States of Matter : Gases & Liquids Test 7

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States of Matter : Gases & Liquids Test 7

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Question 1
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Direction (Q. Nos. 1-11) This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. Helium atom is two times heavier than a hydrogen molecule (H₂). At 298 K, the average kinetic energy of helium atom is

A
two times that of a hydrogen molecule

B
same as that of a hydrogen molecule

C
four times that of a hydrogen molecule

D
half that of a hydrogen molecule

Solution

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.
Question 2
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Mass of gas is 300 gm and its specific heat at constant volume is 750J/kg K. if gas is heated through 75°C at constant pressure of 10⁵ N/m², it expands by volume 0.08 × 10⁶ cm³. find C_P/C_V.

A
1.4

B
1.374

C
1.474

D
1.5

Solution

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)
Question 3
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At what temperature will the collision frequency in N₂ be equal to that in He at 25°C if both gases are at a pressure of 1.00 bar collision diameter = 0.373 nm for N₂ = 0.218 nm for He (Assume equal number of molecules per unit volume.)

A
46K

B
17.80 K

C
29.8 K

D
400K

Solution

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
Question 4
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According to the kinetic theory of gases, in an ideal gas, between two successive collisions a gas molecule travels
[IIT JEE 2003]

A
in a circular path

B
in a wave path

C
in a straight line path

D
with an accelerated velocity

Solution

According to kinetic theory of gases, a gas molecule travels in a straight line path.
Question 5
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As the tem perature is raised from 20°C to 40°C. The average kinetic energy of the neon atoms changes by a factor of which of the following?
[IIT JEE 2004]

A
1/2

B

C

D
2

Solution

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
Question 6
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Which one of the following statements is not true about the effect of an increase in temperature on the distribution of molecular speeds in a gas?
[IIT JEE 2005]

A
The area under the distribution curve remains the same as under the lower temperature

B
The distribution becomes broader

C
The fraction of the molecules with the most probable speed increases

D
The most probable speed increases

Solution

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.
Question 7
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Bond energy of H₂ gas is 104 kcal mol^-1. The temperature at which average kinetic energy of gaseous H₂ molecules is equal to energy required to dissociate the molecules into atoms, is

A
34620 K

B
34893 K

C
31200 K

D
32723 K

Solution

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
Question 8
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Compressibility factor for CO₂ at 400 K and 71.0 bar is 0.8697. Molar volume of CO₂ under these conditions is

A
22.4 L

B
2.24 L

C
0.41 L

D
19.5 L

Solution

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
Question 9
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At what temperature will the hydrogen molecules have the same kinetic energy per mole as nitrogen molecules at 280 K?

A
280 K

B
40 K

C
400 K

D
50 K

Solution

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
Question 10
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Temperature at which average translation energy of a molecule is equal to the kinetic energy of the electron in first Bohr is orbit is

A
2.02 x 10⁵ K

B
1.05 x 10⁵ K

C
273 K

D
298 K

Solution

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
Question 11
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Pressure (p) of an ideal gas is related to kinetic energy (E) per unit volume by equation

A

B

C

D

Solution

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
Question 12
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Direction (Q. Nos. 12 and 13) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.
Q.
Statement I : The pressure exerted by a gas is the result of collisions of the molecules with the walls of the container.
Statement II : Due to collision, momentum is changed and rate of change of momentum is equal to force.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement lI is correct but Statement I is incorrect

Solution

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.
Question 13
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Statement I : As velocity increases, distribution of molecules increases and is maximum at most probable velocity, after which distribution of molecules decreases.
Statement II : Most probable velocity is given by

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement lI is correct but Statement I is incorrect

Solution

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.