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Question-210679




Question Number 210679 by peter frank last updated on 16/Aug/24
Answered by mr W last updated on 16/Aug/24
RH_(16°C) =((273.15+16)/(273.15+20))×((17.5)/(13.6))×36%≈46%
$${RH}_{\mathrm{16}°{C}} =\frac{\mathrm{273}.\mathrm{15}+\mathrm{16}}{\mathrm{273}.\mathrm{15}+\mathrm{20}}×\frac{\mathrm{17}.\mathrm{5}}{\mathrm{13}.\mathrm{6}}×\mathrm{36\%}\approx\mathrm{46\%} \\ $$
Commented by peter frank last updated on 18/Aug/24
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by mr W last updated on 18/Aug/24
is this correct?
$${is}\:{this}\:{correct}? \\ $$
Commented by peter frank last updated on 18/Aug/24
thanks mr W
$$\mathrm{thanks}\:\mathrm{mr}\:\mathrm{W} \\ $$
Commented by peter frank last updated on 18/Aug/24
its correct
$$\mathrm{its}\:\mathrm{correct} \\ $$
Answered by Spillover last updated on 18/Aug/24
Initial temp (T_1 )=20^° C  Final temp (T_2 )=16^°  C  Initia Relative humidity RH_1 =36%  Saturation vapour pressure  at 20°C P_(sat 1) =17.5mmHg  Saturation vapour pressure at 16°C P_(sat 2 ) =13.6mmHg  Actual vapour pressure at 20°C  P_(actual) =RH_1 ×P_(sat1)   P_(actual) =36%×17.5mmHg=6.3mmHg  Relative humidity at 16°C  RH_2 =(P_(actual) /P_(sat 2) )×100  RH_2 =((6.3)/(13.6))×100=46.3%  RH_2 (Relative humidity at 16°C)=46.3%
$${Initial}\:{temp}\:\left({T}_{\mathrm{1}} \right)=\mathrm{20}^{°} {C} \\ $$$${Final}\:{temp}\:\left({T}_{\mathrm{2}} \right)=\mathrm{16}^{°} \:{C} \\ $$$${Initia}\:{Relative}\:{humidity}\:{RH}_{\mathrm{1}} =\mathrm{36\%} \\ $$$${Saturation}\:{vapour}\:{pressure}\:\:{at}\:\mathrm{20}°{C}\:{P}_{{sat}\:\mathrm{1}} =\mathrm{17}.\mathrm{5}{mmHg} \\ $$$${Saturation}\:{vapour}\:{pressure}\:{at}\:\mathrm{16}°{C}\:{P}_{{sat}\:\mathrm{2}\:} =\mathrm{13}.\mathrm{6}{mmHg} \\ $$$${Actual}\:{vapour}\:{pressure}\:{at}\:\mathrm{20}°{C} \\ $$$${P}_{{actual}} ={RH}_{\mathrm{1}} ×{P}_{{sat}\mathrm{1}} \\ $$$${P}_{{actual}} =\mathrm{36\%}×\mathrm{17}.\mathrm{5}{mmHg}=\mathrm{6}.\mathrm{3}{mmHg} \\ $$$${Relative}\:{humidity}\:{at}\:\mathrm{16}°{C} \\ $$$${RH}_{\mathrm{2}} =\frac{{P}_{{actual}} }{{P}_{{sat}\:\mathrm{2}} }×\mathrm{100} \\ $$$${RH}_{\mathrm{2}} =\frac{\mathrm{6}.\mathrm{3}}{\mathrm{13}.\mathrm{6}}×\mathrm{100}=\mathrm{46}.\mathrm{3\%} \\ $$$${RH}_{\mathrm{2}} \left({Relative}\:{humidity}\:{at}\:\mathrm{16}°{C}\right)=\mathrm{46}.\mathrm{3\%} \\ $$$$ \\ $$
Commented by mr W last updated on 18/Aug/24
why is the P_(actual)  the same at 20°C   and at 16°C?
$${why}\:{is}\:{the}\:{P}_{{actual}} \:{the}\:{same}\:{at}\:\mathrm{20}°{C}\: \\ $$$${and}\:{at}\:\mathrm{16}°{C}? \\ $$
Commented by Spillover last updated on 18/Aug/24
The actual vapour pressure remain the same  when the temp changes  (assuming no additional water vapor is added  or removed) becouse amount of water vapor in   the air does not change when temp changes
$${The}\:{actual}\:{vapour}\:{pressure}\:{remain}\:{the}\:{same} \\ $$$${when}\:{the}\:{temp}\:{changes} \\ $$$$\left({assuming}\:{no}\:{additional}\:{water}\:{vapor}\:{is}\:{added}\right. \\ $$$$\left.{or}\:{removed}\right)\:{becouse}\:{amount}\:{of}\:{water}\:{vapor}\:{in}\: \\ $$$${the}\:{air}\:{does}\:{not}\:{change}\:{when}\:{temp}\:{changes} \\ $$
Commented by Spillover last updated on 18/Aug/24
  The actual vapor pressure is a measure of the amount of water vapor present in the air. This value is independent of temperature and is determined by how much water vapor is in the air at any given moment.
$$ \\ $$The actual vapor pressure is a measure of the amount of water vapor present in the air. This value is independent of temperature and is determined by how much water vapor is in the air at any given moment.
Commented by peter frank last updated on 18/Aug/24
thanks spillover
$$\mathrm{thanks}\:\mathrm{spillover} \\ $$

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