Problem 94 - Chemistry Matter and Change

Question

Determine how many liters 8.80 g of carbon dioxide gas would occupy at: a. STP b. 160°C and 3.00 atm c. 288 K and 118 kPa

Step-by-Step Solution

Answer

The volumes of 8.80 g of CO2 under the given conditions are approximately: a) 4.487 L at STP, b) 2.375 L at 160°C and 3.00 atm, and c) 4.083 L at 288 K and 118 kPa.

Step 1: a. Calculate the Volume of CO2 at STP

First, we need to calculate the moles of CO2 given the mass: Molar mass of CO2 = 12.01 g/mol (C) + 2 × 16.00 g/mol (O) = 44.01 g/mol Given mass of CO2: 8.80 g Now, we calculate the number of moles: n = mass / molar mass n = 8.80 g / 44.01 g/mol ≈ 0.2 moles Next, we apply the ideal gas law with the given conditions, at STP (P = 1 atm, T = 273.15 K, R = 0.0821 L atm/mol K): V = nRT / P V ≈ (0.2 mol)(0.0821 L atm/mol K)(273.15 K) / 1 atm V ≈ 4.487 L The volume of 8.80 g of CO2 at STP is approximately 4.487 liters.

Step 2: b. Calculate the Volume of CO2 at 160°C and 3.00 atm

First, convert the given temperature to Kelvin: T = 160°C + 273.15 ≈ 433.15 K Now, apply the ideal gas law with the given conditions (P = 3.00 atm, T = 433.15 K, R = 0.0821 L atm/mol K): V = nRT / P V ≈ (0.2 mol)(0.0821 L atm/mol K)(433.15 K) / 3.00 atm V ≈ 2.375 L The volume of 8.80 g of CO2 at 160°C and 3.00 atm is approximately 2.375 liters.

Step 3: c. Calculate the Volume of CO2 at 288 K and 118 kPa

First, convert the given pressure to atm: P = 118 kPa × (1 atm / 101.3 kPa) ≈ 1.1641 atm Now, apply the ideal gas law with the given conditions (P = 1.1641 atm, T = 288 K, R = 0.0821 L atm/mol K): V = nRT / P V ≈ (0.2 mol)(0.0821 L atm/mol K)(288 K) / 1.1641 atm V ≈ 4.083 L The volume of 8.80 g of CO2 at 288 K and 118 kPa is approximately 4.083 liters.

Key Concepts

STP (Standard Temperature and Pressure)

Standard Temperature and Pressure, commonly referred to as STP, is a set of standard conditions for scientific measurements related to gases. STP is defined as a temperature of 273.15 K (0°C) and a pressure of 1 atm. This standardization allows scientists to compare different experiments and results more consistently.
When dealing with gases, STP provides a common baseline so anyone can predict the behavior of a gas under these specific conditions using the ideal gas law. At STP, one mole of an ideal gas occupies 22.4 liters. This fact often simplifies calculations and helps students and scientists understand the properties of gases more intuitively.

Molar Mass Calculation

Molar mass is critical in converting a substance's mass to the number of moles, which is necessary for many chemical calculations. For carbon dioxide (CO\(_2\)), the molar mass is calculated by adding the atomic masses of one carbon atom and two oxygen atoms:

- Carbon (C): approximately 12.01 g/mol
- Oxygen (O): approximately 16.00 g/mol

The formula for carbon dioxide is CO\(_2\), so its molar mass is:

\[ 12.01 \, \text{g/mol} + 2 \times 16.00 \, \text{g/mol} = 44.01 \, \text{g/mol} \]
This value is used directly in calculations involving the ideal gas law, helping to convert a given mass of CO\(_2\) into moles.

Gas Volume Computation

The gas volume can be computed using the ideal gas law, \( PV = nRT \). Here:

• \( P \) is the pressure
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the ideal gas constant (0.0821 \( \text{L atm/mol K} \) for this context)
• \( T \) is the temperature in Kelvin

To compute gas volume, rearrange the formula to solve for \( V \):

\[ V = \frac{nRT}{P} \]
Using this equation allows us to substitute known values for \( n \), \( R \), \( T \), and \( P \) to directly calculate the volume of gas under various conditions. This is crucial when conditions differ from STP and helps us predict real-world gas behavior.

Temperature Conversion to Kelvin

When performing gas law calculations, all temperatures must be in Kelvin. The Kelvin scale is the standard temperature scale in science because it starts at absolute zero — the point where particles have minimal motion. This is essential in gas law equations like \( PV = nRT \), where the temperature component affects the calculation's accuracy.
To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature:

• For example, 160°C in Kelvin is 160 + 273.15 = 433.15 K

Converting temperatures ensures all variables in your gas law calculations are accurate, allowing for precise results in volume computations and comparisons.