It V is the volume of one molecule of gas under given conditions, the van der Waal’s constant b is

4 V
\(\frac{4V}{N_{0}}\)
\(\frac{N_{0}}{4V}\)
4VN₀

Solution

van der Waals’s constant b = 4 times the actual volume of 1 mole molecules = 4VN₀

At 100°C and 1 atm, if the density of liquid water is 1.0 g cm^–3 and that of water vapour is 0.0006 g cm^–3, then the volume occupied by water molecules in 1 litre of steam at that temperature is

6 cm³
60 cm³
0.6 cm³
0.06 cm³

Solution

Mass of 1 L of vapour = volume × density= 1000 × 0.0006 = 0.6g

V of liquid water = ^mass⁄_density = ^0.6⁄₁ = 0.6 cm³

The ratio ^a⁄_b (a and b being the van der Waal’s constants of real gases) has the dimensions of

atm mol^-1
L mol^-1
atm L mol^-1
atm L mol^-2

Solution

A bottle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are opened simultaneously at both ends the white ammonium chloride ring first formed will be

at the centre of the tube.
near the hydrogen chloride bottle.
near the ammonia bottle.
throughout the length of the tube.

Solution

\(Rate \; of \; diffusion \propto \sqrt{\frac{1}{Molecular \; mass}}\)

∵ Molecularmass of HCl > Molecular mass of NH₃

∴ HCl diffuses at slower rate and white ammonium chloride is first formed near HCl bottle.

Diffusion of helium gas is four times faster than

CO₂
SO₂
NO₂
O₂

Solution

When helium is allowed to expand into vacuum, heating effect is observed. Its reason is that

helium is an ideal gas
helium is an inert gas
the inversion temperature of helium is very low
the boiling point of helium is the lowest amongst the elements

Solution

Since the inversion temperature of helium is very low, hence during the expansion into vacuum, heating effect is observed.

Calculate the total pressure in a 10.0 L cylinder which contains 0.4g helium, 1.6 g oxygen and 1.4 g nitrogen at 27°C.

0.492 atm
49.2 atm
4.52 atm
0.0492 atm

Solution

Given T = 27°C = 27 + 273 = 300 K

V = 10.0 L

Mass of He = 0.4 g

Mass of oxygen = 1.6 g

Mass of nitrogen = 1.4 g

nHe = 0.4/4 = 0.1

n O₂ = 1.6/32 = 0.05

n N₂ = 1.4/28 = 0.05

n total = n He + n O₂+ n N₂ = 0.1 + 0.05 + 0.05 = 0.2

\(P = \frac{nRT}{V} = \frac{0.2 \times 0.082 \times 300}{10} = 0.492 \; atm\)

Consider a real gas placed in a container. If the inter molecular attractions are supposed to disappear suddenly which of the following would happen?

The pressure decreases
The pressure increases
The pressure remains unchanged
The gas collapses

Air at sea level is dense. This is a practical application of

Boyle’s law
Charle’s law
Kelvin’s law
Brown’s law

Solution

dP, ∝ Boyle’s law, (d = ^MP⁄_RT). At sea level pressure is more, hence density of air is more.

Use of hot air balloons in sports and meteorological observations is an application of

Boyle’s law
Charle’s law
Kelvin’s law
Gay-Lussac’s law

Solution

Hot air is lighter due to less density (Charle’s law)

(d = ^MP⁄_RT)

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