What is the momentum of a 5.6 kg ball going 22m/s ?
1 answer:
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Answer :
(a) The value of is 48.6 kJ/mol
(b) The the normal boiling point is 489.2 K
(c) The entropy of vaporization at the boiling point is 99.3 J/K
Explanation :
(a) To calculate of the reaction, we use clausius claypron equation, which is:
where,
= vapor pressure at temperature
= 1.3 kPa
= vapor pressure at temperature
= 5.3 kPa
= Enthalpy of vaporization = ?
R = Gas constant = 8.314 J/mol K
= initial temperature =
= final temperature =
Putting values in above equation, we get:
Therefore, the value of is 48.6 kJ/mol
(b) The clausius claypron equation is:
where,
= vapor pressure at temperature
= 1.3 kPa
= vapor pressure at temperature normal boiling point = 101.3 kPa
= Enthalpy of vaporization = 48.6 kJ/mol
R = Gas constant =
= initial temperature =
= final temperature = ?
Putting values in above equation, we get:
Therefore, the normal boiling point is 489.2 K
(c) Now we have to determine the entropy of vaporization at the boiling point.
where,
= entropy of vaporization = ?
= enthalpy of vaporization = 48.6 kJ/mol
= boiling point = 489.2 K
Now put all the given values in the above formula, we get:
Therefore, the entropy of vaporization at the boiling point is 99.3 J/K