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A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(1.00 \mathrm{g}\) He is placed in a \(1.00-\mathrm{L}\) container at \(27^{\circ} \mathrm{C}\). Calculate the partial pressure of each gas and the total pressure.

Short Answer

Expert verified
The partial pressure of hydrogen is 12.22 atm, the partial pressure of helium is 6.149 atm, and the total pressure inside the container is 18.37 atm.

Step by step solution

01

Convert the Mass of Each Gas to Moles

Using the molar mass of the gases, we can convert each gas's mass to moles by dividing the mass by the molar mass (in grams per mole). For hydrogen, H₂, the molar mass is 2.02 g/mol, and for helium, He, is 4.00 g/mol. Moles of H₂ = mass of H₂ / molar mass of H₂ = \(1.00\text{g} / 2.02\text{g}\text{mol}^{-1}\) = 0.495 moles Moles of He = mass of He / molar mass of He = \(1.00\text{g} / 4.00\text{g}\text{mol}^{-1}\) = 0.250 moles
02

Calculate the Partial Pressure of Each Gas

Using the Ideal Gas Law, \(PV = nRT\), we can determine the partial pressure of each gas. First, convert the temperature to Kelvin. Temperature in Kelvin = 27°C + 273.15 = 300.15 K Next, solve for the partial pressure of each gas using the Ideal Gas Law. Note that the volume and ideal gas constant, R, are the same for both gases. We will use R = 0.0821 L atm / K mol. Partial pressure of H₂ (P_H₂) = \(n_\text{H₂}RT / V\) = \((0.495\text{mol})(0.0821\text{L}\text{atm}\text{K}^{-1}\text{mol}^{-1})(300.15\text{K}) / 1.00\text{L}\) = 12.22 atm Partial pressure of He (P_He) = \(n_\text{He}RT / V\) = \((0.250\text{mol})(0.0821\text{L}\text{atm}\text{K}^{-1}\text{mol}^{-1})(300.15\text{K}) / 1.00\text{L}\) = 6.149 atm
03

Calculate the Total Pressure

To find the total pressure, we can use Dalton's Law of Partial Pressure, which states that the total pressure of a gas mixture equals the sum of the partial pressures of the individual gases: Total pressure (P_total) = P_H₂ + P_He = 12.22 atm + 6.149 atm = 18.37 atm The partial pressure of hydrogen is 12.22 atm, the partial pressure of helium is 6.149 atm, and the total pressure inside the container is 18.37 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 relating the pressure, volume, temperature, and number of moles of an ideal gas. This law is derived from a combination of historical gas laws, including Boyle's Law, Charles's Law, and Avogadro's Law.

Mathematically represented as
\[ PV = nRT \]
where
  • \(P\) is the pressure of the gas,
  • \(V\) is the volume it occupies,
  • \(n\) represents the number of moles of gas,
  • \(R\) is the ideal gas constant, which is 0.0821 L atm K-1 mol-1 when pressure is measured in atmospheres and volume in liters, and
  • \(T\) is the absolute temperature in Kelvin.
This equation is particularly useful in calculating the properties of gases under various conditions. For example, if a gas's temperature or volume changes, we can predict how its pressure will adapt assuming the number of moles remains fixed.
Applying the Ideal Gas Law requires careful conversion of measurements to appropriate units, like temperature to Kelvin and volume to liters. Whenever dealing with gas problems, students should check their units consistently to avoid calculation errors.
Dalton's Law of Partial Pressures
Dalton's Law of Partial Pressures is key when working with mixtures of gases, as often encountered in chemistry and environmental studies. It states that in a mixture, each gas exerts pressure independently of the others as if it were alone in the container. The total pressure exerted by the gas mixture is simply the sum of the partial pressures of individual gases.

Mathematically, \[ P_{\text{total}} = P_1 + P_2 + P_3 + ... + P_n \]
where \(P_{\text{total}}\) is the total pressure of the gas mixture, and \(P_1, P_2, ..., P_n\) are the partial pressures of the individual gas components.

This law is particularly useful when identifying the pressure of a single type of gas within a blend, such as in the given problem involving a mixture of hydrogen and helium. By knowing the individual gas amounts and using the Ideal Gas Law to find their partial pressures, the total pressure inside the container can be accurately determined, which is essential for many applications in both laboratory settings and real-world situations.
Molar Mass
Molar mass stands as a cornerstone concept in chemistry, pivotal when converting between the mass of a substance and the amount of substance (measured in moles). It is defined as the mass of one mole of a substance, typically expressed in units of grams per mole \(g/mol\).

For an element, the molar mass directly corresponds to the atomic weight listed on the periodic table. For compounds, it is the sum of the atomic weights of all atoms present in the chemical formula. For instance, the molar mass of hydrogen (H₂) is 2.02 g/mol, and that of helium (He) is 4.00 g/mol.

Understanding molar mass is key for students as it allows for the conversion between mass and moles, an essential step in stoichiometric calculations and when using the Ideal Gas Law. For any given problem, accurately determining molar mass simplifies the process of solving for other variables, such as in the exercise where it was used to calculate the number of moles of hydrogen and helium from their given masses.

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Most popular questions from this chapter

Calculate the root mean square velocities of \(\mathrm{CH}_{4}(g)\) and \(\mathrm{N}_{2}(g)\) molecules at \(273 \mathrm{K}\) and \(546 \mathrm{K}\).

Metallic molybdenum can be produced from the mineral moIybdenite, MoS \(_{2}\). The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen gas. The balanced equations are $$\begin{array}{l}\operatorname{MoS}_{2}(s)+\frac{7}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{MoO}_{3}(s)+2 \mathrm{SO}_{2}(g) \\\\\mathrm{MoO}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{Mo}(s)+3 \mathrm{H}_{2} \mathrm{O}(l)\end{array}$$ Calculate the volumes of air and hydrogen gas at \(17^{\circ} \mathrm{C}\) and \(1.00\) atm that are necessary to produce \(1.00 \times 10^{3} \mathrm{kg}\) pure molybdenum from MoS \(_{2}\). Assume air contains \(21 \%\) oxygen by volume, and assume \(100 \%\) yield for each reaction.

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