Problem 7 - Chemistry: Principles and Reactions

Question

A sample of \(\mathrm{CO}_{2}\) gas at \(22^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) has a volume of \(2.00 \mathrm{~L}\). Determine the ratio of the original volume to the final volume when (a) the pressure and amount of gas remain unchanged and the Celsius temperature is doubled. (b) the pressure and amount of gas remain unchanged and the Kelvin temperature is doubled.

Step-by-Step Solution

Answer

The volume ratio is approximately 0.931 when the Celsius temperature is doubled, and approximately 0.500 when the Kelvin temperature is doubled.

Step 1: Convert Celsius to Kelvin

The initial temperature is given in Celsius. We need to convert it to Kelvin in order to use the ideal gas law. To do this, add 273.15 to the Celsius temperature: \(T_{1} = 22^{\circ} \mathrm{C} + 273.15 = 295.15\, \mathrm{K}\)

Step 2(a): Find the final temperature in Celsius and Kelvin when Celsius temperature is doubled

We are given that the Celsius temperature is doubled. So, the final Celsius temperature will be: \(T_{2\,C} = 2 \times 22^{\circ} \mathrm{C} = 44^{\circ} \mathrm{C}\) Now, convert the final Celsius temperature to Kelvin: \(T_{2} = 44^{\circ} \mathrm{C} + 273.15 = 317.15\, \mathrm{K}\)

Step 3(a): Calculate the volume ratio for part (a)

We know that the pressure and amount of gas remain constant in this case. Therefore, using the ideal gas law, we can make the following proportion: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) Rearranging for the volume ratio: \(\frac{V_1}{V_2} = \frac{T_1}{T_2}\) Now, plug in the initial and final Kelvin temperatures to find the volume ratio: \(\frac{V_1}{V_2} = \frac{295.15\, \mathrm{K}}{317.15\, \mathrm{K}} \approx 0.931\)

Step 2(b): Find the final temperature in Kelvin when Kelvin temperature is doubled

We are given that the Kelvin temperature is doubled. So, the final Kelvin temperature will be: \(T_{2} = 2 \times 295.15\, \mathrm{K} = 590.3\, \mathrm{K}\)

Step 3(b): Calculate the volume ratio for part (b)

As in part (a), the pressure and amount of gas remain constant. Again, we can use the ideal gas law to create the following proportion: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) Rearranging for the volume ratio: \(\frac{V_1}{V_2} = \frac{T_1}{T_2}\) Now, plug in the initial and final Kelvin temperatures to find the volume ratio: \(\frac{V_1}{V_2} = \frac{295.15\, \mathrm{K}}{590.3\, \mathrm{K}} \approx 0.500\) So, the ratio of the original volume to the final volume is approximately 0.931 when the Celsius temperature is doubled, and approximately 0.500 when the Kelvin temperature is doubled.

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and amount of moles of a gas. Formally, we express this relationship as:

\[\begin{equation}PV = nRT\text{,}\end{equation}\]where P represents the pressure, V stands for the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvins. This equation makes the assumption that the gas behaves ideally, meaning its particles are point masses with no volume of their own and no interactions between them other than elastic collisions.

In practical terms, the ideal gas law allows us to predict how a gas will behave under different conditions. For example, if we know the initial state of a gas (its pressure, volume, and temperature) and we change one of those variables while keeping others constant, we can calculate the new state of the gas. This law is thus crucial for solving many real-world problems, including the one from our exercise where the temperature changes while the pressure and the amount of gas remain unchanged.

Kelvin Temperature Scale

The Kelvin temperature scale is an absolute temperature scale, meaning it starts at absolute zero, the theoretical temperature where particles have minimal vibrational motion. Unlike the Celsius scale, which is based on the properties of water, the Kelvin scale is universal for all types of matter and critical for scientific calculations involving temperature.

One Kelvin (1 K) is defined as the fraction 1/273.16 of the thermodynamic temperature of the triple point of water. By definition, absolute zero is 0 K, and it is equivalent to -273.15°C. To convert from Celsius to Kelvin, one simply adds 273.15. This direct proportionality without offsets makes calculations in physics and chemistry more straightforward, as seen in the exercise where a Celsius temperature was converted to Kelvin to use the ideal gas law.

The Kelvin scale is indispensable in the study of gases because most gas laws, including the ideal gas law, require temperatures to be expressed in Kelvins for accurate predictions.

Volume-Temperature Relationship

The volume-temperature relationship, also known as Charles's Law, states that at a constant pressure, the volume of a gas is directly proportional to its temperature in Kelvins. Mathematically, we can express Charles's Law as:

\[\begin{equation}\frac{V}{T} = k\text{,}\end{equation}\]where V is the volume, T is the absolute temperature, and k is a constant for a given amount of the gas at constant pressure. When the temperature of a gas increases, so does its volume, provided the pressure is held constant.

The relationship is crucial for understanding how gases expand or contract with temperature changes. In our exercise, this principle is demonstrated in both scenarios. When the temperature is doubled in Celsius or Kelvin, the volume-temperature relationship is used to determine the change in volume of the gas. In scenario (a), the volume ratio reflects a change in temperature in Celsius, showing a less straightforward relationship due to the conversion to Kelvin. Meanwhile, scenario (b) shows the direct linear relationship between volume and temperature in Kelvin, resulting in a simpler calculation of the volume ratio.