Problem 10 - Chemistry Matter and Change

Question

Assume that the volume and the amount of gas are constant in the following problems. Challenge If a gas sample has a pressure of 30.7 \(\mathrm{kPa}\) at \(0.00^{\circ} \mathrm{C},\) by how many degrees Celsius does the temperature have to increase to cause the pressure to double?

Step-by-Step Solution

Answer

The temperature must increase by 273.15°C for the pressure to double.

Step 1: Identify the Known and Unknown Variables

We know the initial pressure \(P_1 = 30.7 \ \mathrm{kPa}\), the final pressure is \(P_2 = 2 \times P_1 = 61.4 \ \mathrm{kPa}\). The initial temperature \(T_1 = 0.00^{\circ} \mathrm{C}\), which is \(273.15 \ \mathrm{K}\) when converted into Kelvin. We need to find the final temperature \(T_2\) in Celsius.

Step 2: Use the Pressure-Temperature Relationship

According to Gay-Lussac's Law, \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). Rearrange this to find \(T_2\):\[T_2 = \frac{P_2 \cdot T_1}{P_1}\]

Step 3: Substitute the Known Values

Plug in the known values from Step 1:\[T_2 = \frac{61.4 \cdot 273.15}{30.7}\]

Step 4: Calculate the Final Temperature in Kelvin

Carry out the calculation:\[T_2 = \frac{61.4 \cdot 273.15}{30.7} = 546.3 \ \mathrm{K}\]

Step 5: Convert Kelvin Back to Celsius

To convert \(T_2\) from Kelvin to Celsius, subtract 273.15:\[T_2 = 546.3 - 273.15 = 273.15^{\circ}C\]

Step 6: Determine the Temperature Increase

The increase in temperature is the difference between the final and initial temperatures:\[\Delta T = 273.15 \ - 0 = 273.15^{\circ}C\]

Key Concepts

Pressure-Temperature Relationship

One of the fundamental principles affecting gases is the pressure-temperature relationship, also known as Gay-Lussac's Law. This law directly connects the pressure of a gas with its temperature, provided that the volume and amount of gas remain constant. Essentially, if you increase the temperature of a gas, its pressure will increase too.
See how the relationship unfolds:

• When the temperature of a gas rises, the molecules gain energy and move more rapidly.
• This results in more frequent and forceful collisions against the walls of the container, thus increasing the pressure.
• The law can be expressed with the formula: \[\frac{P_1}{T_1} = \frac{P_2}{T_2} \]where \(P_1\) and \(P_2\) are the initial and final pressures, \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin.

Remember, this law only applies to ideal gases held at a constant volume and amount, making it a useful guide to predict how pressure changes with temperature.

Kelvin to Celsius Conversion

When dealing with the pressure-temperature relationship, it is crucial to use temperatures in Kelvin. This is because Kelvin provides an absolute measurement scale, starting from the coldest possible temperature—absolute zero.
Converting between Kelvin and Celsius is straightforward:

• To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
• For example, if the temperature is \(0\degree C\), the equivalent Kelvin temperature is \(273.15 \, K\).
• Conversely, to convert from Kelvin to Celsius, subtract 273.15 from the Kelvin temperature.
• So, a Kelvin temperature of \(546.3 \, K\) converts to a Celsius temperature of \(273.15^{\circ}C\).

This conversion is important since calculations involving gas behavior (like in Gay-Lussac's Law) require temperature in the absolute scale for accuracy.

Initial and Final Conditions

When problem-solving with the gas laws, it is important to clearly define the initial and final conditions of the system. This ensures accurate application of the formulas and a correct result.
Initial conditions typically include:

• Initial pressure (\(P_1\)), which is how much force the gas exerts before any changes.
• Initial temperature (\(T_1\)), often given in Celsius and needing conversion to Kelvin.

Final conditions capture the state of the gas after some change has occurred:

• Final pressure (\(P_2\)), what the pressure becomes after temperature adjustment.
• Final temperature (\(T_2\)), requiring calculation via the pressure-temperature formula.

For example, if a gas at \(0\degree C\) and \(30.7 \, kPa\) doubles in pressure, calculating accurately depends on both understanding these starting and ending conditions, and converting temperatures to Kelvin.