Which of the following is(are) true about water boiling in a container that is open to the atmosphere? I. dH= 0, II. dS=0, III. dG=0
(A) I. only (B) I. and II. only (C) III. only (D) II. and III. only
Ok, so I know the answer here as it is given in my study guide, which is choice (C) III. only. However this does not make sense to me. How is water boiling in an open system considered to be at equilibrium?? Won\'t the vapor disperse out into the atmosphere- how can it be possible that condensation would be occuring at the same rate that vaporization is occuring in this setting? I got the question and its answer from the American Chemical Society studyguide, so I know it must be legit. Just hoping someone can explain. Thanks in advance.
Solution
Since boiling water is an endothermic process - heat is absorbed by water dH = + sign and
from liqiud state to gaseous state entropy increases dS = + sign.
we can say now the answer is only III. dG = 0 that is in equilibrium, but we need reason
from earth no gas will escape due to high escape velocity Ve = 11.2 km/sec
there is already water vapor exist in the atmosphere that is always in equilibrium with any evaporation process undewent on earth. By definition 1 atmosphere is water is in equilibrium with its vapor also. This process is neither spontaneous nor non-spontaneous since water boiling requires external heat energy as long as we supply heat energy the process is forward once we stop input heat energy the process stopped. Hence we can say dG = 0 only not - (OR) +. condensation <==> vaporization is correct due to outside temperature is lower than boiling temperature otherwise we cannot exist on earth.
Hope this helped you!
Thank You So Much! Please Rate this answer as you wish.(\"Thumbs Up\")
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Which of the following is(are) true about water boiling in a container.docx
1. Which of the following is(are) true about water boiling in a container that is open to the
atmosphere? I. dH= 0, II. dS=0, III. dG=0
(A) I. only (B) I. and II. only (C) III. only (D) II. and III. only
Ok, so I know the answer here as it is given in my study guide, which is choice (C) III. only.
However this does not make sense to me. How is water boiling in an open system considered to
be at equilibrium?? Won't the vapor disperse out into the atmosphere- how can it be possible
that condensation would be occuring at the same rate that vaporization is occuring in this setting?
I got the question and its answer from the American Chemical Society studyguide, so I know it
must be legit. Just hoping someone can explain. Thanks in advance.
Solution
Since boiling water is an endothermic process - heat is absorbed by water dH = + sign and
from liqiud state to gaseous state entropy increases dS = + sign.
we can say now the answer is only III. dG = 0 that is in equilibrium, but we need reason
from earth no gas will escape due to high escape velocity Ve = 11.2 km/sec
there is already water vapor exist in the atmosphere that is always in equilibrium with any
evaporation process undewent on earth. By definition 1 atmosphere is water is in equilibrium
with its vapor also. This process is neither spontaneous nor non-spontaneous since water boiling
requires external heat energy as long as we supply heat energy the process is forward once we
stop input heat energy the process stopped. Hence we can say dG = 0 only not - (OR) +.
condensation <==> vaporization is correct due to outside temperature is lower than boiling
temperature otherwise we cannot exist on earth.
Hope this helped you!
Thank You So Much! Please Rate this answer as you wish.("Thumbs Up")
