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

Q29 jee_main_2026_28_january_evening Mean Free Path

The mean free path of a molecule of diameter

5 \times 10⁻¹⁰ m

at the temperature

41°C

and pressure

1.38 \times 10⁵ Pa

, is given as ____ m.

( Given~kB = 1.38\times 10⁻²³ J / K).

A. $2\sqrt(2) \times 10⁻¹⁰$ $2\sqrt{2} \times 10^{-10}$
B. $10\sqrt(2) \times 10⁻⁸$ $10\sqrt{2} \times 10^{-8}$
C. $2\sqrt(2) \times 10⁻⁸$ $2\sqrt{2} \times 10^{-8}$
D. $2 \times 10⁻⁸$ $2 \times 10^{-8}$

Solution

Related Formula

λ = kB T\sqrt(2) π σ² P

where, $λ$ $\lambda$ = mean free path $kB$ $k_B$ = Boltzmann constant $T$ $T$ = absolute temperature in Kelvin $σ$ $\sigma$ = diameter of the molecule $P$ $P$ = pressure of the gas

Step 1: Extract and Convert Given Data

$d = σ = 5 \times 10⁻¹⁰ m$ $T = 41^ = 273 + 41 = 314 K$ $P = 1.38 \times 10⁵ Pa$ $P = 1.38 \times 10^5 \text{ Pa}$ $kB = 1.38 \times 10⁻²³ J/K$

Step 2: Substitution and Calculation

λ = 1.38 \times 10⁻²³ \times 314\sqrt(2) \times π \times (5 \times 10⁻¹⁰)² \times 1.38 \times 10⁵

Cancel out $1.38$ $1.38$ from numerator and denominator:

λ = 10⁻²³ \times 314\sqrt(2) \times 3.14 \times 25 \times 10⁻²⁰ \times 10⁵

λ = 314 \times 10⁻²³\sqrt(2) \times 3.14 \times 25 \times 10⁻¹⁵

λ = 100 \times 10⁻²³\sqrt(2) \times 25 \times 10⁻¹⁵

λ = 4 \times 10⁻⁸\sqrt(2) = 2\sqrt(2) \times 10⁻⁸ m

Pattern Recognition

The temperature $314 K$ $314 \text{ K}$ and $π \approx 3.14$ $\pi \approx 3.14$ are intentionally matched to cancel out and give a factor of 100. $1.38$ $1.38$ cancels completely. The rest is simply handling powers of 10.

Chapter Mix

Class 11 Physics: Kinetic Theory

Q22 jee_main_2025_02_april_evening Internal Energy of Gas

The internal energy of air in

4m\times 4m\times 3m

sized room at

1

atmospheric pressure will be

_ \times 10⁶J.

(Consider air as diatomic molecule)

Numerical Answer. Answer: 12 to 12

Solution

Related Formula

Ideal Gas Law:

$P V = n R T$ $P V = n R T$

Internal Energy ( $U$ $U$) of a diatomic gas ( $f = 5$ $f = 5$ degrees of freedom):

U = n Cv T = n ((5)/(2) R) T = (5)/(2) P V

Core Logic

Given parameters:

Dimensions of the room = $4 m \times 4 m \times 3 m$
Volume of air in the room $V = 4 \times 4 \times 3 = 48 m³$
Room pressure $P = 1 atm = 10⁵ N/m²$

Step 1: Calculate internal energy

Using the thermodynamic relationship:

U = (5)/(2) P V

Substitute the volume and pressure parameters:

U = (5)/(2) \times 10⁵ N/m² \times 48 m³

U = 5 \times 10⁵ \times 24 = 120 \times 10⁵ J = 12 \times 10⁶ J

Thus, the internal energy is $12 \times 10⁶ ~J$ $12 \times 10^6 \mathrm{~J}$.

Pattern Recognition

Sees: Internal energy of diatomic gas occupying a macroscopic room volume. Trap: Attempting to calculate thermodynamic variables like temperature or density explicitly. Internal energy is completely determined by pressure and volume via degrees of freedom ( $U = (f)/(2)PV$ $U = \frac{f}{2}PV$). Shortcut: A diatomic gas has $f=5$ $f=5$, so $U = 2.5 PV$ $U = 2.5 PV$. Putting in numbers, $U = 2.5 \times 10⁵ \times 48 = 120 \times 10⁵ = 12 \times 10⁶ ~J$ , leaving a coefficient of 12.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases

Q24 jee_main_2025_02_april_morning Specific Heat Capacities of Gases

γA

is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom.

γB

is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If

(γA)/(γB) = (1 + (1)/(n))

then the value of

n

is

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

γ = 1 + (2)/(f)

Core Logic

Let's find the specific heat ratio for each gas based on degrees of freedom:

Monoatomic gas A:
Degrees of freedom, $fA = 3$ $f_A = 3$ (translational only)

γA = 1 + (2)/(3) = (5)/(3)

Polyatomic gas B:
Translational degrees of freedom = 3
Rotational degrees of freedom = 3
Vibrational modes = 1.

Note: Each active vibrational mode has 2 degrees of freedom (kinetic + potential energy terms). This contributes $2 \times 1 = 2$ $2 \times 1 = 2$ degrees of freedom.

Therefore, the total active degrees of freedom is:

fB = 3 + 3 + 2 = 8

The specific heat ratio of B is:

γB = 1 + (2)/(fB) = 1 + (2)/(8) = 1 + (1)/(4) = (5)/(4)

Now, find the ratio of specific heat capacities:

(γA)/(γB) = (5/3)/(5/4) = (4)/(3)

We are given:

(γA)/(γB) = 1 + (1)/(n) (4)/(3) = 1 + (1)/(n) (1)/(n) = (1)/(3) n = 3

Step 1: Final Conclusion

The value of $n$ $n$ is 3.

Pattern Recognition

Always remember that each vibrational mode contributes exactly 2 degrees of freedom because it holds both kinetic and potential energy components ( $fvib = 2 \times modes$ ).

Chapter Mix

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

Q17 jee_main_2025_03_april_evening Gas Laws and Temperature Dependency

Pressure of an ideal gas, contained in a closed vessel, is increased by 0.4% when heated by

1°C

. Its initial temperature must be :

A. $25°C$ $25^{\circ}\mathrm{C}$
B. 2500 K
C. 250 K
D. $250°C$ $250^{\circ}\mathrm{C}$

Solution

Related Formula

For an ideal gas in a closed container, the volume $V$ $V$ remains constant (isochoric process). By Gay-Lussac's Law:

P \propto T \Rightarrow (Δ P)/(P) = (Δ T)/(T)

where $T$ $T$ must be in Kelvin.

Core Logic

Given parameters:

Percent increase in pressure: $(Δ P)/(P) \times 100 = 0.4% \Rightarrow (Δ P)/(P) = 0.004$
Increase in temperature $Δ T = 1°C = 1~K$

Step 1: Calculate initial temperature ( $T$ $T$)

Substitute the values into Gay-Lussac's fractional variance formula:

0.004 = (1)/(T)

T = (1)/(0.004) = 250~K

Pattern Recognition

A standard percentage-increase layout. A change of $0.4%$ $0.4\%$ means $(1)/(250)$ $\frac{1}{250}$ of the original quantity. Hence, a $1~K$ $1\mathrm{~K}$ rise corresponds to an initial temperature of $250~K$ $250\mathrm{~K}$ directly.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases

Q jee_main_2025_07_april_morning Specific Heat Capacity

Match the List-I with List-II

List-I	List-II
A. Triatomic rigid gas	I. $CPCV=(5)/(3)$ $\frac{C_{P}}{C_{V}}=\frac{5}{3}$
B. Diatomic non-rigid gas	II. $CPCV=(7)/(5)$ $\frac{C_{P}}{C_{V}}=\frac{7}{5}$
C. Monoatomic gas	III. $CPCV=(4)/(3)$ $\frac{C_{P}}{C_{V}}=\frac{4}{3}$
D. Diatomic rigid gas	IV. $CPCV=(9)/(7)$ $\frac{C_{P}}{C_{V}}=\frac{9}{7}$

Choose the correct answer from the options given below:

A. A-III, B-IV, C-I, D-II
B. A-III, B-II, C-IV, D-I
C. A-II, B-IV, C-I, D-III
D. A-IV, B-II, C-III, D-I

Solution

Related Formula

The ratio of specific heats $γ$ $\gamma$ is related to degrees of freedom $f$ $f$ by:

γ = (CP)/(CV) = 1 + (2)/(f)

Core Logic Determine the degrees of freedom$ Core Logic Determine the degrees of freedom $ffor each type of gas: Monoatomic gas: Translational only$ for each type of gas: Monoatomic gas: Translational only $\implies f = 3 $ $

γ = 1 + (2)/(3) = (5)/(3) (Matches C-I)

Diatomic rigid gas: Translational (3) + Rotational (2)$ Diatomic rigid gas: Translational (3) + Rotational (2) $\implies f = 5 $ $

γ = 1 + (2)/(5) = (7)/(5) (Matches D-II)

Step 1: Check Remaining Categories Diatomic non-rigid gas: Translational (3) + Rotational (2) + Vibrational (2)$ Step 1: Check Remaining Categories Diatomic non-rigid gas: Translational (3) + Rotational (2) + Vibrational (2) $\implies f = 7 $ $

γ = 1 + (2)/(7) = (9)/(7) (Matches B-IV)

Triatomic rigid gas: Translational (3) + Rotational (3)$ Triatomic rigid gas: Translational (3) + Rotational (3) $\implies f = 6 $ $

γ = 1 + (2)/(6) = 1 + (1)/(3) = (4)/(3) (Matches A-III)

This yields the matching order: A-III, B-IV, C-I, D-II. Pattern Recognition Sees: Degrees of freedom and$ This yields the matching order: A-III, B-IV, C-I, D-II. Pattern Recognition Sees: Degrees of freedom and $\gammavalues. Shortcut: Lower degrees of freedom result in higher$ values. Shortcut: Lower degrees of freedom result in higher $\gammavalues. Order of degrees of freedom: Monoatomic (3) < Diatomic rigid (5) < Triatomic rigid (6) < Diatomic non-rigid (7). Corresponding$ values. Order of degrees of freedom: Monoatomic (3) < Diatomic rigid (5) < Triatomic rigid (6) < Diatomic non-rigid (7). Corresponding $\gamma:$: $\frac{5}{3} > \frac{7}{5} > \frac{4}{3} > \frac{9}{7}$. Chapter Mix Class 11 Physics: Kinetic Theory

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 2

Solution

Related Formula

Step 1: Extract and Convert Given Data

Step 2: Substitution and Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate internal energy

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate initial temperature (T$T$)

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Core Logic

Step 1: Check Remaining Categories

Step 1: Check Remaining Categories

Pattern Recognition

Pattern Recognition

Chapter Mix

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Step 1: Calculate initial temperature ( $T$ $T$)