The density of water is $1000$ kg per meter cube. The density of water vapour at $100$ degree Celsius and $1$ atmospheric pressure is $0.6$ kg per meter cube. The volume of a molecule multiplied by the total number gives what is called molecular volume. Estimate the ratio of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure.

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Solution

The density of water $= 1000 k h / m^{3}$
The density of water vapour $= \frac{100 ° C}{0.6 K g / m^{3}}$
Pressure $= 1 a t p$
The temperature $= 100 ° C$
Estimation of the ratio of the molecular volume to the total volume occupied by the water vapour under the above condition of temperature and pressure.
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
(according to Charle's and Boyle's law)
where $P_{1}$ and $P_{2}$ $ be the density of water and water vapour respectively.
Since pressure and temperature are same and $V_{1}$ and $V_{2}$ be the volume of the water and water vapour respectively.
$P_{1} V_{1} = P_{2} V_{2} 1000 \times V_{1} = 0.6 \times V_{2} \frac{V_{1}}{V_{2}} = \frac{6}{10 \times 1000} = 6 : 10000 = 3 : 5000$

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