Problem 8 - Chemistry Matter and Change

Question

Assume that the volume and the amount of gas are constant in the following problems. The pressure in an automobile tire is 1.88 atm at \(25.0^{\circ} \mathrm{C} .\) What will be the pressure if the temperature increases to \(37.0^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The pressure will be approximately 1.96 atm.

Step 1: Identify the Known Values

First, identify the variables that are given in the problem. We know the initial pressure is \( P_1 = 1.88 \, \text{atm} \). The initial temperature is \( T_1 = 25.0 \, ^{\circ}\text{C} \), which needs to be converted to Kelvin, and the final temperature is \( T_2 = 37.0 \, ^{\circ}\text{C} \).

Step 2: Convert Temperatures to Kelvin

To employ the gas laws, temperatures must be in Kelvin. To convert from Celsius to Kelvin, use the formula: \( T(\text{K}) = T(^{\circ}C) + 273.15 \). Therefore, \( T_1 = 25.0 + 273.15 = 298.15 \, \text{K} \) and \( T_2 = 37.0 + 273.15 = 310.15 \, \text{K} \).

Step 3: Apply Gay-Lussac's Law

Since the volume and the amount of gas remain constant, we can use Gay-Lussac's Law, which states: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). We need to find \( P_2 \), the final pressure. Rearrange the equation to solve for \( P_2 \): \( P_2 = P_1 \times \frac{T_2}{T_1} \).

Step 4: Calculate the Final Pressure

Substitute the known values into the equation: \( P_2 = 1.88 \, \text{atm} \times \frac{310.15 \, \text{K}}{298.15 \, \text{K}} \). Compute this to find \( P_2 \).

Step 5: Solve Equation

Perform the calculation: \( P_2 = 1.88 \times 1.0403 \approx 1.956 \text{ atm} \). The final pressure is approximately 1.96 atm.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry used to describe the state of an ideal gas. It is expressed as \( PV = nRT \), where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume of the gas,
• \( n \) is the number of moles of gas,
• \( R \) is the ideal gas constant, and
• \( T \) is the temperature in Kelvin.

This equation helps us understand how the pressure, volume, and temperature of a gas relate to each other. When two of these factors are held constant (as in our exercise where the volume and amount of gas, \( n \), are constant), changes in one can predict changes in the others using relevant gas laws.

Pressure-Temperature Relationship

The pressure-temperature relationship for gases is described by Gay-Lussac’s Law. It states that when the volume and amount of gas are constant, the pressure of a gas is directly proportional to its temperature in Kelvin. This means if the temperature rises, the pressure rises, provided the other conditions remain unchanged.
In formula terms, this law is expressed as \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), where:

• \( P_1 \) and \( P_2 \) are the initial and final pressures, respectively,
• \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.

Understanding this relationship is crucial for applications like predicting the behavior of air in automobile tires, as seen in the exercise, or even in weather balloons. When temperatures increase or decrease, the pressure follows the same trend, assuming no change in volume or gas quantity.

Temperature Conversion

Temperature conversion is an important step in working with gas laws. Ideally, all temperatures should be converted to the Kelvin scale for calculations involving gas laws. The formula used for conversion from Celsius to Kelvin is simple:
To convert:

• Add 273.15 to the Celsius temperature.

So if you have a temperature of \( 25.0 \, ^{\circ}C \), in Kelvin it would be \( 25.0 + 273.15 = 298.15 \, K \).
This conversion is fundamental because the Kelvin scale starts at absolute zero, where theoretically, a gas would have no kinetic energy, making it ideal for gas laws. The step-by-step problem uses this conversion to ensure accurate calculation of pressure changes with temperature.

Kelvin Scale

The Kelvin scale is a thermodynamic temperature scale that starts at absolute zero, the lowest temperature possible. Zero Kelvin (0 K) is equivalent to -273.15°C, the point at which molecular motion would cease. Kelvin is used in the Ideal Gas Law and other scientific equations because it provides a true zero point, preventing any negative temperature readings.
Unlike Celsius or Fahrenheit, the Kelvin scale is directly correlated to kinetic energy. Therefore, when analyzing gases, converting temperatures to Kelvin ensures calculations remain physically meaningful, avoiding complications that arise with other temperature scales. This is crucial in exercises like the automobile tire problem, where understanding the impact of temperature changes on gas pressure relies on the use of absolute temperatures.