The hydrogen-oxygen fuel cell is described in the text. For the questions below,

Question

The hydrogen-oxygen fuel cell is described in the text. For the questions below, assume the following:

Assume that the air is 20 % O2 by volume and that all the O2 is consumed in the cell. The other components of air do not affect the fuel-cell reactions. Assume ideal gas behavior.

(a) What volume of H2(g), stored at 25.0°C at a pressure of 155 atm, would be needed to run an electric motor drawing a current of 8.50 A for 3.00 hr ? (answer in L) (b) What volume (in liters) of air at 25.0°C and 1.12 atm will have to pass into the cell per minute to run the motor ? (answer in L / min)

Explanation / Answer

a) Current = A = 8.50 A

Time = t = 3.00 h = 3.00*60*60 = 10800 s

Charge = current*time = 8.50*10800 = 91800 C

1 mole equivalent = 96500 C

Number of mole equivalents in 91800 C = 91800/96500 = 0.951 mol eq

For H2, 1 mol = 2 mol equi

So, number of moles in 0.951 mol eq. = 0.951/2 = 0.476 mol

moles = n = 0.476 mol

Temperature = T = 25 oC = 273+25 = 298 K

Pressure = P = 155 atm

R = gas constant = 0.082 L.atm/mol.K

According to ideal gas equation, PV = nRT

V = nRT/P = 0.476*0.082*298/155 = 0.075 L

Volume = 0.075 L

b) Air contains O2. For O2, 1 mol = 4 mol equivalents.

So, from part a, number of mole equivalent = 0.951 mol equi

For O2, number of moles = 0.951/4 = 0.238 mol

pressure = P = 1.12 atm

Volume = nRT/P = 0.238*0.082*298/1.12 = 5.19 L

But the air contains 20 % O2. So, total air volume =

20 % of air = 5.19 L

Volume of air = 5.19/0.20 = 25.95 L

volume of air = 25.95 L

Time taken = 3.00 h = 3.00*60 = 180 min

Volume per min = 25.95/180 = 0.144 L/min

Voume of air needed = 0.144 L/min

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