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Question: The density of water is \(1000kg\) per meter cube. The density of water vapour at \(100{}^\circ C\) ...

The density of water is 1000kg1000kg per meter cube. The density of water vapour at 100C100{}^\circ C and 1 atmospheric pressure1\text{ atmospheric pressure} is 0.6kg0.6kg per meter cube. The volume of a molecule multiplied by the total number will give the quantity called molecular volume. Find out the ratio of the molecular volume to the total volume which is occupied by the water vapour under the above mentioned conditions of temperature and pressure.

Explanation

Solution

According to Boyle’s law and Charle’s law, we can write the ratio of the molecular volume to the total volume by taking the product of density and volume and dividing it by the value of temperature. This will be equal in both situations. Compare them and find out the answer.

Complete answer:

The density of water is given as,

d1=1000kgm3{{d}_{1}}=1000\dfrac{kg}{{{m}^{3}}}

The density of water vapour can be written as,

d2=0.6kgm3{{d}_{2}}=0.6\dfrac{kg}{{{m}^{3}}}

The temperature is mentioned as,

T=100CT=100{}^\circ C

Pressure is given as,

P=1atpP=1\,atp

Now according to Charle’s law and Boyle’s law, let us find out the ratio of the molecular volume to the total volume occupied by the water vapour under the mentioned condition of temperature and pressure.

d1V1T1=d2V2T2\dfrac{{{d}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{d}_{2}}{{V}_{2}}}{{{T}_{2}}}

Where d1{{d}_{1}} and d2{{d}_{2}} be the density of water and water vapour respectively.

Since pressure and temperature are said to be equal, V1{{V}_{1}} and​ V2{{V}_{2}}​ be the volume of the water and water vapour respectively. Therefore we can write that,

V1d1=V2d2{{V}_{1}}{{d}_{1}}={{V}_{2}}{{d}_{2}}

Substituting the values of densities in it,

1000×V1=0.6×V21000\times {{V}_{1}}=0.6\times {{V}_{2}}

From this we can find the ratio of the volume,

$\dfrac{{{V}{1}}}{{{V}{2}}}=\dfrac{0.6}{1000} \\

\Rightarrow \dfrac{{{V}{1}}}{{{V}{2}}} =\dfrac{6}{10000} \\

\therefore \dfrac{{{V}{1}}}{{{V}{2}}} =\dfrac{3}{5000}$

Therefore,the ratio of the molecular volume to the total volume which is occupied by the water vapour under the above mentioned conditions of temperature and pressure is 35000\dfrac{3}{5000}.

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Note:

Boyle’s law says that the pressure of a gas will be varying inversely proportional to the volume of the gas occupied at a constant temperature. Charle’s law says that the volume of the gas occupied will be directly proportional to the temperature at absolute zero at a fixed pressure.