Gaseous State Test

Result

Gaseous State Test - 16

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Weekly Quiz Competition

Question 1
4 / -1

If the mean free path of atoms is doubled then the pressure of gas will become:

A
P/4

B
P/2

C
P/8

D
P

Solution

∴ P ∝ 1/λ i.e by increasing λ two times pressure will become half
Question 2
4 / -1

Ratio of C_p and C_v of a gas 'X' is 1.4. The number of atoms of the gas 'X' present in 11.2 litres of it at N.T.P. is:

A
6.02 × 10²³

B
1.2 × 10²⁴

C
3.01 × 10²³

D
2.01 × 10²³

Solution
Question 3
4 / -1

At o 100°C and 1 atm, if the density of liquid water is 1.0 g cm³ and that of water vapour is 0.0006 g cm³ , then the volume occupied by water molecules in 1 litre of steam at that temperature is:

A
6 cm³

B
60cm³

C
0.6 cm³

D
0.06 cm³

Solution

Volume of steam = 1 lt = 10³ cm³

∵ m = d.V

∴ mass of 10³ cm³ steam = density × Volume

Actual volume occupied by H₂Omolecules is equal to volume of water of same mass

∴ Actual volume of H₂O molecules in 6gm steam

= mass of steam/density of water

= 0.6 gm/1 gm/cm³ ⇒ 0.6 cm³
Question 4
4 / -1

The density of a gas A n is three times that of a gas B. if the molecular mass of A is M, the molecular mass of B is:

A
3 M

B
√3 M

C
M/3

D
M / √3

Solution
Question 5
4 / -1

The density of neon will be highest at:

A
S.T.P.

B
0ºC,2atm

C
273ºC,1atm

D
273ºC,2atm

Solution

The density of neon will be highest at 0°C 2 atm according to d ∝ P/T
Question 6
4 / -1

What is the pressure of 2 mole of NH₃ at 27ºC when its volume is 5 litre in vander Waal’s equation? (a = 4.17, b = 0.03711)

A
10.33 atm

B
9.33 atm

C
9.74 atm

D
9.2 atm

Solution

⇒ 10–0.66 = 9.33 atm
Question 7
4 / -1

For a gas the difference between the two specific heats is 4150 J/kg K. What is the specific heats at constant volume of gas if the ratio of specific heat is 1.4?

A
8475 J/kg - K

B
5186 J/kg - K

C
1660 J/kg - K

D
10375 J/kg - K

Solution

Given C_P – C_V = 4150 ........(i)

By substituting the value of c_p in equation (i) we get 1.4c_v – c_v = 4150
Question 8
4 / -1

The temperature of 5 mole of a gas which was held at constant volume was changed from 100^oC to 120^oC. The change in internal energy was found to be 80 J. The total heat capacity of the gas at constant volume will be equal to?

A
8JK⁻¹

B
0.8 JK⁻¹

C
4JK⁻¹

D
0.4 Jk⁻¹

Solution

At constant volume total energy will be utilised in increasing the temperature of gas

This is the heat capacity of 5 mole gas.
Question 9
4 / -1

The density of a polyatomic gas is standard conditions is 0.795 kgm^–3. The specific heat of the gas at constant volume is:

A
930J-kg^–1 K^–1

B
. 1400J-kg^–1 K^–1

C
1120J-kg^–1 K^–1

D
925J-kg^–1 K^–1

Solution

Ideal gas equation for m gram gas PV = mrT

[where r = Specific gas constant]
Question 10
4 / -1

If two moles of diatomic gas and one mole of monoatomic gas are mixed with then the ratio of specific heats is:

A
7/3

B
5/4

C
19/13

D
15/19

Solution