
Relative humidity of air is 60% and the saturation vapour pressure of water vapour in the air is \[3.6kPa\]. The amount of water vapour present in \[2L\] air at \[300K\] is:
A.\[52g\]
B.\[31.2g\]
C.\[26g\]
D.\[5.2g\]
Answer
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Hint:Vapour pressure is the pressure exerted by a vapour which is in equilibrium with its condensed phase. Relative humidity is the ratio of partial vapour pressure of water and equilibrium vapour pressure of water.
Complete step by step answer:
Vapour pressure is defined as the pressure exerted by a vapour when it is in equilibrium with the condensed form i.e. liquid or solid form, or both. The substance under consideration may exist in three phases under the equilibrium condition.
In other words, vapour pressure is a measure of the tendency of a substance or compound to go into the vapour or gaseous state. The vapour pressure of a material generally increases with temperature.
Relative humidity is defined as the ratio of the partial pressure of water vapour and the equilibrium vapour pressure of water vapour multiplied by \[100\]. It is expressed as
$\phi = \dfrac{{{p_{{H_2}O}}}}{{p_{{H_2}O}^*}} \times 100$ ,
where \[{p_{{H_2}O}}\] is the partial pressure of water vapour and \[p_{{H_2}O}*\] is the equilibrium vapour pressure of water.
From the ideal gas laws, $PV = nRT$, where $P$ is the pressure, \[V\] is the volume, \[n\] is the number of moles of the substance, \[R\] is the gas constant, \[T\] is the temperature.
$PV = \dfrac{w}{M}RT$ , where \[w\], is the weight of the substance and \[M\] is its molar mass.
Given $P = 3.6kPa = 3.6 \times {10^3}Pa$, $V = 2L$, $T = 300K$.
$R = 8.314J/mol/K$ , $M(water) = 18g$ .
Inserting the values,
$w = \dfrac{{3.6 \times {{10}^3} \times 2 \times 18}}{{8.314 \times 300}}$
$w = 51.96g$ .
The relative humidity of air given is \[60\% \] .
Hence the amount of water vapour in \[2L\] air at \[300K\]= \[51.96 \times 0.6 = 31.2g\].
Hence option B is the correct answer, i.e. \[31.2g\].
Note:
Relative humidity is very important for thermal comfort. Humans are comfortable in the humidity range of \[30 - 70\% \]. Low humidity causes respiratory problems and high humidity causes excessive dehydration.
Complete step by step answer:
Vapour pressure is defined as the pressure exerted by a vapour when it is in equilibrium with the condensed form i.e. liquid or solid form, or both. The substance under consideration may exist in three phases under the equilibrium condition.
In other words, vapour pressure is a measure of the tendency of a substance or compound to go into the vapour or gaseous state. The vapour pressure of a material generally increases with temperature.
Relative humidity is defined as the ratio of the partial pressure of water vapour and the equilibrium vapour pressure of water vapour multiplied by \[100\]. It is expressed as
$\phi = \dfrac{{{p_{{H_2}O}}}}{{p_{{H_2}O}^*}} \times 100$ ,
where \[{p_{{H_2}O}}\] is the partial pressure of water vapour and \[p_{{H_2}O}*\] is the equilibrium vapour pressure of water.
From the ideal gas laws, $PV = nRT$, where $P$ is the pressure, \[V\] is the volume, \[n\] is the number of moles of the substance, \[R\] is the gas constant, \[T\] is the temperature.
$PV = \dfrac{w}{M}RT$ , where \[w\], is the weight of the substance and \[M\] is its molar mass.
Given $P = 3.6kPa = 3.6 \times {10^3}Pa$, $V = 2L$, $T = 300K$.
$R = 8.314J/mol/K$ , $M(water) = 18g$ .
Inserting the values,
$w = \dfrac{{3.6 \times {{10}^3} \times 2 \times 18}}{{8.314 \times 300}}$
$w = 51.96g$ .
The relative humidity of air given is \[60\% \] .
Hence the amount of water vapour in \[2L\] air at \[300K\]= \[51.96 \times 0.6 = 31.2g\].
Hence option B is the correct answer, i.e. \[31.2g\].
Note:
Relative humidity is very important for thermal comfort. Humans are comfortable in the humidity range of \[30 - 70\% \]. Low humidity causes respiratory problems and high humidity causes excessive dehydration.
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