When 8.70 kJ of thermal energy is added to 2.50 mol of liquid methanol, it vaporizes. Determine the heat of vaporization in kJ/m
1 answer:
Answer:
The heat of vaporisation of methanol is "3.48 KJ/Mol"
Explanation:
The amount of heat energy required to convert or transform 1 gram of liquid to vapour is called heat of vaporisation
When 8.7 KJ of heat energy is required to vaporize 2.5 mol of liquid methanol.
Hence, for 1 mol of liquid methanol, amount of heat energy required to evaporate the methanol is =
= 3.48 KJ
So, the heat of vaporization
Therefore, the heat of vaporization of methanol is 3.48KJ/Mol
You might be interested in
Answer : The the partial pressure of the nitrogen gas is 0.981 atm.
The total pressure in the tank is 2.94 atm.
Explanation :
The balanced chemical reaction will be:
First we have to calculate the moles of dimethylhydrazine.
Mass of dimethylhydrazine = 150 g
Molar mass of dimethylhydrazine =60.104 g/mole
Now we have to calculate the moles of gas.
From the balanced chemical reaction we conclude that,
As, 1 mole of react to give 3 moles of
gas
So, 2.49 mole of react to give
moles of
gas
Now we have to calculate the partial pressure of nitrogen gas.
Using ideal gas equation :
where,
P = Pressure of gas = ?
V = Volume of gas = 250 L
n = number of moles gas = 7.47 mole
R = Gas constant =
T = Temperature of gas =
Putting values in above equation, we get:
Thus, the partial pressure of the nitrogen gas is 0.981 atm.
Now we have to calculate the total pressure in the tank.
Formula used :
where,
= total pressure = ?
= partial pressure of nitrogen gas = 0.981 atm
= moles of nitrogen gas = 3 mole (from the reaction)
= total moles of gas = (3+4+2) = 9 mole (from the reaction)
Now put all the given values in the above formula, we get:
Thus, the total pressure in the tank is 2.94 atm.