Question

Complete the following table for an ideal gas. $$ \begin{array}{l|lcc|} & P(\mathrm{~atm}) & V(\mathrm{~L}) & n(\mathrm{~mol}) & T \\ \hline \text { a. } & 5.00 & & 2.00 & 155^{\circ} \mathrm{C} \\ \hline \text { b. } & 0.300 & 2.00 & & 155 \mathrm{~K} \\ \hline \text { c. } & 4.47 & 25.0 & 2.01 & \\ \hline \text { d. } & & 2.25 & 10.5 & 75^{\circ} \mathrm{C} \\ \hline \end{array} $$

Short answer

The completed table for the ideal gas is: $$ \begin{array}{l|lcc|} & P(\mathrm{~atm}) & V(\mathrm{~L}) & n(\mathrm{~mol}) & T \\\ \hline \text { a. } & 5.00 & 14.04 & 2.00 & 155^{\circ} \mathrm{C} \\\ \hline \text { b. } & 0.300 & 2.00 & 0.0273 & 155 \mathrm{~K} \\\ \hline \text { c. } & 4.47 & 25.0 & 2.01 & 681.77 \mathrm{~K} \\\ \hline \text { d. } & 131.56 & 2.25 & 10.5 & 75^{\circ} \mathrm{C} \\\ \hline \end{array} $$

Step-by-step solution

Finding The Missing Parameter for case a.

In this case, the missing parameter is volume (V). Given are Pressure (P) = 5 atm, Number of moles (n) = 2 moles, and Temperature (T) = 155 °C. Convert the temperature to Kelvin: T = 155 + 273.15 = 428.15 K. Use the Ideal Gas Law, rearranging for V: \(V = \frac{nRT}{P}\). Using R = 0.0821 L atm/(mol K) (since the units given are atm and L), calculate the volume: \(V = \frac{(2)(0.0821)(428.15)}{5}\) \(V = 14.04 L\) (approximately)

Finding The Missing Parameter for case b.

In this case, the missing parameter is the number of moles (n). Given are Pressure (P) = 0.300 atm, Volume (V) = 2 L, and Temperature (T) = 155 K (already in Kelvin). Use the Ideal Gas Law, rearranging for n: \(n = \frac{PV}{RT}\). Calculate the number of moles: \(n = \frac{(0.300)(2)}{(0.0821)(155)}\) \(n = 0.0273\) moles (approximately)

Finding The Missing Parameter for case c.

In this case, the missing parameter is temperature (T). Given are Pressure (P) = 4.47 atm, Volume (V) = 25 L, and Number of moles (n) = 2.01 moles. Use the Ideal Gas Law, rearranging for T: \(T = \frac{PV}{nR}\). Calculate the temperature in Kelvin: \(T = \frac{(4.47)(25)}{(2.01)(0.0821)}\) \(T = 681.77 K\) (approximately)

Finding The Missing Parameter for case d.

In this case, the missing parameter is pressure (P). Given are Volume (V) = 2.25 L, Number of moles (n) = 10.5 moles, and Temperature (T) = 75 °C. Convert the temperature to Kelvin: T = 75 + 273.15 = 348.15 K. Use the Ideal Gas Law, rearranging for P: \(P = \frac{nRT}{V}\). Calculate the pressure: \(P = \frac{(10.5)(0.0821)(348.15)}{2.25}\) \(P = 131.56\) atm (approximately) The completed table is: $$ \begin{array}{l|lcc|} & P(\mathrm{~atm}) & V(\mathrm{~L}) & n(\mathrm{~mol}) & T \\\ \hline \text { a. } & 5.00 & 14.04 & 2.00 & 155^{\circ} \mathrm{C} \\\ \hline \text { b. } & 0.300 & 2.00 & 0.0273 & 155 \mathrm{~K} \\\ \hline \text { c. } & 4.47 & 25.0 & 2.01 & 681.77 \mathrm{~K} \\\ \hline \text { d. } & 131.56 & 2.25 & 10.5 & 75^{\circ} \mathrm{C} \\\ \hline \end{array} $$

Key concepts

Pressure Volume Temperature Relationship

Understanding the pressure-volume-temperature relationship is fundamental when studying the behavior of gases. This relationship is embodied in the Ideal Gas Law, which is expressed as the equation PV = nRT, where P represents pressure, V the volume, n the number of moles of gas, R the ideal gas constant, and T the absolute temperature in Kelvin. The law asserts that for a fixed amount of gas, the pressure and volume are inversely related when the temperature is constant, and that pressure and temperature are directly related when the volume is constant.

Let's bring this concept to life with a simple example: when heating a sealed container filled with gas, the pressure inside increases. This occurs because the gas particles gain energy and move more rapidly, colliding with the container's walls more often and with greater force. Conversely, if you compress the same amount of gas into a smaller volume without changing the temperature, the pressure increases due to particles hitting the wall more frequently. In our exercise, these principles helped us solve for the unknowns in a table correlating pressure, volume, and temperature for an ideal gas.

Molar Gas Volume

Molar gas volume is a term used to describe the volume one mole of a substance would occupy at a given temperature and pressure. For gases, the molar volume at standard temperature and pressure (STP: 0°C and 1 atmosphere) is 22.4 liters according to Avogadro's law. This value allows chemists to compare the volumes of gases directly and serves as a handy conversion factor in calculations.

In the case of the problems from our exercise, understanding of molar gas volume is crucial when calculating the unknown volume or the number of moles. Suppose you have one mole of an ideal gas at STP; it will occupy 22.4 L. If the conditions change, knowing the molar gas volume lets you predict how the actual volume of gas will change.

Absolute Temperature in Kelvin

The Kelvin scale is an absolute temperature scale that starts at zero, representing the complete absence of thermal energy, known as absolute zero. Unlike Celsius or Fahrenheit, the Kelvin scale is directly proportional to the kinetic energy of the particles in a substance. In the context of the ideal gas law, we always use the temperature in Kelvin, because the law is based on the kinetic theory of gases, which relates temperature to energy.

The relationship between Kelvin and Celsius is given by the formula T(K) = T(°C) + 273.15. This absolute scale ensures that all temperature values in our calculations are positive. During the step by step solutions for our textbook exercise, temperatures in Celsius were converted to Kelvin before applying the ideal gas equation, ensuring accurate calculations.

Avogadro's Law

Avogadro's Law states that equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules. In other words, the volume of a gas is directly proportional to the number of moles of the gas present, as long as the temperature and pressure remain constant. This concept is encompassed in the equation V1/n1 = V2/n2.

For our exercise solutions, Avogadro's Law is implicitly used when manipulating the ideal gas equation. Whenever we calculate the volume per a specific amount of moles, we're taking into consideration that the moles of gas are proportional to the volume. This allows us to determine missing values for volume or moles provided the other variables are held constant.