Within the next decade, hydrogen fuel cell technology may emerge as a popular en

Question

Within the next decade, hydrogen fuel cell technology may emerge as a popular engine alternative within the automotive industry. Hydrogen gas storage technology, which remains a key issue to public safety, is continuing to advance. However, recent research has shown that polymer tanks that are reinforced with carbon fiber can achieve operating pressures above 50 MPa. For this example, consider a hydrogen refueling station that (for safety reasons) dispenses hydrogen gas to vehicles at a constant 20 MPa. Your car has a 45 L fuel tank. When you pull up to the station, your tank is at a pressure of 200 kPa (absolute). You may assume that the temperature of the hydrogen supply (of the refueling station), as well as the initial temperature of your tank are both at 300 K.

(a) If you add a mere 60 grams of hydrogen into your tank, what is the temperature rise inside of your fuel tank?

K

(b) After that little bit of hydrogen is added, what is the new pressure within your tank?

kPa

(c) If instead of just adding a little fuel, you decide to fill up, how much hydrogen will fit in your tank? **Note: "Filling up" means you hold down the handle until no more fuel will flow into your tank. This happens when your tank pressure reaches the pressure of the hydrogen dispenser (20 MPa).

kg

(d) What is the temperature within your tank after the fillup?

K

Explanation / Answer

Step 1:

formula: Ideal gasses formula:

P.V=n.RT

T=temperature, P=pressure V=volume, n= mols, R= 0.082L. atm/ mol.K

  P2.V2=n.RT2

Step 2: Know data:

Constant= 20 MPa

Volume= 45L

Pressure absolute= 200 kPa

Temperature (hydrogen supply and initial temperature)= 300 K

a) add 60g ¿temperature inside the tank?

  P2.V2=n.RT2

T2= P2.V2/n.R

n= m/ PM

n=60g/1.008g/mol =59.52 mol H+

200 Kpa-------x atm

1 Kpa----------0.009869 atm =1.9738 atm

T2= (1.9738 atm).(45L)/ (59.52 mol)(0.082L. atm/ mol.K)

T2= (88.821 L. atm)/(4.88064 mol.L. atm/ mol.K)

T2= 18.19 K

B.

Step 1: Formula:

Gay Lussac's Law:

P1/T1 = P2/T2

P1.T2= P2. T1

P2= P1. T2/ T1

P2= (200 Kpa)(18.19 K/ (300 K)

P2= 12.13 Kpa

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