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 20.0^oC.

         (Answer: P_N2 = 10.1 atm; P_O2 = 4.06 atm; P_He = 5.77 atm; P_total = 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 25^oC, 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

Answer 1

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 = P_N2 = 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 = P_O2 = 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 = P_He = 5.77atm

Total pressure = P_N2 + P_O2 + P_He = 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 = P_N2 + P_H2 = x + 3x = 2.4 atm

Hence, x = 0.6 atm

Therefore, P_N2 = 0.6 atm

               P_H2 = 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.