The helium and argon are put in the flask at the same room temperature (300 K). The ratio of average kinetic energies…

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More Kinetic Theory of Gases Previous-Year Questions — Page 3

Q jee_main_2025_28_jan_morning Rms Speed and Temperature

For a particular ideal gas which of the following graphs represents the variation of mean squared velocity of the gas molecules with temperature?

A.
B.
C. $Graph (3)$ $\text{Graph (3)}$
D. $Graph (4)$ $\text{Graph (4)}$

Solution

Related Formula

Vrms = 3RTM Vrms² = 3RMT

Core Logic

The parameter asked is the mean squared velocity, which corresponds directly to $Vrms²$ $\mathrm{V}_{\mathrm{rms}}^2$.

From the ideal gas kinematics relation, we observe:

Vrms² \propto T

Comparing this format against standard geometric linear templates ( $y = mx$ $y = mx$), the curve must map as a clean straight line originating from absolute zero zero coordinates.

Step 1: Final Conclusion

This linear profile matches Graph (1), selecting option (1).

Pattern Recognition

Watch the vertical ordinate labels carefully: Root-mean-square velocity scales as a sub-linear curve ( $T$ $\sqrt{\mathrm{T}}$), while mean squared metric trends linearly directly ( $y \propto x$ $y \propto x$).

Chapter Mix

Class 11 Physics: Kinetic Theory

Q8 jee_main_2025_04_april_evening Ideal Gas Equation

There are two vessels filled with an ideal gas where volume of one is double the volume of other. The large vessel contains the gas at 8 kPa at 1000 K while the smaller vessel contains the gas at 7 kPa at 500 K. If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at 600 K, at steady state the pressure in the vessels will be (in kPa).

A. 4.4
B. 6
C. 24
D. 18

Solution

Related Formula

Ideal Gas Law:

n = (PV)/(RT)

Conservation of moles: $n₁ + n₂ = nf$ $n_1 + n_2 = n_f$

Core Logic

Let the volume of the smaller vessel be $V₁ = V$ $V_1 = V$, then the volume of the larger vessel is $V₂ = 2V$ $V_2 = 2V$. Initial moles in large vessel:

n₂ = (8 \times 2V)/(R \times 1000) = (16V)/(1000R)

Initial moles in small vessel:

n₁ = (7 \times V)/(R \times 500) = (14V)/(1000R)

Total total initial moles:

ntotal = n₁ + n₂ = (30V)/(1000R)

Step 1: Connect Vessels to Dynamic Equilibrium

When connected, the total final volume is $Vf = V + 2V = 3V$ $V_f = V + 2V = 3V$. The final temperature is $Tf = 600 K$ $T_f = 600\text{ K}$. Using mole conservation:

(30V)/(1000R) = (Pf (3V))/(R \times 600)

(30)/(1000) = (3Pf)/(600) (30)/(1000) = (Pf)/(200)

$Pf = (30 \times 200)/(1000) = 6 kPa$

Pattern Recognition

Connecting chambers preserves the net mass/moles ( $Σ nᵢ = constant$ $\sum n_i = \text{constant}$). Keep everything relative to a common volume multiplier $V$ $V$ to easily cancel terms.

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Class 11 Physics: Kinetic Theory

Q1 jee_main_2025_04_april_morning Mean Free Path and Collision Frequency

The mean free path and the average speed of oxygen molecules at

300~K

and

1~atm

are

3 \times 10⁻⁷~m

and

600~m/s

, respectively. Find the frequency of its collisions.

A. $2 \times 10¹⁰/s$ $2 \times 10^{10}/\mathrm{s}$
B. $9 \times 10⁵/s$ $9 \times 10^{5}/\mathrm{s}$
C. $2 \times 10⁹/s$ $2 \times 10^{9}/\mathrm{s}$
D. $5 \times 10⁸/s$ $5 \times 10^{8}/\mathrm{s}$

Solution

Related Formula

f = (1)/(T) = vavgλ

where:

$f$ $f$ = frequency of collisions
$vavg$ $v_{\text{avg}}$ = average speed of the molecules
$λ$ $\lambda$ = mean free path

Core Logic

Given parameters:

Average speed, $vavg = 600~m/s$ $v_{\text{avg}} = 600\mathrm{~m/s}$
Mean free path, $λ = 3 \times 10⁻⁷~m$ $\lambda = 3 \times 10^{-7}\mathrm{~m}$

Step 1: Calculate Frequency

Substitute the values into the formula:

f = 6003 \times 10⁻⁷ = 2 \times 10⁹~s⁻¹

Hence, the collision frequency is $2 \times 10⁹/s$ $2 \times 10^{9}/\mathrm{s}$.

Pattern Recognition

Collision frequency is simply distance covered per unit time (average velocity) divided by the average distance between consecutive collisions (mean free path).

Chapter Mix

Class 11 Physics: Kinetic Theory

Q4 jee_main_2025_28_jan_evening RMS Velocity

The ratio of vapour densities of two gases at the same temperature is

(4)/(25)

, then the ratio of r.m.s. velocities will be: [cite: 59-61]

A. $(25)/(4)$ $\frac{25}{4}$
B. $(2)/(5)$ $\frac{2}{5}$
C. $(5)/(2)$ $\frac{5}{2}$
D. $(4)/(25)$ $\frac{4}{25}$

Solution

Related Formula

The root-mean-square (r.m.s.) velocity of gas molecules is given by:

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

Since molecular weight $M$ $M$ is directly proportional to the vapour density ( $ρ$ $\rho$), the r.m.s. velocity is inversely proportional to the square root of its vapour density:

vrms1vrms2 = \sqrt((ρ₂)/(ρ₁))

Core Logic

Given the ratio of vapour densities :

(ρ₁)/(ρ₂) = (4)/(25)

Therefore, the ratio of their r.m.s. velocities is:

vrms1vrms2 = \sqrt((25)/(4)) = (5)/(2)

Pattern Recognition

R.M.S. velocity changes inversely with the square root of mass or density. Whenever a density ratio is given, simply invert the fraction and take the square root to immediately find the velocity ratio.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases

The helium and argon are put in the flask at the same room temperature (300 K). The ratio of average kinetic energies (per molecule) of helium and argon is : (Give: Molar mass of helium = 4 g/mol, Molar mass of argon =40~g/mol)$=40~g/mol)$ [cite: 74, 75, 76, 77]

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Kinetic Theory of Gases Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Connect Vessels to Dynamic Equilibrium

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Frequency

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Kinetic Theory of Gases Questions — jee_main_2025_07_april_evening

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The helium and argon are put in the flask at the same room temperature (300 K). The ratio of average kinetic energies (per molecule) of helium and argon is : (Give: Molar mass of helium = 4 g/mol, Molar mass of argon $=40~g/mol)$ $=40~g/mol)$ [cite: 74, 75, 76, 77]