Test: Kinetic Molecular Theory (Old NCERT) - JEE MCQ

Kinetic energy of gas molecule depends on temperature and does not depend upon the nature of gas.

K.E =(3/2)kT.

Thus hydrogen and helium have same K.E at same temperature.

Data given;
ΔT = 75 ^oC = 75 K
ΔV = 0.08 * 10⁶ cm³ = 0.08 m³
C_v  = 750 J kg^-1 K^-1
m = 300 g = 0.300 kg
p  = 10⁵ N m^-2 = 100000 N m^-2
The first law of thermodynamics;
   ΔU = Q – W
      Q = ΔU + W    ---- (1) 
   mC_pΔT = mC_vΔT + pΔV   ----- (2)
Dividing the above expression (2) with C_v , we get;
C_p / C_v = C_v / C_v + pΔV / mΔTC_v
     C_p / C_v = 1 + pΔV / mΔTC_v
     C_p / C_v = 1 + ([100000 N m^-2 . 0.08 m³] /                 [0.300 kg . 75 K .750 J kg^-1K^-1])
      C_p / C_v = 1 + 0.474
      C_p / C_v = 1.474 ; After rounding it off we get,
               = 1.5 (appx)

Collision frequency=RMSvelocity/mean free path= √3RT/M/kT/ √2σ²p

R=gas constant=8.314J/K/mol

T=temperature

M=molecular mass

K=boltzmann constant

P=pressure, σ=collision constant

Collision frequency of nitrogen/ Collision frequency of nitrogen=[√3RT/M]/[kT/√2 σ²p] nitrogen/[√3RT/M]/[kT√2 σ²p]helium

As the collision frequency is same

=>[√3RT/M]/[kT/√2 σ²p] nitrogen=[√3RT/M]/[kT√2 σ²p]helium

=>[√T/M/T/ σ²p] nitrogen=[√T/M/T/ σ²p] helium

=> [σ²/M√T] nitrogen=[[σ²/M√T]helium

=0.373/18x√T=0.218/4x√298

=>T=17.80K

According to kinetic theory of gases, a gas molecule travels in a straight line path.

Average kinetic energy ∝ Temperature in Kelvin

(KE40) / (KE20) = (3/2 nR * 313)/(3/2 nR*293)

KE40 / KE20 = T2/T1 = 40 + 273 / 20 + 273  = 313/293

Distribution of molecular velocities at two different temperature is shown below:-

At higher temperature, moree molecules have higher velocities and less molecules have lower velocities.
From the above figure, it is clear that with the increase in temperature, most probable velocity increases but the fraction of such molecules decreases, 
So option c is correct.

Bond energy is the energy to break bonds and convert the sample into atoms.
We know that, 
K.E of a mol of gas molecule = 3/2 RT
A/Q 3/2 RT = 104 kcal mol^-1
3/2×8.314×T = 104×10³ x 4.18 J
T = 34893 K

We know that 
Z = pV_observed /RT
0.8697   =   71×0.987×V_observed / 0.0821×400
V_observed = 0.8697×0.0821×400/71×0.987
= 0.406 L or 0.41 L

KE = 3/2 nRT
for 1 mole KE=3/2 RT

It is independent of the molecular mass and constant for a particular temperature.

Therefore T will be same 280K

Av. Translational Energy of a molecule= 3/2 KT -------(1)
Kinetic Energy of the electron in first Bohr's orbit= 13.6eV = 13.6×1.6×10-19J--------(2)        
from eqn. 1&2
3/2KT=13.6×1.6×10^-19
T = 1.05×10⁵ K

We know that 
K.E = 3/2RT = 3/2pV
Dividing both sides by V
K.E/V = 3/2p
E = 3/2p     (here E is K.E per unit volume)
Or p = 2/3E

Both statements are correct and statement 2 is the correct explanation of statement 1. The collisions of the molecules among themselves and with the walls of the container are perfectly elastic. Therefore, momentum and kinetic energy of the molecules are conserved during collisions.

Both statements are correct but statement 2 is not the correct explanation of 1.

Velocity distribution follows this graph and not the most probable formula.

At constant temperature, P is inversely proportional to V. So for expansion, V increases and so P decreases. KE is 3/2 kT (for one atoom) or 3/2RT(for 1 mole). And with constant temperature, KE also remains constant. 

According to me, all statements are correct. 
As per the postulates of the kinetic theory of gases, the molecules of gas collide with each other and with the walls of the container. These collisions are perfectly elastic in nature.
The gas molecule with mass m and speed u collides with the walls of the container to transfer the momentum Δp=−2mu. Thus heavier molecules transfer more momentum to the walls of the container. But this is not the postulate of the kinetic theory of gases.
As per Maxwell -Boltzmann distribution of molecular speed, very few molecules move with very high or very low speed. Most of the molecules move with average speed. Hence, only a small number of molecules have very high
velocity.
According to the kinetic theory of gases, a gas molecule moves in straight line with constant velocity. When it collides with another molecule, it changes direction. Between collisions, the molecules move in straight lines with constant velocities.

Volume of single molecule = 4/3π(r)³
= 4/3×π×(200×10^-12)³ m³
V = 3.349×10^-29 m³
Volume of 1 mole (N_A molecules) =  3.349×10^-29 m³×6.022×10²³
= 2.016×10^-5 m³
= 2.016×10^-5×10⁶ cm³ or 20.16 cm³

volume of one molecule of Nitrogen = 4/3πr³
= 4/3 x 3.14 x ( 2 x 10-10)3
= 3.351 x 10-29 m³
volume of one mole of N₂ molecules
= 3.351 x 10^-29 x 6.023 x 10²³ m³
= 2.018 x 10^-5 m³
volume of one mole of N₂ gas = 22.4 Litre at NTP = 22.4 x 10^-3 m³
space occupied by the molecules = 2.018x10^-5/22.4x10³x100
= 0.09 %
Empty space = 100 - 0.09 = 99.91 %

(V_Avg)_a = √(8RT/πM)
On potting the value of condition of A and B, we get
V_H2 = k(12.24)
V_C2H4 = k(5.65)
Thus, molecules in H₂ are moving 2.16 times faster.