Question

Complete the following table for an ideal gas. $$\begin{array}{|lccc|}\hline P(\mathrm{atm}) & V(\mathrm{L}) & n(\mathrm{mol}) & T \\ \hline 5.00 & & 2.00 & 155^{\circ} \mathrm{C} \\\\\hline0.300 & 2.00 & & 155 \mathrm{K} \\ \hline 4.47 & 25.0 & 2.01 & \\\\\hline & 2.25 & 10.5 & 75^{\circ} \mathrm{C} \\\ \hline\end{array}$$

Short answer

The completed table for the ideal gas is: $$\begin{array}{|lccc|}\hline P(\mathrm{atm}) & V(\mathrm{L}) & n(\mathrm{mol}) & T \\\ \hline 5.00 & 14.0 & 2.00 & 428.15 \,\mathrm{K} \\\\\hline0.300 & 2.00 & 0.0296 & 155\, \mathrm{K} \\\ \hline 4.47 & 25.0 & 2.01 & 679.09\,\mathrm{K} \\\\\hline 128.70 & 2.25 & 10.5 & 348.15\, \mathrm{K} \\\ \hline\end{array}$$

Step-by-step solution

Convert temperature to Kelvin

For any row that has temperature given in Celsius, convert it to Kelvin using the formula: \(T_{\mathrm{K}} = T_{^{\circ}\mathrm{C}} + 273.15\) Now that we have the temperatures in Kelvin, use the Ideal Gas Law to solve for the missing values in each row.

Row 1: Solve for the missing volume

We have \(P = 5.00 \, \mathrm{atm}\), \(n = 2.00 \, \mathrm{mol}\), and \(T = 155^{\circ} \mathrm{C} = 428.15 \, \mathrm{K}\). Use the Ideal Gas Law to find the missing volume: \(V = \frac{nRT}{P} = \frac{(2.00 \, \mathrm{mol})(0.0821 \, \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}})(428.15 \, \mathrm{K})}{5.00 \, \mathrm{atm}} = 14.00 \, \mathrm{L}\)

Row 2: Solve for the missing number of moles

We have \(P = 0.300 \, \mathrm{atm}\), \(V = 2.00 \, \mathrm{L}\), and \(T = 155 \, \mathrm{K}\). Use the Ideal Gas Law to find the missing number of moles: \(n = \frac{PV}{RT} = \frac{(0.300\, \mathrm{atm})(2.00 \, \mathrm{L})}{(0.0821 \, \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}})(155 \, \mathrm{K})} = 0.0296 \, \mathrm{mol}\)

Row 3: Solve for the missing temperature

We have \(P = 4.47 \, \mathrm{atm}\), \(V = 25.0 \, \mathrm{L}\), and \(n = 2.01 \, \mathrm{mol}\). Use the Ideal Gas Law to find the missing temperature: \(T = \frac{PV}{nR} = \frac{(4.47 \, \mathrm{atm})(25.0 \, \mathrm{L})}{(2.01 \, \mathrm{mol})(0.0821 \, \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}})} = 679.09 \, \mathrm{K}\)

Row 4: Solve for the missing pressure

We have \(V = 2.25 \, \mathrm{L}\), \(n = 10.5 \, \mathrm{mol}\), and \(T = 75^{\circ} \mathrm{C} = 348.15\, \mathrm{K}\). Use the Ideal Gas Law to find the missing pressure: \(P = \frac{nRT}{V} = \frac{(10.5 \, \mathrm{mol})(0.0821 \, \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}})(348.15\, \mathrm{K})}{2.25 \, \mathrm{L}} = 128.70 \, \mathrm{atm}\) The completed table is: $$\begin{array}{|lccc|}\hline P(\mathrm{atm}) & V(\mathrm{L}) & n(\mathrm{mol}) & T \\\ \hline 5.00 & 14.0 & 2.00 & 428.15 \,\mathrm{K} \\\\\hline0.300 & 2.00 & 0.0296 & 155\, \mathrm{K} \\\ \hline 4.47 & 25.0 & 2.01 & 679.09\,\mathrm{K} \\\\\hline 128.70 & 2.25 & 10.5 & 348.15\, \mathrm{K} \\\ \hline\end{array}$$

Key concepts

Gas Laws Chemistry

Understanding the behavior of gases is foundational in chemistry. The Ideal Gas Law, represented by the equation \( PV = nRT \), is a cornerstone topic within gas laws chemistry that combines the relationships stated in Boyle's, Charles's, Avogadro's, and Gay-Lussac's laws.

In the context of the textbook problem, the Ideal Gas Law allows students to calculate any one of the four variables (pressure \(P\), volume \(V\), number of moles \(n\), or absolute temperature \(T\)) as long as the other three are known. This can open up a whole world of understanding the way gases behave under different conditions. This concept also emphasizes the importance of units, as seen where the value of the gas constant \(R\) must be matched with the pressure in atmospheres, volume in liters, and temperature in Kelvin to get the right answers.

To truly grasp gas laws, experimenting with problems like the one provided helps underline the direct and inverse relationships between pressure, volume, temperature, and amount of gas. For deeper learning, one could explore scenarios at different conditions or look into the real behavior of gases, which diverges from the ideal model at high pressures or low temperatures.

Problem Solving in Chemistry

A methodical approach is crucial for effectively solving problems in chemistry. The textbook exercise is a perfect example of a systematic process. It starts with understanding what is known and what needs to be found.

Next, it is vital to convert all units to a standard system; in this case, temperatures are converted to Kelvin. Understanding what each variable represents and how to reorder the Ideal Gas Law to solve for a specific unknown variable is a skill developed through practice and clear guidance on each step's purpose.

Also, being aware of common pitfalls, such as missing units or math errors, is important in ensuring accurate results. Having a clear roadmap, from analyzing the problem and gathering the relevant equations to executing the solutions and double-checking the results, helps build confidence and competence in tackling complex chemical problems.

Units Conversion Chemistry

Unit conversion is an integral part of chemistry, and errors in this can lead to incorrect answers and misunderstandings. The Ideal Gas Law exercise highlights the importance of converting Celsius to Kelvin before using these temperatures in the computations.

This conversion is essential because the Kelvin scale is an absolute temperature scale used in scientific equations to ensure proportionality and correctness in results. Additionally, careful attention must be given to consistency in units across all the measured values. When converting units, students learn not only about the different scales and measures but also about the broader topic of dimensional analysis, which can be applied to various types of chemical calculations.

Practicing units conversion helps students gain fluency in the language of chemistry, enabling them to effectively communicate and calculate in the global context of scientific research and education.