States of Matter : Gases & Liquids Test 8

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States of Matter : Gases & Liquids Test 8

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Question 1
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Direction (Q. Nos. 1-14) This section contains 14 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

Q. Pressure and temperature needed to confine 1.0 x 10²⁵ gas molecules, each with a mass of 1.0x 10²⁵ kg and root mean square velocity of 1.0x 10³ms^-1 in a 1.0 m³ container are

A
a

B
b

C
c

D
d

Solution

Applying the formula
PV = 1/3(mNV²rms)
Where m = mass of each particle and N = no of particles
We get P = 3.33*10⁵ pascal or 3.3 bar
Applying the formula
V_rms = √(3RT/M)
We get T = 2405 K(approx)
So option a is correct.
Question 2
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At 298 K, which of the following gases has the lowest average molecular speed?

A
CO₂ at 0.20 atm

B
He at 0.40 atm

C
CH₄ at 0.80 atm

D
NO at 1.00 atm

Solution

Under the same conditions the average kinetic energy of the gases should be equal.
speed of gas1/speed of gas 2 = square root of (molar mass of gas2/molar mass of gas 1)
So gases with small molar mass will move quicker
eg speed of N₂/speed of CO₂ = sq. rt (44/28) = 1.25
speed of N₂/speed of F₂ = sq rt. (38/28) = 1.16
So order is N₂, F₂ and CO₂
then we concluded right ans. is (A)
Question 3
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By what ratio will the average velocity of the molecules in a gas change when the temperature is raised from 50°C to 200°C?

A
1.21/1

B
1.46/1

C
2/1

D
4/1

Solution

This is the correct answer. If we take v₂₀₀/v₅₀, then we will get option b. But that is not according to the question.
Question 4
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The root mean square velocity of a monoatomic gas (molar mass = m g mol^-1) is u. Its kinetic energy per mole (E) is related to u by equation

A

B

C

D

Solution

K.E = ½ × mass × V_rms²
Dividing both sides by moles
K.E/Moles = ½ × mass × V_rms² /Mole
K.E/mole = ½ V_rms2 × molar mass
Given, K.E/mole = E, molar mass = m and V_rms = u
E = ½ × m × u²
u² = 2E/m
u = √(2E/m)
Question 5
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Root mean square velocity (u) is dependent on temperature. Value of is

A

B

C

D

Solution

The correct answer is option B
Root mean square is v=under root 3RT/M
so we simply do differentiation with respect to temperature DU/DT=D /DT
So,
= 3R/2M
Question 6
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The pressure needed to confine 1.00 mole of N₂ molecules to 24.8 L flask is 1.0 bar. Thus, root mean square velocity of the gas assuming ideal behaviour is

A
51.5 ms^-1

B
370 ms^-1

C
515 ms^-1

D
37.0 ms^-1

Solution

Vrms = √ 3RT/M = √ 3pV/M
= √ (3×105 × 24.8×10-3)/28
= 51.4 ms^-1
So according to me, option A is correct
Question 7
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The ratio between the root mean square speed of H₂ gas at 50 K and that of O₂ gas at 800 K is
[IIT JEEE 1996]

A
4

B
2

C
1

D
0.25

Solution
Question 8
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The rms velocity of hydrogen gas is √7 times of the rms velocity of nitrogen gas. If T is the temperature of the gas, then
[IIT JEE 2000]

A
T(H₂) = T(N₂)

B
T(H₂) > T(N₂)

C
T(H₂) < T (N₂)

D
T(H₂) = √7 T (N₂)

Solution
Question 9
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For CO₂, given that average velocity at T, is equal to most probable velocity at T₂.
Thus, T₁/T₂ is

A
2.26

B
0.89

C
0.78

D
1.13
Question 10
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For gaseous state, if most probable speed is denoted by C*, average speed by and mean square speed by C, then for a large number of molecules, the ratios of these speeds are
[JEE Main 2013]

A
a

B
b

C
c

D
d

Solution

RMS velocity = √ (3RT/M) = 1.732k
Average velocity = √ (8RT/πM)
Most probable velocity = √ (2RT/M)
=√2:√8/π:√3=1:1.128:1.225

So, C∗:C:C=1:1.128:1.225
Question 11
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Molar mass of a certain gas B is double that of A.
Also RMS velocity of gas A is 200 ms ^-1 at a certain temperature. RMS velocity of gas B at the temperature half that of A is

A
200 ms^-1

B
100 ms^-1

C
300 ms^-1

D
400 ms^-1

Solution

The correct answer is Option B.
Molar mass of B is double that of A.
2 MA = 8 MB
Question 12
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A vessei of capacity 1 dm³ contains 1.0³ x 10²³ H₂ molecules exerting a pressure of 101.325 kPa. Calculate RMS speed and average speed.

A
a

B
b

C
c

D
d

Solution

V = 1dm³ = 10^-3m³
M.w = 2*10^-3g
P = 101.325kPa
= 1.03*10²³ H₂ mol
No of moles = number of molecules/NA
= 1.03/6.023
= 101.325 * 103 * 10-3 = 8.314 * 1.03/6.023 * T
T = 71.27K
u_avg = (8RT/πM)^1/2
= [(8*8.314*71.27)/(3.142*2*10-3)]^1/2
= 868.53 m/s
u_rms = (3RT/M)^1/2
[(3*8.314*71.27)/(2*10-3)]^1/2
= 942.5 m/s
Question 13
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The molecular velocity of any gas is
[AlEEE 2011]

A
inversely proportional to the square root of temperature

B
inversely proportional to absolute temperature

C
directly proportional to square of temperature

D
directly proportional to square root of temperature

Solution

The molecular velocity of any gas whether it is average velocity, root mean square velocity and most probable velocity. It must be proportional to the square to the square root of absolute temperature.
Question 14
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Distribution of molecules with velocity is represented by the curve as shown Select the correct statement(s).

A

B
As u → 0, distribution is maximum

C
As u → ∞, distribution is maximum

D
As u → 0, or u → ∞ distribution is most probable