Problem 33 - Chemistry: The Central Science

Question

Complete the following table for an ideal gas: $$ \begin{array}{llll} \hline P & V & n & T \\ \hline 2.00 \mathrm{~atm} & 1.00 \mathrm{~L} & 0.500 \mathrm{~mol} & ? \mathrm{~K} \\ 0.300 \mathrm{~atm} & 0.250 \mathrm{~L} & ? \mathrm{~mol} & 27^{\circ} \mathrm{C} \\ 650 \text { torr } & ? \mathrm{~L} & 0.333 \mathrm{~mol} & 350 \mathrm{~K} \\ ? \mathrm{~atm} & 585 \mathrm{~mL} & 0.250 \mathrm{~mol} & 295 \mathrm{~K} \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The completed table is: $$ \begin{array}{llll} \hline P & V & n & T \\\ \hline 2.00 \mathrm{~atm} & 1.00 \mathrm{~L} & 0.500 \mathrm{~mol} & 48.7 \mathrm{~K} \\\ 0.300 \mathrm{~atm} & 0.250 \mathrm{~L} & 0.00305 \mathrm{~mol} & 300.15^{\circ} \mathrm{K} \\\ 0.855 \mathrm{~atm} & 10.2 \mathrm{~L} & 0.333 \mathrm{~mol} & 350 \mathrm{~K} \\\ 10.50 \mathrm{~atm} & 0.585 \mathrm{~L} & 0.250 \mathrm{~mol} & 295 \mathrm{~K} \\\ \hline \end{array} $$

Step 1: Find the missing Temperature (T) in the first row

To do this, we will use the ideal gas law: \( PV = nRT \). We need to solve for T, so we will rearrange the equation as: \[ T = \frac{PV}{nR} \] Now substitute the given values into the equation: \( T = \frac{(2.00 \text{ atm})(1.00 \text{ L})}{(0.500 \text{ mol})(0.0821 \text{ L atm/mol K})} \) Finally, solve for T: \( T = 48.7 \text{ K} \).

Step 2: Find the missing number of moles (n) in the second row

To do this, we will again use the ideal gas law: \( PV = nRT \). We need to solve for n, so we rearrange the equation as: \[ n = \frac{PV}{RT} \] Convert the temperature from Celsius to Kelvins: \( T = 27^{\circ} \text{C} + 273.15 = 300.15 \text{ K} \) Now substitute the given values into the equation: \( n = \frac{(0.300 \text{ atm})(0.250 \text{ L})}{(0.0821 \text{ L*atm/mol*K})(300.15 \text{ K})} \) Finally, solve for n: \( n = 0.00305 \text{ mol} \).

Step 3: Find the missing Volume (V) in the third row

For this, we will use the ideal gas law again: \( PV = nRT \). We need to solve for V, so rearrange the equation as: \[ V = \frac{nRT}{P} \] Convert the pressure from torr to atm: \( P = \frac{650 \text{ torr}}{760 \text{ torr/atm}} = 0.855 \text{ atm} \) Now substitute the given values into the equation: \( V = \frac{(0.333 \text{ mol})(0.0821 \text{ L atm/mol K})(350 \text{ K})}{0.855 \text{ atm}} \) Finally, solve for V: \( V = 10.2 \text{ L} \).

Step 4: Find the missing Pressure (P) in the fourth row

To do this, we will use the ideal gas law one last time: \( PV = nRT \). We need to solve for P, so rearrange the equation as: \[ P = \frac{nRT}{V} \] Convert the volume from mL to L: \( V = \frac{585 \text{ mL}}{1000 \text{ mL/L}} = 0.585 \text{ L} \) Now substitute the given values into the equation: \( P = \frac{(0.250 \text{ mol})(0.0821 \text{ L atm/mol K})(295 \text{ K})}{0.585 \text{ L}} \) Finally, solve for P: \( P = 10.50 \text{ atm} \). The completed table is: $$ \begin{array}{llll} \hline P & V & n & T \\\ \hline 2.00 \mathrm{~atm} & 1.00 \mathrm{~L} & 0.500 \mathrm{~mol} & 48.7 \mathrm{~K} \\\ 0.300 \mathrm{~atm} & 0.250 \mathrm{~L} & 0.00305 \mathrm{~mol} & 300.15^{\circ} \mathrm{K} \\\ 0.855 \mathrm{~atm} & 10.2 \mathrm{~L} & 0.333 \mathrm{~mol} & 350 \mathrm{~K} \\\ 10.50 \mathrm{~atm} & 0.585 \mathrm{~L} & 0.250 \mathrm{~mol} & 295 \mathrm{~K} \\\ \hline \end{array} $$

Key Concepts

Pressure Volume Relationship

The pressure-volume relationship in gases is a crucial part of the ideal gas law, represented as \( PV = nRT \). According to this equation, pressure \( P \) and volume \( V \) have an inverse relationship when the amount of gas \( n \) and temperature \( T \) are held constant. In simpler terms, if the volume increases, the pressure decreases and vice versa, provided other factors remain unchanged.

This concept is critical in understanding the behavior of gases under different conditions. For example, let’s imagine you have a balloon. If you squeeze it, you decrease its volume, which increases the pressure of the air inside. This is because the same amount of gas molecules are now constrained in a smaller space, so they collide more often with the walls, increasing pressure.

In the context of the ideal gas law, if any two of these variables are known, it’s possible to calculate the third. This principle makes it very useful in real-world situations, like predicting the behavior of gases in different temperatures and environments.

Mole Calculations

Mole calculations are an essential part of using the ideal gas law. The number of moles \( n \) indicates the amount of substance present and is a central factor in gas calculations. When applying the ideal gas law, knowing the number of moles enables us to find other unknowns like pressure, volume, or temperature, depending on what is given in the problem.

To calculate moles in our exercise, we use the rearranged ideal gas law formula:

• \( n = \frac{PV}{RT} \)

Given a pressure \( P \), volume \( V \), and temperature \( T \), you substitute these values into the formula along with the ideal gas constant \( R \) to find the number of moles. The ideal gas constant \( R \), in the unit of \( \text{L atm/mol K} \), commonly has a value of 0.0821.

In calculations, be mindful that the units must be consistent, particularly the temperature, which should be in Kelvin. Understanding how to compute the number of moles helps solve many practical problems involving gases, from simple lab experiments to large-scale industrial processes.

Temperature Conversion

Temperature conversion is often necessary when working with the ideal gas law, as it requires temperatures in Kelvin. Celsius is more commonly used in everyday life, but Kelvin is the standard unit for scientific calculations involving gases due to its absolute scale.

To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature:

• \( T(\text{K}) = T(\degree \text{C}) + 273.15 \)

This conversion is important because many gas law calculations can produce incorrect results if temperatures are not properly converted. For example, in our exercise, the temperature conversion was necessary to find the number of moles when solving the second row of the table.

Remembering this simple addition can avoid confusion and ensure accurate calculations in any scientific context involving temperatures. This understanding is crucial not only for ideal gas law calculations but also for any other scientific phenomenon where temperature plays a role.