Gaseous States Test

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Gaseous States Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

At constant volume and temperature conditions, the rates of diffusion D_A and D_B of gases A and B having densities ρ_A and ρ_B are related by the expression:

A

B

C

D

Solution

According to Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its density, or simply
Question 2
1 / -0

50 mL of hydrogen diffuses out through a small hole of a vessel, in 20 minutes. The time taken by 40 mL of oxygen to diffuse out is

A
32 minutes

B
64 minutes

C
8 minutes

D
12 minutes

Solution

Volume of hydrogen = 50 mL; Time for diffusion (t) = 20 min and volume of oxygen = 40 mL.

Rare of diffusion of hydrogen (r₁) = 50/20 = 2.5 mL/min

Rate of diffusion of oxygen (r₂) = 40/t mL/min

Since the molecular mass of hydrogen (M₁) = 2 and that of oxygen (M₂) = 32, therefore
Question 3
1 / -0

Under what conditions will a pure sample of an ideal gas not only exhibit a pressure of 1 atm but also a concentration of 1 mole litre^-1?

(R = 0.082 litre atm mol^-1 deg^-1)

A
At STP

B
When V = 22.4 litres

C
When 7 = 12 K

D
Impossible under any conditions

Solution
Question 4
1 / -0

Which of the following indicates the value of gas constant

A
1.987cal K−mole

B
8.3cal deg mole

C
0.0821lit deg mole

D
1.987 Joules deg mole

Solution

In a gas equation PV = nRT the value of universal gas constant depends upon the units of measurement. Its values are 8.314 J/mol/K, 0.0821L.atm/mol/K , 1.987cal/mol/K
Question 5
1 / -0

The kinetic theory of gases predicts that total kinetic energy of a gaseous assembly depends on

A
pressure of the gas

B
temperature of the gas

C
Volume of the gas

D
Pressure, volume and temperature of the gas

Solution

The total kinetic energy of a gaseous assembly depends on absolute temperature of the gas KE ∝ T.

The average kinetic energy of a gas molecule = 3/2 kT
Question 6
1 / -0

The densities of two gases are in the ratio of 1:16. The ratio of their rates of diffusion is:

A
16:1

B
4:1

C
1:4

D
1:16

Solution

Hint: According to Graham's law of gas diffusion, rate of diffusion ∝

Step: Determination of the ratio of rates of diffusion:

Let, the rate of diffusion be ′ r ′ and density be ′ d ′.

So, by Graham's law of gas diffusion -

By question,

The densities of the two gases are in the ratio of 1:16.
Question 7
1 / -0

The ratio among most probable velocity, mean velocity and root mean square velocity is given by:

A
1:2:3

B
1:√2:√3

C
√2:√3:√8/π

D
√2:√8/π:√3

Solution

Most probable velocity,
Question 8
1 / -0

If there unreactive gases having partial pressure P_A,P_B and P_C and there moles are 1,2 and 3 respectively then their total pressure will be :

A
P = P_A + P_B +P_C

B

C

D
None of these

Solution

According to Dalton's law of partial pressure,

Total pressure = sum of partial pressures of individual gases

P_total = P_a + P_b +P_c
Question 9
1 / -0

According to Graham's law, at a given temperature the ratio of the rates of diffusion r_A/r_B of gases A and B is given by: (where p and M are pressures and molecular weights of gases A and B respectively)

A

B

C

D

Solution

According to the Graham's law of diffusion, the rates of diffusion of two gases are directly proportional to their partial pressures and inversely proportional to their molecular weights.

The expression for the relative rates of diffusion of two gases is

Here, r_A and r_B are the rates of diffusion of two gases A and B respectively,

P_A and P_B are their partial pressures and M_A and M_B are their molecular weights respectively.
Question 10
1 / -0

What is the relationship between the average velocity (v), root mean square velocity (u) and most probable velocity (α)

A
α:v:u::1:1.128:1.224

B
α:v:u::1.128:1:1.224

C
α:v:u::1.128:1.224:1

D
α:v:u::1.124:1.228:1

Solution

The correct option is B

α : v : u :: 1 : 1.128 : 1.224

V_av: V_rms: V_{most probabale}= V : U : α