- #1
Seibtsantos
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- Homework Statement
- The calorific value of food can be determined in a bomb calorimeter, which consists of a hermetically sealed stainless steel container, in which the sample is burned in the presence of pure oxygen and the resulting heat measured by the temperature variation of a water bath surrounding the container. Due to the thickness and mass of the container, the water bath undergoes a few degrees variation, however at the time of combustion the internal temperature can reach 1200°C.
In determining the heat of combustion of a glucose sample (C6H12O6), 1.8 g of this substance was placed in a calorimetric pump and the container (1 L capacity) was pressurized to 30 atm with pure oxygen (1.23 moles ) and the ignition perpetrated.
Consider: The adiabatic system, a complete combustion at 1200 °C and that the gases generated are ideal gases. Data: R = 0.082 atm.L.K-1.mol-1; Molar Mass (C6H12O6) = 180 g/mol.
The internal pressure of the container at the exact moment of the sample combustion is:
a) 124 atm.
b) 145 atm.
►c) 152 atm.
d) 163 atm.
e) 173 atm.
- Relevant Equations
- Ideal gas law
First, I calculated the number of moles of glucose.
n = m / M
n = 1.8 / 180
n = 0.01 moles of glucose
So I checked the combustion reaction.
1 C6H12O6 + 6 O2 -> 6 CO2 + 6 H20
1 + 6 -> 6 + 6
0.01 + 0.06 -> 0.06 + 0.06
I considered the number of moles at the end of the reaction.
I subtracted the number of moles from the initial oxygen by the oxygen consumed.
1.23 - 0.06 = 1.17 moles
I calculated the total number of moles.
0.06 + 0.06 + 1.17 = 1.29 moles
So I used it in the equation.
pV = nRT
p = (nRT) / V
p = (1.29 * 0.082 * 1473) / 1 (considering 1200 ° C = 1473K)
p = 155.8 atm
I didn't get to the result and I don't know what is missing
Thank you!
n = m / M
n = 1.8 / 180
n = 0.01 moles of glucose
So I checked the combustion reaction.
1 C6H12O6 + 6 O2 -> 6 CO2 + 6 H20
1 + 6 -> 6 + 6
0.01 + 0.06 -> 0.06 + 0.06
I considered the number of moles at the end of the reaction.
I subtracted the number of moles from the initial oxygen by the oxygen consumed.
1.23 - 0.06 = 1.17 moles
I calculated the total number of moles.
0.06 + 0.06 + 1.17 = 1.29 moles
So I used it in the equation.
pV = nRT
p = (nRT) / V
p = (1.29 * 0.082 * 1473) / 1 (considering 1200 ° C = 1473K)
p = 155.8 atm
I didn't get to the result and I don't know what is missing
Thank you!