An entropy 3.00 L Steel vessel is fluid with 1.00 atm of CH4(g) and 4.00 atm of
ID: 861504 • Letter: A
Question
An entropy 3.00 L Steel vessel is fluid with 1.00 atm of CH4(g) and 4.00 atm of O2(g) at 300 degree C. A spark causes the CH4 to burn completely, according to the equation: CH4(g)+2O2(g) right arrow CO2(g) + 2H2O (g) Delta H=-802 kJ a) What mass of CO2 (g) is produced in reaction? b) What is the final temperature inside the vessel after combustion, assuming that the street vessel has a mass of 14.500 kg, the mixture of gases has an average molar heat capacity of 21 J/(mol. Degree C ), and the heat capacity of a steel is 0.449 J/kg. degree C)? c) What is the partial pressure of CO2 (g) in the vessel after Combustion?Explanation / Answer
CH4 + **3O2** ------> CO2 + 2H20
You end up with one mole CO2, 1 mole O2, and 2 moles H2O for every mole CH4 you had to start with
heat available = 802,000 J x no. moles CH4 to start with
So you need to get moles CH4 using PV = nRT
every degree increase in T absorbs 14,500 x 0.449 J heating the vessel, and 4n x 21 J heating the gases. (Make sure you understand where that came from)
So divide total heat released by heat absorbed per degree to get the temp increase.
I hope I am able to help you!