Question

The lab focuses on CO2 and its chemistry. Therefore it is useful to think about our role in producing CO2 The ∆H for the combustion of methane is -882.0 kJ/mol. Methane can be a useful model for natural gas since natural gas is about 95% methane and 5% other compounds. Therefore we can use the combusion of methane as a substitute for natural gas. a. Recently, the T.B Simon Power Plant on MSU's campus began burning natural gas as its primary fuel source. The heat produced from fuel burned for electricity production runs the campus's heating system, which is used to heat the residence halls and other campus buildings. If a typical dorm room is 12.2 ft by 11.8 ft by 9.5 ft, what is the volume of the empty room in Liters? (Recall that 1cm=1mL.)

a) How much energy (in kJ) is required to raise the temperature of the air in the empty room from 11 degrees Celsius to 25 degrees Celsius? (Specific heat for air is 0.030 kJ/(mol K) )

b) What mass of natural gas (in kg) is required to provide that much energy? (Formula mass of natural gas =16.042 g/mol)

c) Given that the efficiency of heating your room is roughly 35%, what mass of natural gas is actually necessary to heat your room from 11 to 25 degrees C?

d) Making the assumption that natural gas is 100% methane, what mass of CO2 (in kg) would be produced from heating up the air in your room?

Answer 1

Room volume = 38727 L

a) 726.2 kJ

b) 0.013 kg

c) 0.037 kg

d) 0.103 kg

Step-by-step explanation:

The total volume of the room is the multiplication of its dimensions:

12.2 * 11.8 *9.5 = 1367.62 ft³

1 ft³ = 28.3168 L

So, the volume of the room is 38727 L

a) The volume of the air in the room is the room volume: 38727 L. If we consider the air as an ideal gas, the volume of 1 mol of it at 1 atm must be 22.4 L. Let's admit that in the temperature range given the volume will not change. So, the number of moles of it is:

n = 38727/22.4 = 1729 moles

The heat to raise the temperature is given by the equation:

Q = n*c*ΔT

Where c is the specific heat, and ΔT the temperature variation (final - initial).

Q = 1729*0.030*(25 - 11)

Q = 726.2 kJ

b) The enthalpy of combustion of the gas indicates how much heat per mol it provides. Thus, the number of moles of the gas that provides 726.2 kJ of heat is:

n = 726.2/882

n = 0.8233 mol

The mass is the number of moles multiplied by the molar mass:

m = 0.8233 * 16.042

m = 13.21 g = 0.013 kg

c) Let's make a simple indirect three rule (if the efficiency decreases, the heat must increase):

100% efficience -------- 726.2 kJ

35% ------ x

35x = 72620

x = 2074.86 kJ

So, the number of moles of the methane will be:

n = 2074.86/882

n = 2.35 moles

And the mass:

m = 2.35 * 16.042

m = 37.74 g = 0.037 kg

d) The combustion reaction of methane is:

CH4 + 2O2 --> CO2 + 2H2O

So, 1 mol of methane produces 1 mol of CO2. The molar mass of CO2 is 44 g/mol, thus transforming the stoichiometric relation to mass relation:

1mol of CH4 * 16.042 ----------- 1 mol of CO2 * 44

16.042 g of CH4 --------------- 44 g of CO2

37.74 g --------------- x

By a simple direct three rule:

16.042x = 1660.56

x = 103.51 g

x = 0.103 kg of CO2