Question

What volume does a mixture of \(14.2 \mathrm{~g}\) of \(\mathrm{He}\) and \(21.6 \mathrm{~g}\) of \(\mathrm{H}_{2}\) occupy at \(28^{\circ} \mathrm{C}\) and \(0.985 \mathrm{~atm} ?\)

Short answer

The mixture of He and H2 will occupy a volume of approximately \(349\,L\) at \(28^{\circ}C\) and \(0.985\,atm\).

Step-by-step solution

Calculate the number of moles of helium and hydrogen

In order to calculate the number of moles of He and H2, we need their molar masses, which are approximately \(\mathrm{4.00 \, g/mol}\) for He and \(\mathrm{2.02 \, g/mol}\) for H2. For Helium, Number of moles (n) = mass/molar mass, so \(n_{He} = \frac{14.2 \, g}{ 4.00 \, g/mol} = 3.55 \, mol\) For Hydrogen, Number of moles (n) = mass/molar mass, so \(n_{H2} = \frac{21.6 \, g}{ 2.02 \, g/mol} = 10.7 \, mol\)

Calculate the total number of moles in the mixture

Since both gases are mixed together, we can add their number of moles to find the total amount of the mixture. Total number of moles, \(n_{total} = n_{He} + n_{H2} = 3.55 \, mol + 10.7 \, mol = 14.25 \, mol\)

Convert the temperature to Kelvin

The Ideal Gas Law requires the temperature to be in Kelvin. To convert from Celsius to Kelvin, add 273.15. Temperature in Kelvin, \(T = 28^{\circ}C + 273.15 = 301.15\,K\)

Apply the Ideal Gas Law to find the volume

We have the total number of moles (14.25 mol), temperature (301.15 K), and pressure (0.985 atm), and now we can use the Ideal Gas Law, \(PV = nRT\), to find the volume of the mixture. Here, the gas constant R = 0.0821 L atm/mol K. Rearranging the Ideal Gas Law to obtain V: \(V = \frac{nRT}{P} = \frac{(14.25\, mol)(0.0821\, L\, atm/mol\, K)(301.15\, K)}{0.985\, atm}\) Now solve for V: \(V \approx 349\,L\) So, the mixture of He and H2 will occupy a volume of approximately 349 liters at 28°C and 0.985 atm.

Key concepts

Understanding Molar Mass

Molar mass serves as a bridge between the mass of a substance and the number of particles or moles of that substance. It is defined as the mass of one mole of a particular chemical element or chemical compound. Given in units of grams per mole (g/mol), it corresponds to the mass of Avogadro's number of atoms or molecules of the element or compound. For instance, helium (He) has a molar mass of approximately 4.00 g/mol, meaning one mole of helium atoms weighs 4.00 grams.

To find the molar mass, one typically looks to the periodic table where the atomic weight of an element, rounded to the nearest tenth, gives the molar mass for that element. In chemistry exercises, molar mass is crucial because it allows you to convert between mass and moles, an essential step in many stoichiometric calculations.

Moles Calculation Simplified

The moles calculation is a fundamental aspect of chemistry that quantifies the amount of a substance. One mole represents Avogadro’s number of particles (\( 6.022 \times 10^{23} \text{particles/mol} \)) and is used as a counting unit for atoms, ions, or molecules in a sample. To calculate the number of moles (\( n \)) from a given mass (\( m \)), you divide the mass by the substance’s molar mass (\( M \)): \[ n = \frac{m}{M} \
The calculation enables us to measure the amount of a substance in a way that connects the macroscopic mass we can measure with the microscopic count of particles, a critical step for understanding the proportions and reactions in chemistry. This process is essential for balancing chemical equations and for the steps in the Ideal Gas Law to find the volume of gaseous substances.

Volume Calculation for Gas Mixtures

When dealing with gas mixtures in chemistry, we often need to determine the volume occupied by the mixture. The Ideal Gas Law (\( PV=nRT \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P ), where \( P \( P \( P \( P \( P \( P \( P \( P \) \( P \( P \( P \( P P \( P P \( P P \( P \( P \( P P \( P P \( P P \( P P \( P P P P P P P P P P P P P P P P P P P P P P P P P P P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P P P P P P P P P P P P P P P P P P P P P P P P P P \( P P \( P P \( P P \( P P \( P P \( P P P \( P P P \( P P P \( P P P P P P P P P P P P P P P P P P P P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P P P P P P P P P P P P P P \( P P \( P P \( P P \( P \( P P \( P P \( P P \( P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P \( P P P P P P P P P P P P P P P P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P \( P P P P P P P P P P P \( P P \( P P \( P P \) \( P \( P P \( P P P \( P P P \( P P P \( P \( P P \( P P P P P P P P \( P P \( P P \( P P \)mixtures can’t be directly added together. It’s only after determining the total number of moles of the gases that the volume can be calculated.

To compute the volume of the gas mixture, you add the moles for each gas and then apply the Ideal Gas Law with the combined moles, along with the known temperature and pressure conditions. This approach ensures accurate determination of how much space the gas mixture will take up, which is a critical piece of information in many chemical processes and experimental setups.

Essentials of Temperature Conversion

Temperature conversion is necessary in many areas of chemistry, including when working with the Ideal Gas Law. Celsius and Kelvin are two scales often used. The Celsius scale, based on the freezing and boiling points of water, is commonly used in daily life and laboratory settings. On the other hand, Kelvin is the base unit of temperature in the International System of Units (SI) and is favored in scientific calculations because it starts at absolute zero, the theoretical point where all kinetic motion stops.

To convert Celsius to Kelvin, one adds 273.15 to the Celsius temperature. For example, a room temperature of 25°C would be 298.15K. This simple calculation is vital for accurately applying the Ideal Gas Law, as the law requires that all temperatures be in Kelvin to ensure proper measurement and behavior of gases under varying thermal conditions.