Question

Calculate A rigid plastic container holds 1.00 L of methane gas at 660 torr pressure when the temperature is 22.0ºC. How much pressure will the gas exert if the temperature is raised to 44.6ºC?

Short answer

When the temperature is raised to 44.6ºC, the gas will exert approximately 709.69 torr of pressure.

Step-by-step solution

Convert the temperatures to Kelvin

Convert the initial and final temperatures from Celsius to Kelvin by adding 273.15 to each temperature value: Initial temperature, T1 = 22.0ºC + 273.15 = 295.15 K Final temperature, T2 = 44.6ºC + 273.15 = 317.75 K

Convert the initial pressure to atm

To make the calculations easier, convert the given initial pressure from torr to atm: Initial pressure, P1 = 660 torr 1 atm = 760 torr P1 = (660 torr)/(760 torr/atm) = 0.868421 atm

Apply the Gay-Lussac's Law formula

With the known initial pressure in atm, initial temperature, and final temperature, we can apply the Gay-Lussac's Law formula, which is given by: P1/T1 = P2/T2 where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature, respectively.

Calculate the final pressure

Rearrange the Gay-Lussac's Law formula to solve for the final pressure (P2): P2 = P1 * (T2/T1) Then substitute the given values into the equation: P2 = 0.868421 atm * (317.75 K / 295.15 K) Calculate the final pressure: P2 = 0.868421 atm * 1.07656 ≈ 0.934059 atm

Convert the final pressure to torr

Convert the final pressure back to torr by multiplying by the conversion factor for torr to atm: Final pressure, P2 = 0.934059 atm * (760 torr/atm) ≈ 709.69 torr The gas will exert approximately 709.69 torr of pressure when the temperature is raised to 44.6ºC.

Key concepts

Temperature Pressure Relationship

Understanding the relationship between temperature and pressure in gases is crucial for studying their behavior under various conditions. At the core of this relationship lies the direct proportionality, which states that for a fixed amount of gas at a constant volume, the pressure exerted by the gas is directly proportional to its absolute temperature (measured in Kelvin). This principle is a key aspect of Gay-Lussac's Law, one of the fundamental gas laws governing the behavior of ideal gases.

In practical terms, if you increase the temperature of a gas that is contained within a rigid container (where volume can't change), the pressure of the gas will also increase. Conversely, if the temperature drops, the pressure will decrease, assuming no gas particles are added or removed during the process. This phenomenon is common in everyday life; for example, it is the reason why a car's tire pressure may increase during a hot day.

When it comes to solving problems involving Gay-Lussac's Law, such as the given textbook example, it's imperative to use Kelvin for temperature measurements because Kelvin is the SI unit for thermodynamic temperature and reflects the direct relationship more clearly than other scales like Celsius or Fahrenheit.

Gas Laws

The gas laws are a set of fundamental principles that describe the behavior of ideal gases. Aside from Gay-Lussac's Law, which focusses on the temperature-pressure relationship at constant volume, there are several other important laws within this group:

• Boyle's Law: Describes the inverse relationship between pressure and volume when temperature is kept constant.
• Charles's Law: States that volume and temperature are directly proportional to each other at constant pressure.
• Avogadro's Law: Suggests that at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas present.

Combined, these laws can be unified into the ideal gas law, represented as PV = nRT, which is a cornerstone in the study of thermodynamics and fluid mechanics. By mastering the individual gas laws, students are better prepared to understand and apply the ideal gas law in various chemical calculations and real-world applications.

Chemistry Calculations

Chemistry calculations often involve converting measurements to ensure consistency and accuracy, as demonstrated in the textbook example. Converting temperatures to Kelvin and pressures to atmospheres plays a vital role in simplifying the calculation process within gas law applications.

When you encounter a gas law problem, like the one given, it is crucial to:

• Identify the variables and constants in the problem.
• Ensure that all units are consistent (pressure in atm, volume in liters, and temperature in Kelvin).
• Apply the correct gas law formula.
• Rearrange the formula, if necessary, to solve for the unknown variable.
• Carry out the calculations accurately, keeping track of significant figures.

To improve the learnability of chemistry calculations, it can be beneficial to practice with examples from daily life, and to approach problems methodically by breaking them down into smaller, more manageable steps. Furthermore, understanding the theoretical underpinnings of the laws can provide a stronger foundation for solving these problems with confidence.