Which of these occupies the greatest volume? (a) \(0.2 \mathrm{~mol}\) of chlorine gas at \(48^{\circ} \mathrm{C}\) and \(80 \mathrm{~cm} \mathrm{Hg}\) (b) \(4.2 \mathrm{~g}\) of ammonia at \(0.65 \mathrm{~atm}\) and \(-11^{\circ} \mathrm{C}\) (c) \(21 \mathrm{~g}\) of sulfur trioxide at \(55^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\)

When it comes to understanding how to calculate the volume of gas, we rely on the Ideal Gas Law. This principle helps us understand the relationship between pressure (P), volume (V), temperature (T), and the number of moles (n) in a gas. The equation is: \[ PV = nRT \] Here, \( R \) is the ideal gas constant. Its value is \( 0.0821 \, L·atm/mol·K \) for calculations involving pressure in atmospheres and volume in liters.
Let's walk through an example by calculating the volume of chlorine gas.
We are given the moles \( n = 0.2 \, mol \), pressure \( P = 80 \, cm Hg \), and temperature \( T = 48^{\text{°C}} \).
First, we convert the temperature to Kelvin: \( 48 + 273 = 321 \, K \).
Convert pressure to atmospheres: \( 80 \, cm Hg = 80/76 \, atm \). Substituting these values into the Ideal Gas Law, we get: \[ V = \frac{nRT}{P} = \frac{0.2 \times 0.0821 \times 321}{80/76} \ \ \text{This gives us } V \ \ V \approx 5.00 \, L. \] This constant interplay between these variables helps us in predicting how gases behave under different conditions.

Understanding unit conversions is crucial for solving problems in chemistry, especially when dealing with gas laws. We often need to convert units of pressure, temperature, and amounts to their standard forms. For instance, temperature should always be in Kelvin (K) when using the Ideal Gas Law. To convert from Celsius to Kelvin, just add 273: \[ \text{K} = \text{°C} + 273 \] Converting pressure often involves knowing that 1 atmosphere (atm) equals 760 mm Hg or 76 cm Hg. If you have pressure in cm Hg, you convert by dividing by 76: \[ \text{Pressure} = \frac{\text{cm Hg}}{76} \text{ atm} \] For mass conversions, the moles (n) can be calculated from the mass (g) and molar mass (g/mol) using the equation: \[ n = \frac{\text{mass}}{\text{molar mass}} \] Remembering these conversions makes calculations seamless and accurate.
For example, converting 80 cm Hg to atm: \[ 80 \, cm Hg = \frac{80}{76} \, atm \] This is essential to ensure correctness in our calculations.

Understanding molecular mass and moles is fundamental for calculating gas volumes and chemical reactions. The molecular mass (also known as molar mass) is the weight of one mole of a substance, expressed in grams per mole (g/mol).
To find the number of moles, you use the formula: \[ n = \frac{\text{mass}}{\text{molar mass}} \]
Here's how it works: If we have 4.2 g of ammonia (NH3), and we know the molar mass of NH3 is 17 g/mol, we can calculate the number of moles as follows: \[ n = \frac{4.2 \, g}{17 \, g/mol} = 0.247 \, mol \] Once we have the number of moles, we can link this to the volume calculation using the Ideal Gas Law. In every chemistry exercise involving gases, calculating moles accurately is a key step to obtaining the right answer.
This approach not only helps in understanding gas behavior but also in predicting outcomes in various chemical reactions.