Question

A.) A mixture of oxygen and krypton gases, in a 6.68 L flask at 42 °C, contains 9.38 grams of oxygen and 12.0 grams of krypton. The partial pressure of krypton in the flask is ___ atm and the total pressure in the flask is ___ atm.

B.) A mixture of helium and methane gases is maintained in a 7.52 L flask at a pressure of 1.92 atm and a temperature of 25 °C. If the gas mixture contains 0.693 grams of helium, the number of grams of methane in the mixture is ___ g.

C.) The stopcock connecting a 4.98 L bulb containing nitrogen gas at a pressure of 6.78 atm, and a 8.69 L bulb containing carbon dioxide gas at a pressure of 1.18 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is ___ atm.

Answer 1

A.) A mixture of oxygen and krypton gases, in a 6.68 L flask at 42 °C, contains 9.38 grams of oxygen and 12.0 grams of krypton. The partial pressure of krypton in the flask is ___ atm and the total pressure in the flask is ___ atm.

Solution :-

Lets first calculate the moles of gas

Moles = mass / molar mass

Moles of oxygen = 9.38 g / 32 g per mol = 0.293 mol O2

Moles of krypton = 12.0 g /83.80 g per mol = 0.143 mol Kr

Total moles = 0.293 mol + 0.143 mol = 0.436 mol

Now using the total moles we can find the total pressure of the flask using the ideal gas law formula

V= 6.68 L , T= 42 C +273 = 315 K , R= 0.08206 L atm per mol K

PV= nRT

P= nRT/ V

P=0.436 mol *0.08206 L atm per mol K * 315 K / 6.68 L

P = 1.69 atm

Now lets calculate the partial pressure of krypton

Partial pressure of krypton = mole fraction of krypton * total pressure

                                           = (0.143 mol / 0.436mol)*1.69 atm

                                           = 0.554 atm

Therefore the partial pressure of krypton is 0.554 atm

Total pressure is 1.69 atm

B.) A mixture of helium and methane gases is maintained in a 7.52 L flask at a pressure of 1.92 atm and a temperature of 25 °C. If the gas mixture contains 0.693 grams of helium, the number of grams of methane in the mixture is ___ g.

Solution :-

Lets first calculate the total moles of gases in the flask

T= 25 C+273 = 298 K

PV= nRT

PV/RT=n

[1.92 atm * 7.52 L] / [0.08206 L atm per mol K * 298 K] = n

0.5904 mol = n

Now lets calculate the moles of helium

Moles of helium = 0.693 g / 4.0026 g per mol = 0.1731 mol helium

Now using the moles of helium and total moles we can find moles of methane

Moles of methane = total moles – moles of helium

                             = 0.5904 mol – 0.1731 mol

                            = 0.4173 mol CH4

Now lets convert moles of methane to its mass

Mass= moles x molar mass

Mass of methane = 0.4173 mol * 16.04 g per mol

                             = 6.70 g methane

Hence mass of methane is 6.70 g

C.) The stopcock connecting a 4.98 L bulb containing nitrogen gas at a pressure of 6.78 atm, and a 8.69 L bulb containing carbon dioxide gas at a pressure of 1.18 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is ___ atm.

Solution :-

After opening the stopcock the gases will mix and the volume becomes the additive.

Therefore total volume after mixing = 4.98 L + 8.69 L = 13.67 L

Now using the Boyels law formula we can find the new partial pressure of each gas

P1V1=P2V2

Calculating the new partial pressure of nitrogen

P2= P1V1/V2

   = 6.78 atm * 4.98 L / 13.67 L

   = 2.47 atm

Calculating the new partial pressure of carbon dioxide

P2= P1V1/V2

   = 1.18 atm * 8.69 L / 13.67 L

    = 0.750 atm

Now lets calculate the total pressure

Total pressure = partial pressure of N2 + partial pressure of CO2

                           = 2.47 atm + 0.750 atm

                          = 3.22 atm

Hence the final pressure in the flask after mixing is 3.22 atm