Question

Complete the following table for an ideal gas: $$\begin{array}{llll} P & V & n & T \\ \hline 303.98 \mathrm{kPa} & 3.00 \mathrm{~L} & 1.500 \mathrm{~mol} & ? \mathrm{~K} \\ 50.663 \mathrm{kPa} & 0.750 \mathrm{~L} & ? \mathrm{~mol} & 300 \mathrm{~K} \\\ 101.33 \mathrm{kPa} & ? \mathrm{~L} & 3.333 \mathrm{~mol} & 300 \mathrm{~K} \\\ ? \mathrm{kPa} & .750 \mathrm{~L} & 0.750 \mathrm{~mol} & 298 \mathrm{~K} \\ \hline \end{array}$$

Short answer

The completed table is: $$ \begin{array}{llll} P & V & n & T \\ \hline 303.98 \mathrm{kPa} & 3.00 \mathrm{~L} & 1.500 \mathrm{~mol} & 728.24 \mathrm{~K} \\ 50.663 \mathrm{kPa} & 0.750 \mathrm{~L} & 0.2536 \mathrm{~mol} & 300 \mathrm{~K} \\ 101.33 \mathrm{kPa} & 8.21 \mathrm{~L} & 3.333 \mathrm{~mol} & 300 \mathrm{~K} \\ 24.7 \mathrm{kPa} & .750 \mathrm{~L} & 0.750 \mathrm{~mol} & 298 \mathrm{~K} \\ \hline \end{array} $$

Step-by-step solution

Row 1: Find T

Given values: P = 303.98 kPa, V = 3.00 L, n = 1.500 mol, R = 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)). Solve for T. Using ideal gas law equation: 303.98 kPa × 3.00 L = 1.500 mol × 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)) × T Solve for T: T = \(\frac{303.98 \times 3.00}{1.500 \times 0.0831}\) = 728.24 K

Row 2: Find n

Given values: P = 50.663 kPa, V = 0.750 L, T = 300 K, R = 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)). Solve for n. Using ideal gas law equation: 50.663 kPa × 0.750 L = n × 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)) × 300 K Solve for n: n = \(\frac{50.663 \times 0.750}{0.0831 \times 300}\) = 0.2536 mol

Row 3: Find V

Given values: P = 101.33 kPa, n = 3.333 mol, T = 300 K, R = 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)). Solve for V. Using the ideal gas law equation: 101.33 kPa × V = 3.333 mol × 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)) × 300 K Solve for V: V = \(\frac{3.333 \times 0.0831 \times 300}{101.33}\) = 8.21 L

Row 4: Find P

Given values: V = 0.750 L, n = 0.750 mol, T = 298 K, R = 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)). Solve for P. Using ideal gas law equation: P × 0.750 L = 0.750 mol × 0.0831 (\(\frac{\mathrm{kPa \cdot L}}{\mathrm{mol \cdot K}}\)) × 298 K Solve for P: P = \(\frac{0.750 \times 0.0831 \times 298}{0.750}\) = 24.7 kPa The completed table is: $$\begin{array}{llll} P & V & n & T \\\ \hline 303.98 \mathrm{kPa} & 3.00 \mathrm{~L} & 1.500 \mathrm{~mol} & 728.24 \mathrm{~K} \\\ 50.663 \mathrm{kPa} & 0.750 \mathrm{~L} & 0.2536 \mathrm{~mol} & 300 \mathrm{~K} \\\ 101.33 \mathrm{kPa} & 8.21 \mathrm{~L} & 3.333 \mathrm{~mol} & 300 \mathrm{~K} \\\ 24.7 \mathrm{kPa} & .750 \mathrm{~L} & 0.750 \mathrm{~mol} & 298 \mathrm{~K} \\\ \hline \end{array}$$

Key concepts

Gas Laws

The Ideal Gas Law is a fundamental principle in thermodynamics that combines several individual gas laws into one comprehensive equation: \( PV = nRT \). This formula relates the pressure \( P \), volume \( V \), and temperature \( T \) of an ideal gas, with \( n \) representing the number of moles and \( R \) being the ideal gas constant. This equation is essential in understanding how gases behave under different conditions.

**Understanding the Variables**

• **Pressure (P)**: Measured in units such as kPa, it indicates the force exerted by the gas molecules against the walls of their container.


• **Volume (V)**: Measured in liters (L), it represents the space occupied by the gas.


• **Temperature (T)**: Always measured in Kelvin (K) for such equations, it reflects the average kinetic energy of gas molecules.


• **Moles (n)**: This is the measure of the quantity of gas present. One mole equals Avogadro’s number of particles (approximately \( 6.022 \times 10^{23} \)).


• **Ideal Gas Constant (R)**: A proportionality constant with values depending on the units used, commonly \( 0.0831 \frac{\text{kPa} \cdot \text{L}}{\text{mol} \cdot \text{K}} \).

When applying the ideal gas law, remember it's based on the assumption that the gas behaves ideally, meaning its particles occupy no space and do not interact with each other except through elastic collisions. Real gases behave approximately ideally at high temperatures and low pressures.

Mole Calculation

Calculating the number of moles for a gas is essential in mastering the Ideal Gas Law and involves the relationship \( n = \frac{PV}{RT} \). Moles bridge the macroscopic and molecular worlds by quantifying molecular-scale processes in bulk substances.

• **Determining Number of Moles**: By substituting known values of pressure, volume, and temperature into the gas law equation, one can solve for the moles \( n \).


• **Conversion to Grams**: Often in chemistry, you need to convert moles to grams using the molar mass (gram per mole) of the gas in question. This conversion requires multiplying the number of moles by the substance’s molar mass.

Understanding mole calculations is crucial as it allows precise control and prediction of chemical reactions and processes, such as those involving gases in controlled environments under the assumption that the gas behaves ideally.

Thermodynamic Equations

Thermodynamics involves the study of energy changes accompanying physical and chemical processes, where the Ideal Gas Law plays a crucial role. It helps in various calculations related to energy changes, efficiency, and work done by or on the gaseous systems.

**Application of Thermodynamic Concepts**

• **Energy Calculations**: In processes involving gases, such as expanstion or compression, using the Ideal Gas Law allows for the calculation of work done (e.g., \( W = P\Delta V \)).


• **Understanding State Functions**: Gases are often used to illustrate state functions like enthalpy and entropy. Changes in moles or temperature directly affect these functions.


• **Predicting Behavior**: Knowing the conditions a gas will face, you can predict how its volume, temperature, and pressure will change using thermodynamic equations.

Applying thermodynamics with the aid of the Ideal Gas Law not only provides insights into classical physics but also aids in modern engineering applications such as designing engines and other machinery where gases perform critical functions. Proper grasp of these concepts empowers one to tackle more advanced chemical and physical scenarios with confidence.