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Helga [31]
3 years ago
10

A 25-liter sample of steam at 100°c and 1.0 atm is cooled to 25°c and expanded until the pressure is 19.71 mmhg. if no water con

denses, calculate the final volume of the water vapor. 770 l 240 l 1.3 x 10-3 l 1.0 l
Chemistry
2 answers:
qwelly [4]3 years ago
8 0
I will use [pV/T] in the state 1 = [pV/T] in the state 2.

State 1:
p = 1.0 atm
V = 25 liter
T = 100 + 273.15 = 373.15 K

State 2:

p = 19.71 mmHg * 1.atm / 760 mmHg = 0.0259atm
V= ?
T = 25 + 273.15 = 298.15 K

Application of the formula

1.0 atm * 25 liter / 373.15 k = 0.0259 atm * V / 298.15 K =>
V = [1.0atm * 25 liter / 373.15 K]*298.15K/0.0259atm = 771 liter

 
abruzzese [7]3 years ago
6 0

the correct answer is 770 L. I had this question this morning.

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How much ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.
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Answer:

An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of −17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.

The answer to the above question is

Therefore 0.1133 kg  ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C

Explanation:

To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows

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