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Test: Gaseous State - 4 - Chemistry MCQ

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20 Questions MCQ Test - Test: Gaseous State - 4

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Test: Gaseous State - 4 - Question 1

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Maximum number of electrons in a subshell with l = 3 and n = 4 is

A.
10
B.
12
C.
14
D.
16

Detailed Solution for Test: Gaseous State - 4 - Question 1

n = 4, l =3, which denotes 4f subshell.
In f subshell, there are 7 orbitals and each orbital can accommodate a maximum of two electrons, so, maximum no. of electrons in 4f subshell = 7 × 2 = 14

Formula to find out the no. of electrons = 4l+ 2
∵ l= 3
⇒ 4 × 3 + 2 = 14

Thus, there are 14 electrons.

Test: Gaseous State - 4 - Question 2

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Among the following curves, which is not according to Charle’s law:

Detailed Solution for Test: Gaseous State - 4 - Question 2

According to Charles Law, V ∝ T.
i.e. V=kT, where k is proportionality constant.

For A: Charles law is in the format of y=mx
Therefore it is a correct representation.
For B: on taking log on both sides in Charles law we get, log V= log T + log k
i.e. y = mx + c, where m = 1 = tan 45º
Therefore it is a correct representation.
For C: on multiplying with T on both sides we get, VT ∝ T²
Therefore it is a correct representation.
For D: ∵ V/T = constant
Therefore it is an incorrect representation.

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Test: Gaseous State - 4 - Question 3

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The electronic configuration of a dipositive metal M²⁺ is 2, 8, 14 and its atomic weight is 56 a.m.u. The number of neutrons in its nuclei would be

A.
56
B.
36
C.
98
D.
30

Detailed Solution for Test: Gaseous State - 4 - Question 3

Test: Gaseous State - 4 - Question 4

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Which in not correct curve for gay-lussac’s law:

Detailed Solution for Test: Gaseous State - 4 - Question 4

According to Gay-Lussac's Law, P ∝ T.
i.e. P =kT, where k is proportionality constant.

For A: On taking log on both sides in Gay-Lussac's Law we get, log P= log T + log k
i.e. y = mx + c, where m = 1 = tan 45º
Therefore it is a correct representation.
For B: ∵ P/T = constant
Therefore it is an incorrect representation.
For C: ∵ P/T = constant
Therefore it is a correct representation.
For D: Gay-Lussac's Law is in the format of y=mx
Therefore it is a correct representation.

Test: Gaseous State - 4 - Question 5

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Which is incorrect curve for Boyle’s law:

Detailed Solution for Test: Gaseous State - 4 - Question 5

According to Boyle's Law, P ∝ 1/V.

For A: Boyle's law is in the format of y=m/x
Therefore it is a correct representation.
For B: On taking log on both sides in Boyle's law we get, log P= - log V + log k
i.e. y = mx + c, where m = -1 = tan 135º
Therefore it is a correct representation.
For C: Boyle's law is in the format of y=m/x
Therefore it is an incorrect representation.
For D: ∵ V/T = constant
Therefore it is a correct representation.

Test: Gaseous State - 4 - Question 6

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The energy of an electron in first Bohr orbit of H-atom is –13.6 eV. The possible energy value of electron in the excited state of Li²⁺ is

A.
– 122.4 eV
B.
30.6 eV
C.
-30.6 eV
D.
13.6 eV

Detailed Solution for Test: Gaseous State - 4 - Question 6

where

E₁=Energy of first Bohr orbit =−13.6eV

n= no of orbit

Z=Atomic no

For Li²⁺, the excited state, n=2 and Z=3

Test: Gaseous State - 4 - Question 7

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Which of the following curve does not represent gay lusacc’s law:

Detailed Solution for Test: Gaseous State - 4 - Question 7

P = KT
P = K(t^∘C+273)

Test: Gaseous State - 4 - Question 8

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For the gaseous reaction, the rate is often expressed in terms of dp/dt instead of dC/dt or dn/dt. (where c is the concentration and n the number of mol). What is the relationship among these three expression:

A.
B.
C.
D.
None of the above is correct

Detailed Solution for Test: Gaseous State - 4 - Question 8

∵ C = n/V
on differentiating both sides,
According to Ideal gas Equation, PV = nRT
Substituting value of n gives,

Test: Gaseous State - 4 - Question 9

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A gas can be liquefied most suitably at:

A.
T = T_c and P < P_c
B.
T < T_c and P> P_c
C.
T < T_c and P = P_c
D.
T > T_c and P > P_c

Detailed Solution for Test: Gaseous State - 4 - Question 9

For liquefaction of gas temperature should be lower than critical temperature. Also higher is the P in comparison to critical pressure easier is liquefaction .

Test: Gaseous State - 4 - Question 10

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For a closed (not rigid) container containing n = 10 moles of an ideal gas fitted with a movable, frictionless, weightless piston operating such that pressure of gas remains constant at 0.821 atm, which graph represents the correct variation of log V vs log T where V is in litre and T in Kelvin:

Detailed Solution for Test: Gaseous State - 4 - Question 10

According to Charles Law, V ∝ T.
i.e. V=kT, where k is proportionality constant.
On taking log on both sides in Charles law we get, log V= log T + log k
i.e. y = mx + c, where m = 1 = tan 45º.
On comparing with the Ideal gas equation,
k = V/T = nR/P
substituting values,
k = (10 * 0.0821)/0.821 = 1

⇒ log V= log T + log 1
i.e. log V= log T

Test: Gaseous State - 4 - Question 11

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At a definite temperature (T), the distribution of velocities is given by the curve. The curve that indicates that the velocity corresponding to points A, B and C are:

A.
Most probable, root mean square and average
B.
Average, root mean square and most probable
C.
Root mean square, average and most probable
D.
Most probable, average and root mean square

Detailed Solution for Test: Gaseous State - 4 - Question 11

V_rms = √3kT/m
V_avg= √8kT/πm
V_mp = √2kT/m
√3 > √8\π > √2 (V_rms > V_avg > V_mp)

Speed increases from left to right on the x-axis. Therefore, the root means square speed is farthest to the right on the graph (means it is largest) and the most probable speed is farthest to the left (means that is small).

Test: Gaseous State - 4 - Question 12

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A graph is plotted between P (atm) vs t°C for 10 mol of an ideal gas as follows:

Then slope of curve and volume of container (L) respectively, is

A.
0.1, 8.21
B.
8.21, 0.1
C.
27.3, 8.21
D.
8.21, 27.3

Detailed Solution for Test: Gaseous State - 4 - Question 12

y−axis =P (atm), x−axis =T (ºC)
Equation of the line,
P = mT + C
at T = 0, P = 27.3 atm
⇒ C = 27.3
∴ P = mT + 27.3 ...(i)

Ideal gas equation (T in K): PV = nRT
Volume at 0 ºC,
27.3 × V = 10 × 0.0821 × 273
⇒ V = 8.21 Litres

Ideal gas equation in (T in ºC):
PV = nR (T+273)
n = 10, R = 0.0821, V = 8.21
P = (nR/V) T + (nR/T) × 273
Substituting values,
P = 0.1 T + 27.3 ...(ii)
Comparing (ii) with (I)
Slope = m = 0.1

Test: Gaseous State - 4 - Question 13

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For two samples A and B of ideal gas following curve is plotted between n vs V (volume) of container) at 16.42 atm pressure as follows, then temperature of A and B respectively are:

Test: Gaseous State - 4 - Question 14

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Match the Column-I and Column-II:

A.
A-P, B-S, C-R, D-Q
B.
A-R, B-S, C-P, D-Q
C.
A-Q, B-S, C-R, D-P
D.
A-Q, B-S, C-P, D-R

Detailed Solution for Test: Gaseous State - 4 - Question 14

Correct Answer :- D

Explanation : A-Q, B-S, C-P, D-R

Test: Gaseous State - 4 - Question 15

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Match the correct from List-1 to List-2 on the basic of following Andrews isotherm of Real gas

A.
A-QS, B-R, C-PT, D-T
B.
A-PT, B-R, C-QS, D-T
C.
A-QS, B-PT, C-R, D-T
D.
A-R, B-QS, C-T, D-PT

Detailed Solution for Test: Gaseous State - 4 - Question 15

Refer to the following diagram and compare

Test: Gaseous State - 4 - Question 16

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Match the Colum –I and Column-II

Fraction of the gas particles

A.
A-S, B-P, C-S, D-QR
B.
A-P, B-S, C-P, D-QR
C.
A-S, B-QR, C-P, D-S
D.
A-S, B-P, C-QR, D-S

Test: Gaseous State - 4 - Question 17

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Initial temperature of an ideal gas is 75°C. At what temperature, the sample of neon gas would be heated to double its pressure, it the initial volume of gas is reduced by 15%:

A.
60°C
B.
592°C
C.
128°C
D.
319°C

Detailed Solution for Test: Gaseous State - 4 - Question 17

Test: Gaseous State - 4 - Question 18

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At constant volume for a fixed number of moles of a gas, the pressure of the gas increases with the rise in temperature due to:

A.
Increases in average molecular speed
B.
Increase in rate of collisions amongst molecules
C.
Increase in molecule attraction
D.
Increase in mean free path

Detailed Solution for Test: Gaseous State - 4 - Question 18

Pressure on the walls of the container is equal to the change of momentum per unit time per unit area.
At constant volume, for a fixed number of moles of a gas, the pressure increases with rise in temperature due to an increase in average molecular speed.
This increases the change in momentum during collisions.

Test: Gaseous State - 4 - Question 19

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Three flasks of equal volumes contain CH₄, CO₂, and Cl₂ gases respectively. They will contain equal number of molecules if:

A.
The mass of all the gases is same
B.
The moles of all the gas is same but temperature is different
C.
Temperature and pressure of all the flasks are same
D.
Temperature, pressure and masses same in the flasks

Detailed Solution for Test: Gaseous State - 4 - Question 19

Avogadro's Law
At same condition of temperature and pressure, equal volumes of different gases contain an equal number of molecules.

Test: Gaseous State - 4 - Question 20

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At 0°C and one atm pressure, a gas occupied 100 cc. If the pressure is increased to one and a halftime and temperature is increased by one-third of absolute temperature, then final volume of the gas will be:

A.
80 cc
B.
88.9 cc
C.
66.7 cc
D.
100 cc

Detailed Solution for Test: Gaseous State - 4 - Question 20

Initial pressure P₁ = 1atm
Initial volume V₁ = 100cc
Final temperature T₂ = (1 + 1 / 3) × 273.15 = 364.2K
Final pressure P₂ = 1½ × 1 = 1.5atm
Final volume V₂ = ??
P₁V₁ / T₁ = T₂P₂ / V₂
1atm × 100cc / 273.15K = 1.5atm × V₂ / 364.2K
V₂ = 88.9cc
Hence, the final volume of the gas will be 88.9 cc.

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