Question

If \(18.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\) gas has a temperature of \(350 \mathrm{~K}\) and a pressure of \(550 \mathrm{mmHg}\), what is its volume?

Short answer

The volume is approximately 22.3 liters.

Step-by-step solution

Convert the Pressure to Atmospheres

The pressure is given in mmHg. Convert it to atmospheres since the ideal gas law uses pressure in atm. Use the conversion factor: \(1 \text{ atm} = 760 \text{ mmHg}\). Thus, \(P = \frac{550}{760} \text{ atm}\). Calculate this to get approximately \(P \approx 0.724 \text{ atm}\).

Convert the Mass of \(\mathrm{O}_2\) to Moles

First, find the molar mass of \(\mathrm{O}_2\). Since the atomic mass of oxygen is approximately \(16.0 \text{ g/mol}\), the molar mass of \(\mathrm{O}_2\) is \(32.0 \text{ g/mol}\). Now, convert grams to moles: \(\text{moles of } \mathrm{O}_2 = \frac{18.0 \text{ g}}{32.0 \text{ g/mol}}\). This yields approximately \(0.5625 \text{ mol}\).

Apply the Ideal Gas Law

Use the ideal gas law, \(PV = nRT\), where \(P\) is the pressure in atm, \(V\) is the volume in liters, \(n\) is the number of moles, \(R\) is the ideal gas constant \(0.0821 \text{ L atm/mol K}\), and \(T\) is the temperature in Kelvin. Substitute the known values: \(0.724 \text{ atm} \cdot V = 0.5625 \text{ mol} \cdot 0.0821 \text{ L atm/mol K} \cdot 350 \text{ K}\). Simplify the right side first: \(0.5625 \times 0.0821 \times 350 = 16.1446875\).

Solve for Volume

Rearrange the equation from Step 3 to solve for \(V\): \(V = \frac{16.1446875}{0.724}\). Compute this to find \(V \approx 22.3 \text{ L}\).

Key concepts

Pressure Conversion

Pressure conversion is an essential step when dealing with gas laws, especially when calculations involve different units. In this exercise, the pressure was given in millimeters of mercury (mmHg), a common unit in chemistry that can contribute to confusion if you need to use the ideal gas law.
To correctly employ the ideal gas law, pressures must be converted to atmospheres (atm) because the gas constant (R) is often given in ATM-related units. To convert from mmHg to atm, use the conversion factor where

• 1 atm = 760 mmHg

This is a simple division problem: take the pressure in mmHg and divide it by 760. For example, a pressure of 550 mmHg converted to atm would be \[P = \frac{550 \text{ mmHg}}{760 \text{ mmHg/atm}} \approx 0.724 \text{ atm}\].
This conversion ensures that the use of the ideal gas constant later in problem-solving is consistent and yields accurate results.

Moles Calculation

Understanding how to calculate moles from the mass of a substance is critical in chemistry. The mole relates the mass of a material to the substance's quantity, considering its atomic or molecular mass. In this scenario, you are given 18.0 grams of \(\mathrm{O}_{2}\) gas.
To find the number of moles, you'll first need to know the molar mass of \(\mathrm{O}_{2}\). Since oxygen's atomic mass is roughly 16.0 g/mol, the molar mass of molecular oxygen (\(\mathrm{O}_2\)) is 32.0 g/mol because \(\mathrm{O}_2\) consists of two oxygen atoms. Thus, the calculation of moles is done by dividing the mass by the molar mass:

• Number of moles = mass / molar mass
• \(\text{moles of } \mathrm{O}_2 = \frac{18.0 \text{ g}}{32.0 \text{ g/mol}}\) which is approximately 0.5625 mol

This step is crucial as it converts the problem from working with mass to using an amount that can be directly plugged into the ideal gas law formula.

Ideal Gas Constant

The ideal gas constant, \(R\), is a crucial factor in the ideal gas law equation \(PV=nRT\). Its value varies depending on the pressure, volume, and temperature units used within the equation.
For most calculations involving standard atmospheric pressure (atm), volume in liters (L), and temperature in Kelvin (K), \(R\) is used as \(0.0821 \text{ L atm/mol K}\). This specific value ensures that when you apply it within the ideal gas law equation, the units properly cancel out, and the solution is coherent.
The ideal gas law is beneficial because it allows you to solve for one property of a gas when you have the others. Here, once you have \(n\) (moles), \(P\) (pressure), and \(T\) (temperature), along with \(R\), you rearrange the equation to solve for the unknown variable, \(V\) (volume). This systematic use of the ideal gas constant simplifies the computation of various properties of gases under ideal conditions.