**5.23 ** Explain the physical significance of van der Waals parameters.

The equation of van der Waals after taking into account the corrections for pressure and volume,
Where a and b are called van der Waals constants or parameters.
Here the significance of a and b is important:
Value of 'a' is a measure of the magnitude of intermolecular attractive forces within the gas and is independent of temperature and pressure.
And the value of 'b' is the volume occupied by...

**5.22 ** Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why?

The critical temperature we know is the highest temperature at which liquid exists above it is gas.
Given that the critical temperature of Carbon dioxide and Methane are.
Higher is the critical temperature of a gas, easier is its liquefaction.
So, as the critical temperature increases the gas is now easier to liquefaction.
That means the intermolecular forces of attraction between the...

**5.21 ** In terms of Charles’ law explain why –273 °C is the lowest possible temperature.

Charles' Law states that pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature.
Charles found that for all gases, at any given pressure, the graph of volume vs temperature (in Celsius) is a straight line and on extending to zero volume, each line intercepts the temperature axis at – 273.15 ° C.
We can see that the volume of the...

**5.20** What would be the unit for the quantity ?

Given quantity, ,
The SI units of each factors are,
For pressure p is .
For Volume V is .
For Temperaute T is.
For number of moles, n is mol.
Therefore we have for the quantity
The SI unit is

**5.19** A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen.
So, let us assume the weight of dihydrogen in a total mixture weight of be and dioxygen be .
Then, the number of moles of will be, and the number of moles of , .
And given that the total pressure of the mixture is .
Then we have a partial pressure of ,
Therefore, the partial pressure of is .

**5.18 ** 2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?

Let the molar mass of the gas be ,
Then, given that of a gas at occupied the same volume as
of at
So, we have the relation, and
Therefore
wihch gives,
Or,
Or,

**5.17** Calculate the volume occupied by 8.8 g of at 31.1°C and 1 bar pressure. .

Given the mass of carbon dioxide is 8.8grams at and at 1 bar pressure.
So, the number of moles of
Now, the pressure of
Given
also, the
The Ideal gas equation;
So volume occupied by 8.8 g of carbon dioxide is 5.048L

**5.16 ** Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C.

(Density of and ).

The payload can be defined as:
(Mass of the displaced air - Mass of the balloon)
Given the radius of the balloon, r = 10 m.
Mass of the balloon, m = 100 kg.
Therefore, the volume of the balloon will be:
Now, the volume of the air displaced:
The mass of the air displaced :
Let be the mass of helium gas filled into the balloon, then
Or,
approximately.
The balloon is filled with He...

**5.15** Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of at 27°C. .

Given the mass of oxygen gas and mass of hydrogen gas.
Molar mass of
Therefore
Molar mass of
Therefore
Total number of moles of mixture
and
So, Ideal gas equation;
or
Therefore Total pressure is

**5.14** How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second ?

Given that grains are distributed each second.
Time taken to distribute grains = 1second.
Time taken to distribute one Avogadro number of wheat grains.

**5.13** Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Dinitrogen has a molar mass .
Given 1.4g of gas.
That means
number of molecules.
And as 1 molecule of has 14 electrons.
So, molecules of has electrons.

**5.12** Calculate the temperature of of a gas occupying at 3.32 bar.

We have the ideal gas equation,
Pressure, p=3.32bar
Volume, V=5
Number of moles, n=4 mol
Gas constant R= 0.083
Temperature, T=?
so, temperature, T=50K

**5.11 ** A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out?

Assume the round-bottomed flask has volume
so, the volume of air in the flask at .
To find out how much air has been expelled out, we use Charle's Law:
Therefore
Or, Fraction of air expelled

**5.10** of phosphorus vapour weighs 0.0625 g at and 0.1 bar pressure. What is the molar mass of phosphorus?

Given,
The volume of phosphorus vapour which weights about .
Temperature
Pressure
Hence we apply the ideal gas equation,
, where the number of moles will be
Therefore the molar mass is

**5.9** Density of a gas is found to be at at 2 bar pressure. What will be its density at STP ?

Given the density of a gas is equal to at and at 2bar pressure.
Density , for the same gas at different temperatures and pressures we can apply,
Here, , ,
then at STP, we have
or

**5.8 ** What will be the pressure of the gaseous mixture when of at bar and of dioxygen at bar are introduced in a 1L vessel at ?

Pressure of the gas mixture will be the sum of partial pressure of gas and partial pressure of gas.
So, calculating the partial pressures of each gas.
Partial pressure of in a 1L volume vessel.
,
As the temperature remains constant,
or or
Now, calculating the partial pressure of gas in 1L vessel.
or
Therefore the total pressure

**5.7** What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a flask at ?

Given a mixture of 3.2g of methane and 4.4g of carbon dioxide contained a flask at .
So, the pressure exerted by the mixture will be
The pressure exerted by the Methane gas,
The pressure exerted by the Carbon dioxide gas,
So, the total pressure exerted
And in terms of SI units, we get,

**5.6** The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminum reacts?

We have the chemical reaction:
(Where dihydrogen is being produced.)
So, at and pressure, the volume of gas that will be released when of aluminium reacts.
As we can see from the reaction equation that 2 moles of aluminium produce 3 moles of dihydrogen.
i.e., reacts to give
At STP condition
and ,
of Al gives of .
Therefore of Al will give of
i.e., of .
At STP condition:
...

**5.5** Pressure of 1 g of an ideal gas A at is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

Given Pressure of 1g of an ideal gas A at is .
When 2g of another ideal gas B is introduced in the same flask at the same temperature,
The pressure becomes .
.
We can assume the molecular masses of A and B be respectively.
The ideal gas equation,
So, we have and
Therefore,
or,
Hence the relation between the two gases is .

**5.4 ** At , the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

Given the condition :
Temperature ,
for oxide gas and for dinitrogen.
As Density of gas is represented by,
And given that density is the same for both the gases at the same temperature condition.
So, we have (as R is constant}
or,
or,

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