Homes in rural areas where natural gas service is not available often rely on pr

Question

Homes in rural areas where natural gas service is not available often rely on propane to fuel kitchen ranges. The propane is stored as a liquid, and the gas to be burned is produced as the liquid evaporates. Suppose an architect has hired you to consult on the choice of a propane tank for such a new home. The propane gas consumed in 1.0 hour by a typical range burner at high power would occupy roughly 165 L at 25°C and 1.0 atm, and the range chosen by the client will have 4 burners. If the tank under consideration holds 400 gallons of liquid propane, what is the minimum number of hours it would take for the range to consume an entire tank of propane? The density of liquid propane is 0.5077 kgL 3436 3hours

Explanation / Answer

Ans. Step 1: Given, propane gas consumed in 1.0 hr by 1 burner-

            Volume, V = 165 L

            Temperature, T = 25.00C = 298.15 K

            Pressure, P = 1.00 atm

Its assumed that propane gas behave as ideal gas.

Using Ideal gas equation:    PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol-1K-1

            T = absolute temperature (in K) = (0C + 273.15) K

Putting the values in above equation-

            1.0 atm x 165.0 L = n x (0.0821 atm L mol-1K-1) x 298.15 K

            Or, n = 165.0 atm L / (24.478115 atm L mol-1)

            Hence, n = 6.7407 mol

Therefore, 1 burner consumes 6.7407 mol propane in 1 hr.

Or, rate of propane consumption = 6.7407 mol burner-1 hr-1

# Step 2: Given, there are 4 burner.

So,

Total rate of propane consumption for 4 burner =

Rate of propane consumption per burner x number of burner

= (6.7407 mol burner-1 hr-1) x 4 burners

                                    = 26.9628 mol hr-1                         ; [mass = Moles x Mw]

                                    = (26.9628 mol x 44.1 g mol-1) hr-1

                                    = 1189.05948 g hr-1

                                    = 1.189 kg hr-1

# Step 3: Given, volume of propane = 400 gallons

= 400 x (3.78541 L)

= 1514.164 L

# Mass of propane = Volume x Density

                                    = 1514.164 L x (0.5077 kg/ L)

                                    = 768.7410628 kg

Now,

Minimum time of consumption of entire tank =

Total mass of propane / Total rate of propane consumption

= 768.7410628 kg / (1.189 kg hr-1)

= 646.544 hr

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