Gaseous States Test

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Gaseous States Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following expressions is true for an ideal gas?

A

B

C

D

Solution

For an ideal gas

△U = nC_v△T

PV=nRT

Now
Question 2
1 / -0

A pre-weighed vessel was filled with oxygen at NTP and weighed. It was then evacuated, filled with SO₂ at the same temperature and pressure, and again weighed. The mass of oxygen will be?

A
The same as that of SO₂

B
Half that of SO₂

C
Twice that of SO₂

D
One-fourth of SO₂

Solution

At same temperature & pressure

∴ The no. of moles of different gases are same.

∴ At NTP

mole of O₂= mole of SO₂

Suppose each gas occupies 1 mole

1 mole O₂=1 mole SO₂

32 gm=64 gm

∴ The mass of oxygen will occupy half of SO₂
Question 3
1 / -0

A gas behaves most like an ideal gas under conditions of:

A
high pressure and low temperature

B
high temperature and high pressure

C
low pressure and high temperature

D
low pressure and low temperature

Solution

From van der Waals equation

At low pressure and high temperature, intermolecular forces become weak and gas behaves like an ideal gas according to the equation.
Question 4
1 / -0

The compressibility factor of a gas is less than unity at STP. Therefore,

A
V_m > 22.4 litres

B
V_m < 22.4 litres

C
V_m = 22.4 litres

D
V_m = 44.8 litres

Solution
Question 5
1 / -0

Equal masses of ethane and hydrogen are mixed in an empty container at 25^oC. The fraction of the total pressure exerted by hydrogen is?

A
1:2

B
1:1

C
1:16

D
15:16

Solution
Question 6
1 / -0

There are 6.02×10²² molecules each of N₂,O₂and H₂ which are mixed together at 760 mm and 273 K. The mass of total mixture in grams is

A
6.2 gm

B
4.12 gm

C
3.09 gm

D
7 gm

Solution
Question 7
1 / -0

Vapour density of a gas is 11.2. The volume occupied by 11.2 g of this gas at STP is:

A
22.4l

B
11.2l

C
1l

D
2.25l

Solution

Vapour density of a gas is 11.2.

Molecular weight is twice its vapour density.

It is 2 × 11.2 = 22.4 g/mol.

At STP 1 mole of gas = 22.4 L.

At STP 0.5 mole of gas = 0.5 × 22.4=11.2 L.
Question 8
1 / -0

When an ideal gas undergoes unrestrained expansion, no cooling occurs because of the molecules:

A
are above the inversion temperature.

B
exert no attractive forces on each other.

C
do work equal to loss in kinetic energy.

D
collide without loss of energy.

Solution

According to postulates of kinetic theory, there is no intermolecular attractions or repulsions between the molecules of ideal gases.

Thus when an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules do not exert any attractive forces on each other.
Question 9
1 / -0

An ideal gas obeying kinetic gas equation :

A
can be liquefied if its temperature is more than critical temperature

B
can be liquefied at any value of any value T and P

C
can not be liquefied under any value of T and P

D
can be liquefied if its pressure is more than critical pressure

Solution

An ideal gas is meant to have no intermolecular force of attraction

That's why a real gas has to reach, lowest pressure and highest temperature enough to overcome the intermolecular force of attraction and behave like an ideal gas.

So ideal gas can't be liquefied at any T and P.
Question 10
1 / -0

A bottle of cold drink has 200 mL liquid in which CO₂ is 0.1 molar. If CO₂behaves as ideal gas the volume of C0₂ at S.T.P. solution of cold drink is

A
0.224 litres

B
0.448 litres

C
22.4 litres

D
2.24 litres

Solution