please help out guys!!!
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Hey there!!
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Water will be in equilibrium with its vapour when the vapour gets saturated. In this case, the pressure of vapour is equal to the saturation vapour pressure = 15 mm of mercury.
= (15 * 10^-3 m) (13600 kg m ^-3) (9.8 m s^-2)
= 2000 N m ^-2
Using gas law,
pV = m/M(RT)
therefore,
m = MpV/ RT
= [ (18 g mol ^-1)(2000 N m^-2) (76 m^-3) ] / [( 8.3 J K ^-1 mol ^-1) (288 K)]
= 1145 g
= 1.14 kg
Thus, 1.14 kg of water will evapourate .
====================================================
HOPE IT HELPS!!!
===================
Water will be in equilibrium with its vapour when the vapour gets saturated. In this case, the pressure of vapour is equal to the saturation vapour pressure = 15 mm of mercury.
= (15 * 10^-3 m) (13600 kg m ^-3) (9.8 m s^-2)
= 2000 N m ^-2
Using gas law,
pV = m/M(RT)
therefore,
m = MpV/ RT
= [ (18 g mol ^-1)(2000 N m^-2) (76 m^-3) ] / [( 8.3 J K ^-1 mol ^-1) (288 K)]
= 1145 g
= 1.14 kg
Thus, 1.14 kg of water will evapourate .
====================================================
HOPE IT HELPS!!!
brainly567:
thank you
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