Chapter 5: Problem 68
\(\Lambda 100 \mathrm{~mL}\) sample of a gas at \(73^{\circ} \mathrm{C}\) and 2 atmospheres is heatcd to \(127^{\circ} \mathrm{C}\) and the pressure is reduced to \(0.5\) atmosphere. What will be the final volume? (1) \(8000 \mathrm{~mL}\) (2) \(800 \mathrm{~mL}\) (3) \(400 \mathrm{~mL}\) (4) \(4000 \mathrm{~mL}\)
Short Answer
Expert verified
The final volume is closest to 400 mL.
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gas Laws
Gas laws are fundamental in understanding how gases behave under different conditions. The combined gas law, for instance, merges Boyle's, Charles's, and Gay-Lussac's laws. It shows the relationship between pressure, volume, and temperature in a closed system.
The combined gas law formula is:\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]This means the initial pressure and volume, divided by the initial temperature, equals the final pressure and volume divided by the final temperature.
This equation helps to predict how a gas will react when subjected to changes in pressure, volume, and temperature.
When solving these problems, it's important to ensure the units are consistent. Temperatures must be in Kelvin, as Celsius can complicate calculations.
By understanding and applying the combined gas law, one can solve real-world gas problems easily.
The combined gas law formula is:\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]This means the initial pressure and volume, divided by the initial temperature, equals the final pressure and volume divided by the final temperature.
This equation helps to predict how a gas will react when subjected to changes in pressure, volume, and temperature.
When solving these problems, it's important to ensure the units are consistent. Temperatures must be in Kelvin, as Celsius can complicate calculations.
By understanding and applying the combined gas law, one can solve real-world gas problems easily.
Temperature Conversion
Before using the gas laws, you must convert temperatures from Celsius to Kelvin. This is because gas law equations are derived based on Kelvin (K). To convert Celsius (°C) to Kelvin (K), use this simple formula:
\[T (in \, K) = T (in \, °C) + 273.15\]
For example, an initial temperature of 73°C is converted to Kelvin like this:
\[T_1 = 73 + 273.15 = 346.15 \, K\]
Similarly, a final temperature of 127°C converts to:
\[T_2 = 127 + 273.15 = 400.15 \, K\]
This conversion is crucial for accurately applying temperature in gas law calculations. Kelvin ensures that we are working with an absolute temperature scale.
Always remember to convert temperature first when tackling gas law problems.
\[T (in \, K) = T (in \, °C) + 273.15\]
For example, an initial temperature of 73°C is converted to Kelvin like this:
\[T_1 = 73 + 273.15 = 346.15 \, K\]
Similarly, a final temperature of 127°C converts to:
\[T_2 = 127 + 273.15 = 400.15 \, K\]
This conversion is crucial for accurately applying temperature in gas law calculations. Kelvin ensures that we are working with an absolute temperature scale.
Always remember to convert temperature first when tackling gas law problems.
Pressure and Volume Calculations
Pressure and volume are also key factors in gas law problems. Changing the pressure and volume of a gas will affect its other properties. Here's how you use the combined gas law to find new volume \(V_2\):
The equation is rearranged to solve for \(V_2\):
\[V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}\]
Plugging in the values given in the problem:
\[V_2 = \frac{(2 \, atm)(100 \, mL)(400.15 \, K)}{(346.15 \, K)(0.5 \, atm)}\]
Simplifying the equation, we calculate:
\[V_2 = \frac{2 \times 100 \times 400.15}{346.15 \times 0.5}\]
\[V_2 \right) \frac{80030}{173.075}\]
\[V_2 \right) 462.07 \, mL\]
The final volume \(V_2\) is approximately 462.07 mL.
Since this isn't one of the provided choices, you select the closest value, which is 400 mL.
Pressure and volume calculations become simple and intuitive with practice and understanding.
The equation is rearranged to solve for \(V_2\):
\[V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}\]
Plugging in the values given in the problem:
\[V_2 = \frac{(2 \, atm)(100 \, mL)(400.15 \, K)}{(346.15 \, K)(0.5 \, atm)}\]
Simplifying the equation, we calculate:
\[V_2 = \frac{2 \times 100 \times 400.15}{346.15 \times 0.5}\]
\[V_2 \right) \frac{80030}{173.075}\]
\[V_2 \right) 462.07 \, mL\]
The final volume \(V_2\) is approximately 462.07 mL.
Since this isn't one of the provided choices, you select the closest value, which is 400 mL.
Pressure and volume calculations become simple and intuitive with practice and understanding.
Physics in Chemistry
The study of gas laws is where physics beautifully intersects with chemistry. Gases are part of both fields, influencing countless chemical reactions and physical processes.
Gas molecules move randomly and exert pressure when they collide with container walls. This motion is explained through kinetic molecular theory—a basic physics concept.
By understanding this motion, you can see why temperature, pressure, and volume are interconnected. Heating a gas increases the kinetic energy of its molecules, causing them to move faster and collide more often. This is why temperature changes impact gas behavior.
Concepts from physics, like energy and molecular motion, are essential in explaining how gases behave in chemistry. They help chemists predict and control reactions and processes.
So, mastering gas laws isn't just solving equations; it's about appreciating the physics in chemistry.
Gas molecules move randomly and exert pressure when they collide with container walls. This motion is explained through kinetic molecular theory—a basic physics concept.
By understanding this motion, you can see why temperature, pressure, and volume are interconnected. Heating a gas increases the kinetic energy of its molecules, causing them to move faster and collide more often. This is why temperature changes impact gas behavior.
Concepts from physics, like energy and molecular motion, are essential in explaining how gases behave in chemistry. They help chemists predict and control reactions and processes.
So, mastering gas laws isn't just solving equations; it's about appreciating the physics in chemistry.
