Answer : The normal boiling point of ethanol will be, 
or 
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of ethanol at 
= 98.5 mmHg
= vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
= temperature of ethanol = 
= normal boiling point of ethanol = ?
= heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the normal boiling point of ethanol will be, or
<span>To solve this we assume that the gas inside the balloon is an ideal </span>gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant volume pressure and number of moles of the gas
the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:
T1/P1 = T2/P2
P2 = T2 x P1 / T1
P2 = 25 x 29.4 / 75
P2 = 9.8 kPa
