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Introductory Physics Homework Help
Pressure in a bomb calorimeter at the moment of combustion
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[QUOTE="Seibtsantos, post: 6439627, member: 685824"] [B]Homework Statement:[/B] 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. [B]Relevant Equations:[/B] 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! [/QUOTE]
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Pressure in a bomb calorimeter at the moment of combustion
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