a liquid mixture of a and b contains 1 mole of each ,if the pressure is reduced at what pressure does the last part of liquid disappear .? p°a=400 mm hg p°b=100 mm hg
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Given:
A liquid mixture of a and b contains 1 mole of each.
To find:
If the pressure is reduced at what pressure does the last part of liquid disappear .? p°a=400 mm hg p°b=100 mm hg
Solution:
From given, we have,
A liquid mixture of a and b contains 1 mole of each.
The last part of the liquid will disappear when the composition of the vapour phase become χa=0.5 and χb=0.5.
The pressure at which this occurs can be calculated as follows:
1/P = χa/Pa∘ + χb/Pb∘
1/P = 0.5/400 + 0.5/100
1/P = 0.25
P = 4 mmHg
Therefore, the pressure at which the last part of liquid disappear is 4 mmHg
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