- 25.0-L tank contains 295 g N 2 , 135 g O 2 , and 24.0 He gases. Calculate the

Question

- 25.0-L tank contains 295 g N2, 135 g O2, and 24.0 He gases. Calculate the partial pressure of each gas and the total gas pressure at 20.0oC.

         (Answer: PN2 = 10.1 atm; PO2 = 4.06 atm; PHe = 5.77 atm; Ptotal = 19.9 atm)

- A 5.0-gallon gas tank contains N2 and H2 gas, such that the partial pressure of H2 is three times that of N2 gas. If the total gas pressure is 2.4 atm at 25oC, how many grams of N2 and H2, respectively, are present? (Answer: 13 g of N2 and 2.8 g of H2)

Thank you for your your help, if you can help me showing me the process to learn about it at least of one I will appreciate

Explanation / Answer

1.

molar mass of N2 = 28 g/mol

mass of N2 present in the tank = 295 g

no. of moles of N2 = (295 g)/(28 g/mol) = 10.54 mol

volume of the tank = V = 25.0 L

temperature, T = 20.0 oC = 20.0 +273.15 K = 293.15 K

Using ideal gas equation, PV = nRT,

P = nRT/V = (10.54 mol)(0.0821 L.atm/K.mol)(293.15 K)/(25.0 L) = 10.14 atm

partial pressure of nitrogen = PN2 = 10.14 atm

molar mass of O2 = 32 g/mol

mass of O2 present in the tank = 135 g

no. of moles of O2 = (135 g)/(32 g/mol) = 4.22 mol

Using ideal gas equation, PV = nRT,

P = nRT/V = (4.22 mol)(0.0821 L.atm/K.mol)(293.15 K)/(25.0 L) = 4.06 atm

partial pressure of nitrogen = PO2 = 4.06 atm

molar mass of He= 4.00 g/mol

mass of He present in the tank = 24 g

no. of moles of He = (24 g)/(4.00 g/mol) = 6 mol

Using ideal gas equation, PV = nRT,

P = nRT/V = (6 mol)(0.0821 L.atm/K.mol)(293.15 K)/(25.0 L) = 5.77 atm

partial pressure of nitrogen = PHe = 5.77atm

Total pressure = PN2 + PO2 + PHe = 10.1 atm + 4.06 atm + 5.77 atm = 19.9 atm (upto 3 significant figures)

2. Let partial pressure of N2 be x atm. Then, partial pressure of H2 would be 3x atm.

Total pressure = PN2 + PH2 = x + 3x = 2.4 atm

Hence, x = 0.6 atm

Therefore, PN2 = 0.6 atm

               PH2 = 1.8 atm

Volume of the tank, V = 5.0 gallon = 5(3.78 L)= 18.9 L

T = 25 C = 25 + 273.15 K = 298.15 K

From ideal gas equation, PV = nRT

or, n = PV/RT

For nitrogen gas: n = (0.6 atm)(18.9 L)/(0.0821 L.atm/K.mol)(298.15 K) = 0.46 mol

molar mass of N2 = 28 g/mol

Mass of nitrogen present in the tank = (0.46 mol)(28 g/mol) = 12.88                                 (take upto 2 sig. fig)

Hence, 13 g of nitrogen is present in the tank.

For hydrogen gas: n = (1.8 atm)(18.9 L)/(0.0821 L.atm/K.mol)(298.15 K) = 1.4 mol

molar mass of H2 = 2 g/mol

Mass of hydrogen present in the tank = (1.4 mol)(2 g/mol) = 2.8

Hence, 2.8 g of hydrogen is present in the tank.

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