at surface = 27 = 300 , K Standard combined gas law application. Always remember to convert Celsius to Kelvin and carefully compute gauge pressure plus atmospheric pressure for the submerged state. Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Mechanical Propertie

An air bubble of volume 2.9 , cm3 rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17 , . The volume of the bubble when it reaches the surface, where the water temperature is 27 , , is ____ cm3 . (g = 10 , m/s2 , density of water = 103 , kg/m3 , and 1 atm pressure is 105 , Pa )

For an air bubble rising in water, the number of moles of gas remains constant. P1 = Pressure at bottom = Patm + gh T1 = Temperature at bottom = 17 = 290 , K V1 = 2.9 , cm3 P2 = Pressure at surface = Patm T2 = Temperature at surface = 27 = 300 , K Standard combined gas law application. Always remember to convert Celsius to Kelvin and carefully compute gauge pressure plus atmospheric pressure for the submerged state. Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Mechanical Properties of Fluids

An air bubble of volume 2.9 , cm3 rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17 , . The volume of the bubble when it reaches the surface, wher

For an air bubble rising in water, the number of moles of gas remains constant. P1 = Pressure at bottom = Patm + gh T1 = Temperature at bottom = 17 = 290 , K V1 = 2.9 , cm3 P2 = Pressure at surface = Patm T2 = Temperature at surface = 27 = 300 , K Standard combined gas law application. Always remember to convert Celsius to Kelvin and carefully compute gauge pressure plus atmospheric pressure for the submerged state. Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Mechanical Propertie

An air bubble of volume 2.9 , cm3 rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water…

Year	2026	2025	2024	Total
Questions	6	11	9	26

Q50 jee_main_2024_27_jan_morning Kinetic Energy and Temperature

The average kinetic energy of a monatomic molecule is

0.414 eV

at temperature:

A. $3000 K$ $3000\text{ K}$
B. $3200 K$ $3200\text{ K}$
C. $1600 K$ $1600\text{ K}$
D. $1500 K$ $1500\text{ K}$

Solution

Related Formula

Kavg = (3)/(2) kB T

Core Logic

Given energy is in electron-volts ( $1 eV = 1.6 \times 10⁻¹⁹ J$ ), we isolate $T$ $T$:

T = 2 Kavg3 kB

Substitute constants ( $kB = 1.38 \times 10⁻²³ J/K$ $k_B = 1.38 \times 10^{-23}\text{ J/K}$):

Step 1: Compute value

T = 2 \times 0.414 \times 1.6 \times 10⁻¹⁹3 \times 1.38 \times 10⁻²³

T = 1.3248 \times 10⁻¹⁹4.14 \times 10⁻²³ = 0.32 \times 10⁴ = 3200 K

Pattern Recognition

Converting eV energy properties straight to structural SI standard Joules reveals highly cleanly simplified scalar components when paired with Boltzmann values.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases

Q50 jee_main_2024_29_jan_morning Ideal Gas Equation

Two vessels A and B are of the same size and are at same temperature. A contains

1 ~g

of hydrogen and B contains

1 ~g

of oxygen.

PA

and

PB

are the pressures of the gases in A and B respectively, then

(PA)/(PB)

is:

A. 16
B. 8
C. 4
D. 32

Solution

Related Formula

From the Ideal Gas Law equation:

P V = n R T P = (n R T)/(V)

Core Logic

Given that both vessels possess the same volume ( $VA = VB$ $V_A = V_B$) and identical temperature states ( $TA = TB$ $T_A = T_B$), the ratio of pressure reduces directly to:

(PA)/(PB) = (nA)/(nB)

where $nA$ $n_A$ and $nB$ $n_B$ are the number of moles of Hydrogen and Oxygen respectively.

Step 1: Calculate the Number of Moles

For Hydrogen ( $H₂$ $H_2$, molar mass = $2 ~g/mol$ $2 \mathrm{~g/mol}$):

nA = (1)/(2)

For Oxygen ( $O₂$ $O_2$, molar mass = $32 ~g/mol$ $32 \mathrm{~g/mol}$):

nB = (1)/(32)

Step 2: Find the Pressure Ratio

Substituting mole counts into the direct ratio:

(PA)/(PB) = (1/2)/(1/32) = (32)/(2) = 16

Therefore, the pressure ratio $(PA)/(PB)$ $\frac{P_A}{P_B}$ is $16$ $16$.

Pattern Recognition

For gas mixtures or vessel comparisons under constant volume and temperature, pressure matches the molar abundance directly ( $P \propto n$ $P \propto n$). Remember that standard elementary gases ( $H₂, O₂, N₂$ $H_2, O_2, N_2$) exist as diatomic configurations when defining molar values.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases

Q46 jee_main_2024_30_january_evening Mixture of Gases

If three moles of monoatomic gas

(γ = (5)/(3))

is mixed with two moles of a diatomic gas

(γ = (7)/(5))

, the value of adiabatic exponent

γ

for the mixture is:

A. $1.75$ $1.75$
B. $1.40$ $1.40$
C. $1.52$ $1.52$
D. $1.35$ $1.35$

Solution

Related Formula

fmixture = (n₁ f₁ + n₂ f₂)/(n₁ + n₂)

γmixture = 1 + 2fmixture

Core Logic

For a monoatomic gas, degrees of freedom $f₁ = 3$ $f_1 = 3$. Moles $n₁ = 3$ $n_1 = 3$. For a diatomic gas, degrees of freedom $f₂ = 5$ $f_2 = 5$. Moles $n₂ = 2$ $n_2 = 2$. We can compute the equivalent degrees of freedom for the mixture using a weighted average.

Step 1: Calculate Equivalent Degrees of Freedom

fmixture = (n₁ f₁ + n₂ f₂)/(n₁ + n₂)

fmixture = (3(3) + 2(5))/(3 + 2) = (9 + 10)/(5) = (19)/(5)

Step 2: Calculate Adiabatic Exponent

γmixture = 1 + 2fmixture

γmixture = 1 + (2)/((19)/(5)) = 1 + (10)/(19) = (29)/(19) \approx 1.52

Pattern Recognition

Alternatively, you can compute $Cv$ $C_v$ and $Cₚ$ $C_p$ for the mixture: $Cv,mix = n₁ Cv1 + n₂ Cv2n₁ + n₂$ , and $γmix = Cp,mixCv,mix$ . Both methods yield identical results rapidly.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics

Q49 jee_main_2024_30_jan_morning RMS Velocity of Gases

At which temperature the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at

47°C

?

A. $80 ~K$ $80 \mathrm{~K}$
B. $-73 ~K$ $-73 \mathrm{~K}$
C. $4 ~K$ $4 \mathrm{~K}$
D. $20 ~K$ $20 \mathrm{~K}$

Solution

Related Formula

vrms = \sqrt((3RT)/(M))

Core Logic

For the RMS velocities to be equal, the ratio of temperature to molar mass ( $T/M$ $T/M$) must be identical for both gases.

Step 1: Set Up Equivalency

3RTH₂MH₂ = 3RTO₂MO₂

TH₂MH₂ = TO₂MO₂

Step 2: Substitute Values

$TO₂ = 47^ = 47 + 273 = 320 ~K$ $MH₂ = 2 ~g/mol$ $M_{H_2} = 2 \mathrm{~g/mol}$ $MO₂ = 32 ~g/mol$ $M_{O_2} = 32 \mathrm{~g/mol}$

TH₂2 = (320)/(32)

TH₂ = 2 \times 10 = 20 ~K

Pattern Recognition

$vrms$ $v_{\text{rms}}$ scales strictly as $\sqrt(T/M)$ $\sqrt{T/M}$. Remember to always convert Celsius to Kelvin before substituting.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases

Q38 jee_main_2024_31_jan_evening Internal Energy of a Gas Mixture

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

A. $29 RT$ $29 \text{ RT}$
B. $20 RT$ $20 \text{ RT}$
C. $27 RT$ $27 \text{ RT}$
D. $21 RT$ $21 \text{ RT}$

Solution

Related Formula

$U = n Cv T$ $U = n C_v T$ where $Cv = (f)/(2) R$ $C_v = \frac{f}{2} R$ and $f$ $f$ is the degree of freedom.

Core Logic

Argon (Ar) is monatomic $f₁ = 3 Cv₁ = (3R)/(2)$ . Oxygen ( $O₂$ $O_2$) is diatomic $f₂ = 5$ $\implies f_2 = 5$ (neglecting vibrational modes) $Cv₂ = (5R)/(2)$ $\implies C_{v_2} = \frac{5R}{2}$. Total internal energy $U = U₁ + U₂ = n₁ Cv₁ T + n₂ Cv₂ T$ .

Step 1: Compute Total Energy

U = 8 \times ((3R)/(2)) T + 6 \times ((5R)/(2)) T

U = 4(3RT) + 3(5RT)

$U = 12RT + 15RT$ $U = 12RT + 15RT$ $U = 27 RT$ $U = 27 RT$

Pattern Recognition

Internal energy is strictly additive. Immediately map Monatomic $\to 3/2$ $\to 3/2$ and Diatomic $\to 5/2$ $\to 5/2$. Plug linearly: $8(1.5) + 6(2.5) = 12 + 15 = 27$ $8(1.5) + 6(2.5) = 12 + 15 = 27$.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Pressures

Step 2: Apply Ideal Gas Law

Pattern Recognition

Chapter Mix

Reference Study Guides

More Kinetic Theory of Gases Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Compute value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the Number of Moles

Step 2: Find the Pressure Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Equivalent Degrees of Freedom

Step 2: Calculate Adiabatic Exponent

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set Up Equivalency

Step 2: Substitute Values

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Total Energy

Pattern Recognition

Chapter Mix

More Kinetic Theory of Gases Questions — jee_main_2026_23_january_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back