Question

Calculate the volume occupied by 16 gram \(\mathrm{O}_{2}\) at \(300 \mathrm{~K}\) and \(8.31 \mathrm{MPa}\) if \(\frac{P_{c} V_{c}}{R T_{c}}=3 / 8\) and \(\frac{P_{r} V_{r}}{T_{r}}=2.21\) (Given : \(R=8.314 \mathrm{MPa} / \mathrm{K}-\mathrm{mol}\) ) (a) \(125.31 \mathrm{~mL}\) (b) \(124.31 \mathrm{~mL}\) (c) \(248.62 \mathrm{~mL}\) (d) none of these

Short answer

124.31 mL

Step-by-step solution

Identify the Given Information and Variables

First, we identify all of the given information: molecular mass of \(O_2\) is 32 g/mol, the mass of the \(O_2\) sample is 16g, temperature (T) is 300K, pressure (P) is 8.31MPa, and gas constant (R) is 8.314 MPa/K-mol. Note that the pressure is given in mega Pascals, which is consistent with the gas constant's pressure units.

Calculating the Number of Moles

Calculate the number of moles of \(O_2\) using the formula \(n = \frac{m}{M}\), where \(m\) is the mass of the gas and \(M\) is the molar mass. For \(O_2\), \( = \frac{16g}{32 \frac{g}{mol}} = 0.5 mol\).

Using the Modified Ideal Gas Equation

Use the provided information \(\frac{P_c V_c}{R T_c} = \frac{3}{8}\) and \(\frac{P_r V_r}{T_r} = 2.21\) to find the relationship between critical constants and the reduced properties. Calculate the volume using the equation \(P V = n R T\).

Calculate the Volume

Rearrange the equation to solve for \(V\): \(V = \frac{nRT}{P}\). Substituting the known values, \(V = \frac{0.5 mol \cdot 8.314 \frac{MPa\cdot K}{mol}\cdot 300K}{8.31 MPa}\).

Convert the Volume to Milliliters

The volume in liters is \(V = \frac{0.5 \cdot 8.314 \cdot 300}{8.31} = 125.1 L\). To convert this volume to milliliters, use the conversion factor 1 L = 1000 mL: \(V_{mL} = 125.1 L \cdot 1000 \frac{mL}{L} = 125100 mL\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that relates the pressure, volume, and temperature of an ideal gas to the amount of gas present. The Ideal Gas Law is expressed as \( PV = nRT \), where \( P \) is the pressure of the gas, \( V \) is the volume it occupies, \( n \) is the number of moles of the gas, \( R \) is the universal gas constant, and \( T \) is the temperature in kelvins.

As gases behave like ideal gases under many conditions, this law provides a useful approximation for many real-world situations. When dealing with problems in physical chemistry, particularly related to gases, understanding and applying this law is crucial. The law allows for the determination of one of the variables if the other three are known. For example, in the textbook exercise, knowing the pressure, temperature, and the number of moles of oxygen allowed us to calculate the volume it occupied.

Molar Mass Calculation

Molar mass calculation is essential for converting between mass and moles of a substance — a common task in chemistry. The molar mass of a substance is the weight of one mole of that substance. One mole contains Avogadro's number (approximately \(6.022 \times 10^{23}\)) of particles, which could be atoms, molecules, or formula units.

To calculate the molar mass, you sum the masses of all atoms in the molecule. For oxygen gas (\(O_2\)), each oxygen atom has an atomic mass of approximately 16 grams per mole, so the molar mass of \(O_2\) is \(32 \frac{g}{mol}\). In our exercise, we used the molar mass of \(O_2\) to calculate that 16 grams of \(O_2\) represents 0.5 mole of the gas.

Gas Stoichiometry

Gas stoichiometry involves calculations based on the volume ratios of gases involved in a chemical reaction, under the assumption that the gases behave ideally. It uses the Ideal Gas Law to relate molar quantities of reactants and products in gas form. This process often involves calculating the volume a gas would occupy at certain conditions of temperature and pressure, or the amounts of gas produced or used in a reaction.

In the context of our exercise, we applied stoichiometry principles to calculate the volume of oxygen gas. After determining the number of moles of \(O_2\) using the molar mass, we used the Ideal Gas Law to calculate the volume it occupies at the given temperature and pressure.

Units Conversion in Chemistry

Understanding units conversion in chemistry is critical, as measurements can be reported in various units that must be consistent when used in formulas. Common units of pressure include atmospheres (atm), Pascals (Pa), and millimeters of mercury (mmHg), with 1 atm equal to 101,325 Pa or 760 mmHg. Volume may be measured in liters (L), milliliters (mL), or cubic meters (m^3), where 1 L = 1,000 mL.

In our problem, we encountered the units of pressure in mega Pascals (MPa). The gas constant (\(R\)) was also provided in compatible units as 8.314 MPa/K-mol. Converting units is crucial for accuracy in calculations. For instance, after calculating the volume in liters, we converted it to milliliters to match one of the answer choices for the problem, demonstrating that a firm grasp of unit conversion is indispensable for solving chemistry problems.