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# Partial Pressure Questions

The molecules of air frequently collide with the vessel’s walls. These impacts exert energy on the vessel’s walls, which is referred to as pressure. The gas’s pressure is defined as the force per unit area. As we all know, the air is a mixture of gases such as nitrogen, oxygen, and others. The pressure contributed by each gas is referred to as its partial pressure.

Partial pressure can thus be defined as the pressure applied if the gas has engaged the volume on its own. As a result, the total pressure of air will be the sum of the partial pressures of oxygen, nitrogen, and so on.

 Definition: According to Dalton’s law of partial pressures, the total pressure of a gas mixture is equal to the sum of the partial pressures of the constituent gases: PTotal = Pgas 1 + Pgas 2 + Pgas 3 + ….. Dalton’s law can also be expressed in terms of a gas’s mole fraction, x. Pgas 1 = x1PTotal

## Partial Pressure in Chemistry Questions with Solutions

Q1. Which of the following gas mixtures at room temperature obeys Dalton’s law of partial pressures?

a.) NO and O2

b.) CO and CO2

c.) NH3 and HCl

d.) SO2 and O2

Correct Answer– (b.) CO and CO2

Q2. A closed container at a constant temperature contains equal molar quantities of hydrogen gas and oxygen gas. Which of the following quantities for the two gases will be the same?

a.) partial pressure

b.) average kinetic energy

c.) average molecular velocity

d.) All of the above

Correct Answer– (a.) partial pressure and (b.) average kinetic energy

Q3. The partial pressure of the gaseous component divided by the total vapour pressure of the mixture will be equal to which of the following option?

a.) mass of components

b.) mole fraction of the component

c.) mass% of the component

d.) molecular mass of the component

Correct Answer– (b.) mole fraction of the component.

Q4. For a dilute solution, Raoult’s law states that:

a.) The relative lowering of vapour pressure is equal to the mole fraction of solute

b.) The relative lowering of vapour pressure is equal to the mole fraction of solvent

c.) The relative lowering of vapour pressure is proportional to the amount of solute in solution

d.) The vapour pressure of the solution is equal to the mole fraction of the solvent

Correct Answer– (a.) The relative lowering of vapour pressure is equal to the mole fraction of solute

Q5. At a total pressure of 10 atm, 56 g of nitrogen and 96 g of oxygen are mixed isothermally. The ratio of oxygen and nitrogen partial pressures is:

a.) 3:2

b.) 2:3

c.) 3:1

d.) 2:1

Explanation:

Moles of nitrogen is 56/28 = 2 mol

Moles of oxygen is 96/32 = 3 mol

Total moles = 5 mol

Mole fraction of nitrogen is ⅖ = 0.4

Mole fraction of oxygen is 1 – 0.4 = 0.6

Partial pressure of oxygen is 0.6 × 10 = 6 atm

Partial pressure of nitrogen is 0.4 × 10 = 4 atm

Therefore, the ratio will be 6:4 = 3:2.

Q6. State Dalton’s law of partial pressure.

Answer. Dalton’s partial pressure law states that when two or more gases that do not react chemically are enclosed in a vessel, the total pressure equals the sum of their partial pressures.

Q7. What is the partial pressure in the atmosphere?

Answer. At sea level, with an atmospheric pressure of 760 mm Hg, the partial pressures of the various gases can be estimated to be 593 mm Hg for nitrogen, 160 mm Hg for oxygen, and 7.6 mm Hg for argon.

Q8. What factors influence partial pressure?

Answer. The partial pressure of a single gas is equal to the total pressure multiplied by its mole fraction. The Ideal Gas Equation applies equally well to mixtures of gases as it does to pure gases because it is based solely on the number of particles and not the identity of the gas.

Q9. Hydrogen, helium, neon, and argon are all components of a gas mixture. The mixture has a total pressure of 93.6 kPa. Helium, neon, and argon have partial pressures of 15.4 kPa, 25.7 kPa, and 35.6 kPa, respectively. What is the pressure extended by the hydrogen?

Answer. Using Dalton’s Law of Partial Pressure:

PTotal = Phydrogen + Phelium + Pneon + Pargon

93.6 kPa = Phydrogen + 15.4 kPa + 25.7 kPa + 35.6 kPa

Phydrogen = 16.9 kPa

Q10. Three gases are held in a container: oxygen, carbon dioxide, and helium. The three gases have partial pressures of 2.00 atm, 3.00 atm, and 4.00 atm, respectively. How much is the total pressure inside the container?

Answer. Using Dalton’s Law of Partial Pressure:

PTotal = Poxygen + Pcarbondioxide + Phelium

PTotal = 2 + 3 + 4 = 9 atm

Q11. At 255K, a 2L mixture of helium, nitrogen, and neon has a total pressure of 815 mmHg. What mass of neon is present in the mixture if the partial pressure of helium is 201 mmHg and the partial pressure of nitrogen is 351 mmHg?

PTotal = 815 mmHg

PHe = 201 mmHg

PN = 351 mmHg

PTotal = PHe + PN + PNe

PNe = 815 – 201 – 351 = 263 mmHg = 0.346 atm

Using ideal gas law PV = nRT

$$\begin{array}{l}n = \frac{PV}{RT}\end{array}$$
$$\begin{array}{l}n = \frac{0.346\ atm \times 2 L}{0.08206\frac{L.atm}{mol.K}\times 255K}\end{array}$$

n = 0.03307 moles of Ne

Converting moles into grams:

0.3307 moles of Ne × (20.18 g / 1 mole of Ne) = 0.667g

Q12. At 25°C, a 4.0 L container is filled with 2.0 g neon and 8.0 g helium. What is the total pressure of the mixture?

The number of moles of neon:

nneon = 2/20.2 = 0.099 mol

The number of moles of helium:

nhelium = 8/4.003 = 2 mol

Total moles = 0.099 + 2 = 2.1 mol

Using ideal gas law to calculate the total partial pressure:

PV = nRT

$$\begin{array}{l}P=\frac{2.1\ mol\times 0.08206\frac{atm.L}{mol.K}\times 298.15K}{4L}\end{array}$$

PTotal = 12.8 atm ≅ 13 atm.

Q13. A container containing two gases, helium and argon, contains 30.0% helium by volume. If the total pressure inside the container is 4.00 atm, calculate the partial pressures of helium and argon.

Answer. Since the temperature and pressure remain constant, according to Avogadro’s Hypothesis:

Two equal-volume of gas at the same temperature and pressure contain the same number of molecules.

Therefore, 30% volume of helium = 30% molecules

Hence, mole fraction of helium = 0.3

Partial pressure of helium will be 4 atm × 0.3 = 1.2 atm

Partial pressure of argon:

Pargon = 4 – 1.2 = 2.80 atm.

Q14. A flask contains 14.0 g of hydrogen, 84.0 g of nitrogen, and 2.00 moles of oxygen. What is the total pressure in the flask if the partial pressure of oxygen is 78.00 mm of mercury?

Mass of hydrogen = 14 g

Mass of nitrogen = 84 g

Moles of oxygen = 2 mol

Partial pressure of oxygen = 78 mm Hg

The number of moles of H is:

$$\begin{array}{l}n_{H}=\frac{14g}{1.00\ g/mol}=13.8 mol\end{array}$$

The number of moles of N is:

$$\begin{array}{l}n_{N}=\frac{84g}{28.014\ g/mol}=2.9 mol\end{array}$$

Total number of moles = 13.8 + 2.9 + 2 = 18.7 mol

Mole fraction of each gas:

Gas Moles Mole fraction
Hydrogen 13.8 13.8/18.7 = 0.73
Nitrogen 2.9 2.9/18.7 = 0.15
Oxygen 2 2/18.7 = 0.1

Determining partial pressure of hydrogen and nitrogen by using the given information about oxygen by ratio and proportion method:

Hydrogen:

$$\begin{array}{l}\frac{0.1}{78\ mmHg}=\frac{0.73}{x}\end{array}$$

x = 569.4 mmHg

Nitrogen:

$$\begin{array}{l}\frac{0.1}{78\ mmHg}=\frac{0.15}{y}\end{array}$$

y = 117 mmHg

Total partial pressure = 569.4 + 117 + 78 = 764.4 mmHg

Q15. To avoid serious problems, such as “the bends,” deep-sea divers must use special gas mixtures in their tanks rather than compressed air. Divers are subjected to a pressure of approximately 10 atm at depths of approximately 350 ft. A typical gas cylinder used for such depths has a volume of 10.0 L and contains 51.2 g of O2 and 326.4 g of He. At 20°C, what is the partial pressure of each gas, and what is the total pressure in the cylinder?

Mass of helium = 326.4 g, mass of oxygen = 51.2g

Temperature = 20°C = 293.1K

The number of moles of He is:

$$\begin{array}{l}n_{He}=\frac{326.4g}{4.003\ g/mol}=81.54 mol\end{array}$$

The number of moles of O2 is:

$$\begin{array}{l}n_{O_{2}}=\frac{51.2g}{32\ g/mol}=1.60 mol\end{array}$$

Using ideal gas law to calculate the partial pressure of each gas:

PV = nRT

For helium gas,

$$\begin{array}{l}P_{He}=\frac{n_{He}RT}{V}\end{array}$$
$$\begin{array}{l}P_{He}=\frac{81.54\ mol\times 0.08206\frac{atm.L}{mol.K}\times 293.1K}{10.0L}\end{array}$$

PHe = 196.2 atm

For oxygen gas,

$$\begin{array}{l}P_{O_{2}}=\frac{n_{O_{2}}RT}{V}\end{array}$$
$$\begin{array}{l}P_{O_{2}}=\frac{1.60\ mol\times 0.08206\frac{atm.L}{mol.K}\times 293.1K}{10.0L}\end{array}$$

PO2 = 3.85 atm

The total pressure = Sum of the two partial pressures

PT = PHe +PO2 = 196.2 + 3.85 = 200.1 atm.

## Practice Questions on Partial Pressure

Q1. Dalton’s Law of partial pressure is not applicable for which of the following?

a.) reactive gases

b.) non-reacting gases

c.) solids

d.) All of the above

Q2. In a closed system, a sample of gas A evaporates over water. What is the pressure of gas A if the total pressure is 780 torr and the pressure of water vapour is 1 atm?

a.) 1 atm

b.) 0.03 atm

c.) 0.3 atm

d.) 3.0 atm

Q3. At 50.0 °C, 80.0 litres of oxygen are collected over water. The room has an atmospheric pressure of 96.00 kPa. Determine the partial pressure of oxygen.

Q4. A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0°C and the total pressure in the cylinder.

Q5. What is the pressure of the resulting gas mixture if 3.00 mol of N2 and 4.00 mol of O2 are placed in a 35.0 L container at 25.0 °C?

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