Question:

The relative humidity on a day when partial pressure of water vapour is $0.012 \times 10^{5} Pa$ at $12^{\circ} C$ is (Take vapour pressure of water at this temperature as $0.016 \times 10^{5}Pa)$

Updated On: Jan 18, 2023
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The Correct Option is C

Solution and Explanation

Given,
Partial pressure of water vapour $=0.125 \times 10^{5} Pa$
Vapour pressure of water $=0.016 \times 10^{5} Pa$
Relative humidity at a given temperature $(R)$
$=\frac{\text { Partial pressure of } \text { water } \text { vapour }}{\text { vapour pressure of water }}$
$=\frac{0.125 \times 10^{5}}{0.016 \times 10^{5}}$
$=0.75$
$=75 \%$
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Concepts Used:

Partial Pressure

Partial Pressure is defined as if a container filled with more than one gas, each gas exerts pressure. The pressure of anyone gas within the container is called its partial pressure.

Dalton’s Law of Partial Pressure:

According to Dalton’s law of partial pressures, the total pressure exerted by the mixture of gases is the sum of the partial pressure of every existing individual gas, and every gas is assumed to be an Ideal gas.

Ptotal = P1 + P2 + P3

Where P1, P2, P3 are the partial pressures of gas 1, gas 2, and gas 3. Since every gas has an independent behavior, the ideal gas law is used to find the pressure of that gas if its number of moles, the volume of container and temperature is known.