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A deep-sea diver uses a gas cylinder with a volume of 10.0 \(\mathrm{L}\) and a content of 51.2 \(\mathrm{g}\) of \(\mathrm{O}_{2}\) and 32.6 \(\mathrm{g}\) of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is \(19^{\circ} \mathrm{C}\) .

Short Answer

Expert verified
The partial pressure of O2 is 3.8108 atm, the partial pressure of He is 19.7326 atm, and the total pressure inside the gas cylinder is 23.5434 atm.

Step by step solution

01

Convert masses to moles

We need to convert the mass of O2 and He gases given in grams to moles. To do this, we will use the molar masses of O2 and He. The molar mass of O2 = 32 g/mol. The molar mass of He = 4 g/mol. Moles of O2 = (mass of O2) / (molar mass of O2) Moles of O2 = (51.2 g) / (32 g/mol) = 1.6 mol Moles of He = (mass of He) / (molar mass of He) Moles of He = (32.6 g) / (4 g/mol) = 8.15 mol
02

Convert temperature to Kelvin

The given temperature is in Celsius. We need to convert it to Kelvin. Temperature in Kelvin = Temperature in Celsius + 273.15 Temperature = 19°C + 273.15 = 292.15 K
03

Find partial pressures of O2 and He

Now we will use the Ideal Gas Law to find the partial pressures of O2 and He. PV = nRT P = (nRT) / V Partial pressure of O2, P_O2 = (n_O2 * R * T)/V P_O2 = (1.6 mol * 0.0821 L·atm/mol·K * 292.15 K) / 10.0 L P_O2 = 3.8108 atm Partial pressure of He, P_He = (n_He * R * T)/V P_He = \( (8.15 \text{ mol} \times 0.0821 \text{ L·atm/mol·K} \times 292.15 \text{ K})/(10.0 \text{ L})\) P_He = 19.7326 atm
04

Find total pressure

To find the total pressure, we will add the partial pressures of O2 and He. Total pressure, P_total = P_O2 + P_He P_total = 3.8108 atm + 19.7326 atm = 23.5434 atm So, the partial pressure of O2 is 3.8108 atm, the partial pressure of He is 19.7326 atm, and the total pressure inside the gas cylinder is 23.5434 atm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, moles, and temperature of a gas using the equation: \[ PV = nRT \]where
  • \(P\) is the pressure of the gas,
  • \(V\) is the volume,
  • \(n\) represents the number of moles,
  • \(R\) is the ideal gas constant, typically 0.0821 \, \text{L}·\text{atm}/\text{mol}·\text{K},
  • and \(T\) is the temperature in Kelvin.
To find the pressure of a gas given the other variables, we rearrange the equation to solve for \(P\): \[ P = \frac{nRT}{V} \]This formula is essential for determining both partial and total pressures in mixtures of gases, as we see in deep-sea diving scenarios where gases are stored under pressure for breathing.
Molar Mass
Molar mass is a vital concept in chemistry that denotes the mass of one mole of a substance, allowing us to convert between grams and moles. Different elements and compounds have unique molar masses, often derived from the atomic masses on the periodic table.

Calculating Moles from Mass

To calculate the number of moles from a given mass, use the formula:\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]For example:
  • Oxygen \((\text{O}_2)\): Molar mass = 32 \, \text{g/mol},
  • Helium \((\text{He})\): Molar mass = 4 \, \text{g/mol}.
These conversion processes enable us to use gases in scientific calculations such as determining the number of molecules in a gas sample. This conversion was crucial for our diver example to find the moles of \(\text{O}_2\) and \(\text{He}\) used to calculate partial pressures.
Temperature Conversion
Temperature conversion is another important step in gas calculations because the ideal gas law works with temperatures in Kelvin. The conversion from Celsius to Kelvin is simple and necessary for accuracy.The conversion formula is:\[ T_{\text{K}} = T_{\text{°C}} + 273.15 \]With this conversion, a temperature like 19°C becomes 292.15 \(\text{K}\). Kelvin is an absolute temperature scale that starts from absolute zero, making it suitable for physics and chemistry calculations. Using Kelvin ensures consistent results when calculating the properties of gases.
Moles Calculation
The process of calculating moles is foundational in gas law applications. It connects the mass of a substance to the number of molecules or atoms it contains. Moles allow for consistent measurements in reactions and gas properties predictions.

Using Moles in the Ideal Gas Law

Once we determine the moles of a gas, we use this number to calculate the pressure using the ideal gas law. Knowing the moles of each component in a gas mixture as with the diver's oxygen and helium mixtures helps compute the partial pressure. The partial pressure for each gas is calculated by its moles' proportion in the total gas volume. This step-by-step process facilitates the understanding and application of gas laws in practical scenarios.

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Most popular questions from this chapter

On a single plot, qualitatively sketch the distribution of molecular speeds for (a) \(\operatorname{Kr}(g)\) at \(-50^{\circ} \mathrm{C},(\mathbf{b}) \mathrm{Kr}(g)\) at \(0^{\circ} \mathrm{C},\) (c) \(\operatorname{Ar}(g)\) at \(0^{\circ} \mathrm{C} .[\) Section 10.7\(]\)

A mixture containing 0.765 mol \(\mathrm{He}(g), 0.330 \mathrm{mol} \mathrm{Ne}(g),\) and 0.110 \(\mathrm{mol} \mathrm{Ar}(g)\) is confined in a \(10.00-\mathrm{L}\) vessel at \(25^{\circ} \mathrm{C}\) . (a) Calculate the partial pressure of each of the gases in the mixture. (b) Calculate the total pressure of the mixture.

The atmospheric concentration of \(\mathrm{CO}_{2}\) gas is presently 407 \(\mathrm{ppm}(\) parts per million, by volume; that is, 407 \(\mathrm{L}\) of every \(10^{6} \mathrm{L}\) of the atmosphere are \(\mathrm{CO}_{2}\) . What is the mole fraction of \(\mathrm{CO}_{2}\) in the atmosphere?

Carbon dioxide, which is recognized as the major contributor to global warming as a "greenhouse gas," is formed when fossil fuels are combusted, as in electrical power plants fueled by coal, oil, or natural gas. One potential way to reduce the amount of \(\mathrm{CO}_{2}\) added to the atmosphere is to store it as a compressed gas in underground formations.Consider a 1000 -megawatt coal-fired power plant that produces about \(6 \times 10^{6}\) tons of \(\mathrm{CO}_{2}\) per year. (a) Assuming ideal-gas behavior, 1.00 atm, and \(27^{\circ} \mathrm{C},\) calculate the volume of \(\mathrm{CO}_{2}\) produced by this power plant. (b) If the \(\mathrm{CO}_{2}\) is stored underground as a liquid at \(10^{\circ} \mathrm{C}\) and 120 \(\mathrm{atm}\) and a density of \(1.2 \mathrm{g} / \mathrm{cm}^{3},\) what volume does it possess?(c) If it is stored underground as a gas at \(30^{\circ} \mathrm{C}\) and \(70 \mathrm{atm},\) what volume does it occupy?

(a) What are the mole fractions of each component in a mixture of 15.08 \(\mathrm{g}\) of \(\mathrm{O}_{2}, 8.17 \mathrm{g}\) of \(\mathrm{N}_{2},\) and 2.64 \(\mathrm{g}\) of \(\mathrm{H}_{2}\) ? (b) What is the partial pressure in atm of each component of this mixture if it is held in a \(15.50-\mathrm{L}\) vessel at \(15^{\circ} \mathrm{C} ?\)

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