Answer:
V = 12.93 L
Explanation:
Given data:
Number of moles = 0.785 mol
Pressure of balloon = 1.5 atm
Temperature = 301 K
Volume of balloon = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will put the values.
V = nRT/P
V = 0.785 mol × 0.0821 atm.L/ mol.K × 301 K / 1.5 atm
V = 19.4 L /1.5
V = 12.93 L
Flammable liquid,gasoline, oil, and etc
Hydrogen Bonding will effect the boiling point the most. Let's take an example Butane a four carbon unsaturated organic compound with molecular formula C₄H₁₀ and boiling point -1 °C.
H₃C-CH₂-CH₂-CH₃
Now, replace one hydrogen on terminal carbon with -OH group and convert it into Butanol.
H₃C-CH₂-CH₂-CH₂-OH
The Boiling point of Butanol is 117.7 °C. This increase in boiling point is due to formation of hydrogen bondings between the molecules of Butanol.