Chapter 8: Problem 65
What is the effect on the volume of a gas if you simultaneously: (a) Halve its pressure and double its Kelvin temperature? (b) Double its pressure and double its Kelvin temperature?
Short Answer
Expert verified
(a) Volume becomes four times the original; (b) Volume remains unchanged.
Step by step solution
01
Identify the Gas Law
The exercise involves understanding the relationship between pressure, volume, and temperature of a gas. According to the combined gas law, the relationship is given by:\[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]where \( P \) is pressure, \( V \) is volume, and \( T \) is temperature in Kelvin.
02
Apply Changes for Part (a)
For part (a) of the question, the pressure is halved and the temperature is doubled, so we express these changes using the initial conditions:- Pressure: \( P_2 = \frac{1}{2}P_1 \)- Temperature: \( T_2 = 2T_1 \)Plug these into the gas law equation:\[ \frac{P_1V_1}{T_1} = \frac{\frac{1}{2}P_1 \cdot V_2}{2T_1} \]
03
Solve the Equation for Part (a)
Simplify the equation from Step 2:\[ \frac{P_1V_1}{T_1} = \frac{1}{4} \cdot \frac{P_1V_2}{T_1} \]This simplifies to:\[ V_2 = 4V_1 \]Thus, when pressure is halved and temperature is doubled, the volume becomes four times the original volume.
04
Apply Changes for Part (b)
For part (b) of the question, both the pressure and temperature are doubled, so we express these changes:- Pressure: \( P_2 = 2P_1 \)- Temperature: \( T_2 = 2T_1 \)Substitute into the gas law equation:\[ \frac{P_1V_1}{T_1} = \frac{2P_1 \cdot V_2}{2T_1} \]
05
Solve the Equation for Part (b)
Simplify the equation from Step 4:\[ \frac{P_1V_1}{T_1} = \frac{P_1V_2}{T_1} \]This implies:\[ V_2 = V_1 \]Thus, when both pressure and temperature are doubled, the volume remains unchanged.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Pressure, Volume, and Temperature Relationship
The relationship between pressure, volume, and temperature of a gas can be understood using the combined gas law. This fundamental chemistry principle can be expressed with the formula: \[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]where:
- \(P\) stands for pressure.
- \(V\) is the volume.
- \(T\) is the temperature, but it must always be in Kelvin.
- If pressure increases, volume decreases if temperature remains the same, illustrating Boyle's Law.
- If temperature increases, volume increases, showing Charles's Law in action.
- Pressure and temperature are directly proportional, indicated by Gay-Lussac's Law.
Gas Laws Problem Solving
Applying the combined gas law to solve problems involves understanding changes in conditions such as pressure, volume, and temperature. Careful examination of what is being altered or unchanged helps in setting up the right equation. For example, consider part (a) in the original exercise. The problem states both pressure is halved, and temperature is doubled. The combined gas law helps us find the new volume by substituting these changes into our equation:- For example, in our change setup: \[ P_2 = \frac{1}{2}P_1 \] \[ T_2 = 2T_1 \]- Now, it becomes: \[ \frac{P_1V_1}{T_1} = \frac{\frac{1}{2}P_1 \cdot V_2}{2T_1} \] After simplifying, \( V_2 \) becomes four times \( V_1 \). In part (b), when both pressure and temperature are doubled:- Here, replace with: \[ P_2 = 2P_1 \] \[ T_2 = 2T_1 \]- And our law then is: \[ \frac{P_1V_1}{T_1} = \frac{2P_1 \cdot V_2}{2T_1} \] Simplifying gives \( V_2 = V_1 \), meaning no volume change. This structured substitution method in the combined gas law clarifies the scenarios.
Kelvin Temperature Effect on Gases
Understanding how Kelvin temperature affects gases is crucial in gas laws. The Kelvin scale begins at absolute zero, which is \(-273.15 \ ^\circ\text{C}\), and it forms a zero point commonly used in sciences to avoid negative temperature when doing these calculations.In any gas law calculation:
- The temperature's absolute value makes it straightforward when using the formula for transformations.
- Doubling the Kelvin temperature in calculations directly impacts volume—it equates to doubling kinetic energy, expanding volume if pressure is constant.