Chapter 13: Problem 24
What pressure (in atmospheres) is required to compress \(1.00 \mathrm{~L}\) of gas at \(760 . \mathrm{mm} \mathrm{Hg}\) pressure to a volume of \(50.0 \mathrm{~mL} ?\)
Short Answer
Expert verified
A pressure of 20 atm is required to compress 1.00 L of gas at 760 mmHg pressure to a volume of 50.0 mL.
Step by step solution
01
Convert pressure from mmHg to atmospheres
In this problem, our given initial pressure is in mmHg. However, we need the final answer in atmospheres, so let's first convert the given pressure from mmHg to atmospheres.
1 atm = 760 mmHg
Given initial pressure = 760 mmHg
Therefore, initial pressure in atmospheres = \( \frac{760 \ mmHg}{760 \ mmHg/atm} = 1 \ atm\).
02
Convert volume from L to mL
Next, we need to convert the given initial and final volume of the gas to the same unit. In this case, we will convert both to milliliters (mL) since the final volume is given in mL.
1 L = 1000 mL
Initial volume in mL = 1 L * 1000 \( \frac{mL}{L} \) = 1000 mL
03
Use Boyle's Law to find the new pressure
Now that we have the initial pressure in atmospheres and both volumes in mL, we can solve for the new pressure using Boyle's Law.
Boyle's Law: \(P_1V_1 = P_2V_2\)
Initial Pressure (P1) = 1 atm
Initial Volume (V1) = 1000 mL
New Volume (V2) = 50 mL
Now, we need to find the new pressure (P2).
1 atm × 1000 mL = P2 × 50 mL
Now, we isolate P2:
P2 = \(\frac{1 \ atm × 1000 \ mL}{50 \ mL}\)
04
Calculate the new pressure
Now that we have an equation for P2, we can solve for the pressure required to compress the gas:
P2 = \(\frac{1 \ atm × 1000 \ mL}{50 \ mL}\) = 20 atm
Therefore, a pressure of 20 atmospheres is required to compress 1.00 L of gas at 760 mmHg pressure to a volume of 50.0 mL.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Atmospheric Pressure
Atmospheric pressure is a fundamental concept in understanding various physical phenomena including gas laws. It is the force per unit area exerted by the weight of the atmosphere. At sea level, it typically measures about 760 millimeters of mercury (mmHg), which is also equivalent to 1 atmosphere (atm). Given this, atmospheric pressure serves as the point of reference in the problem, where the initial gas pressure equaling 760 mmHg is equivalent to 1 atm. At different altitudes, atmospheric pressure varies significantly, which can influence gas behavior and compression.
For problems involving gases, it's essential to measure pressure relative to atmospheric pressure, as it impacts how gases expand or compress. When a problem requires changing units of pressure, as in the provided exercise, understanding the equivalence between mmHg and atm becomes critical. This conversion ensures accuracy when applying formulas like Boyle's Law.
For problems involving gases, it's essential to measure pressure relative to atmospheric pressure, as it impacts how gases expand or compress. When a problem requires changing units of pressure, as in the provided exercise, understanding the equivalence between mmHg and atm becomes critical. This conversion ensures accuracy when applying formulas like Boyle's Law.
Gas Compression
Gas compression is the process of decreasing the volume of a gas by applying pressure. Gases are compressible because the particles within a gas have a significant amount of space between them, unlike solids or liquids. When we compress a gas, we are essentially forcing the particles closer together, reducing the volume the gas occupies. The problem at hand deals with gas compression by asking what pressure is required to compress a certain volume of gas to a smaller volume.
It is vital to recognize that during compression, the temperature and amount of gas remain constant for Boyle’s Law to be applicable. Additionally, when a gas is compressed, the energy often increases in the system, which might raise the temperature if the gas is not allowed to cool. This aspect, although not directly addressed in this problem, is an important consideration in real-world applications of gas compression.
It is vital to recognize that during compression, the temperature and amount of gas remain constant for Boyle’s Law to be applicable. Additionally, when a gas is compressed, the energy often increases in the system, which might raise the temperature if the gas is not allowed to cool. This aspect, although not directly addressed in this problem, is an important consideration in real-world applications of gas compression.
Pressure-Volume Relationship
The pressure-volume relationship of gases is beautifully captured in Boyle's Law, a cornerstone in the study of gas behavior under varying conditions. Boyle’s Law states that for a given mass of gas at constant temperature, the volume of the gas is inversely proportional to its pressure. Mathematically, this is represented as \( P_1V_1 = P_2V_2 \), where \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( P_2 \) and \( V_2 \) are the new pressure and volume after the change.
In the workout example given, the student is required to apply Boyle's Law to solve for the new pressure after decreasing the volume of the gas. The direct, inverse relationship implies that when the volume decreases, the pressure must increase proportionally, and vice versa. By re-arranging the equation, you can isolate and solve for the unknown variable, which in this case is the new pressure exerted on the gas. The law is a perfect illustration of how interconnected pressure and volume are for a gas held at constant temperature.
In the workout example given, the student is required to apply Boyle's Law to solve for the new pressure after decreasing the volume of the gas. The direct, inverse relationship implies that when the volume decreases, the pressure must increase proportionally, and vice versa. By re-arranging the equation, you can isolate and solve for the unknown variable, which in this case is the new pressure exerted on the gas. The law is a perfect illustration of how interconnected pressure and volume are for a gas held at constant temperature.