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As discussed in this section. Boyle's law states that at a constant temperature the pressure \(P\) exerted by a gas is related to the volume \(V\) by the equation \(P=k / V\) (a) Find the appropriate units for the constant \(k\) if pressure (which is force per unit area) is in newtons per square meter \(\left(\mathrm{N} / \mathrm{m}^{2}\right)\) and volume is in cubic meters \(\left(\mathrm{m}^{3}\right)\) (b) Find \(k\) if the gas exerts a pressure of \(20.000 \mathrm{N} / \mathrm{m}^{2}\) when the volume is 1 liter \(\left(0.001 \mathrm{m}^{3}\right)\) (c) Make a table that shows the pressures for volumes of \(0.25,0.5,1.0 .1 .5,\) and 2.0 liters. (d) Make a graph of \(P\) versus \(V\)

Short Answer

Expert verified
Units for \(k\) are N\cdot m, \(k=20\) N\cdot m, pressures: (0.25, 80000), (0.5, 40000), (1.0, 20000), (1.5, 13333.33), (2.0, 10000).

Step by step solution

01

Understanding Boyle's Law

Boyle's Law is given by the formula \( P = \frac{k}{V} \), where \( P \) is the pressure, \( V \) is the volume, and \( k \) is a constant that depends on the specific gas and conditions of the system such as temperature.
02

Finding Units for k

Given that pressure \( P \) is in \( \mathrm{N}/\mathrm{m}^2 \) and volume \( V \) is in \( \mathrm{m}^3 \), rearrange the equation \( P = \frac{k}{V} \) to solve for \( k \):\[k = P \cdot V \tag{1}\]Substitute the units for \( P \) and \( V \):\[k = \left( \mathrm{N}/\mathrm{m}^2 \right) \cdot \mathrm{m}^3 = \mathrm{N}\cdot \mathrm{m}otag\]Thus, the units for \( k \) are newton-meters (\( \mathrm{N}\cdot \mathrm{m} \)).
03

Calculating k for Given Conditions

Given \( P = 20000 \mathrm{N}/\mathrm{m}^2 \) and \( V = 0.001 \mathrm{m}^3 \), use equation \( (1) \) to find \( k \):\[k = 20000 \cdot 0.001 = 20 \mathrm{N}\cdot \mathrm{m} otag\]Therefore, \( k = 20 \mathrm{N}\cdot \mathrm{m} \).
04

Creating a Pressure Table

Use the derived value of \( k = 20 \) to find the pressures for different volumes. For each given volume \( V \), calculate pressure \( P \) using the equation \( P = \frac{k}{V} \):- For \( V = 0.25 \mathrm{L} = 0.00025 \mathrm{m}^3 \), \( P = \frac{20}{0.00025} = 80000 \mathrm{N}/\mathrm{m}^2 \)- For \( V = 0.5 \mathrm{L} = 0.0005 \mathrm{m}^3 \), \( P = \frac{20}{0.0005} = 40000 \mathrm{N}/\mathrm{m}^2 \)- For \( V = 1.0 \mathrm{L} = 0.001 \mathrm{m}^3 \), \( P = \frac{20}{0.001} = 20000 \mathrm{N}/\mathrm{m}^2 \)- For \( V = 1.5 \mathrm{L} = 0.0015 \mathrm{m}^3 \), \( P = \frac{20}{0.0015} \approx 13333.33 \mathrm{N}/\mathrm{m}^2 \)- For \( V = 2.0 \mathrm{L} = 0.002 \mathrm{m}^3 \), \( P = \frac{20}{0.002} = 10000 \mathrm{N}/\mathrm{m}^2 \)
05

Graphing Pressure vs Volume

Plot pressure \( P \) on the y-axis and volume \( V \) on the x-axis. Use the points calculated in the previous step:- \( (0.00025, 80000) \)- \( (0.0005, 40000) \)- \( (0.001, 20000) \)- \( (0.0015, 13333.33) \)- \( (0.002, 10000) \)This will show an inverse relationship between pressure and volume, consistent with Boyle's Law.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure and Volume Relationship
Boyle's Law elegantly explains the inverse relationship between the pressure and volume of a gas. This means that when you decrease the volume occupied by the gas, the pressure exerted by the gas increases, provided the temperature remains constant. This relationship is expressed by the equation \( P = \frac{k}{V} \), where \(P\) is pressure, \(V\) is volume, and \(k\) is a constant. In simple terms, if you press a gas into a smaller space, it pushes back harder.

Let's break it down further:
  • When volume \(V\) gets smaller, \(P\) increases because \(V\) is in the denominator. Less space means molecules collide more frequently, so pressure rises.
  • If volume \(V\) increases, \(P\) decreases, as there is more space for molecules to move and cause fewer collisions.
The constant \(k\) embodies the specific properties of the gas and the environmental conditions like temperature. By understanding this fundamental principle, you can analyze various gas-related phenomena in real-world situations.
Unit Conversion in Physics
Converting units in physics is crucial for ensuring accuracy and consistency. In the context of Boyle's Law, we need to ensure that units for pressure and volume are compatible.

Let's explore why unit conversion is important:
  • Pressure is measured in \(\frac{\mathrm{N}}{\mathrm{m}^2}\) (newtons per square meter), which means force per unit area.
  • Volume is generally calculated in \(\mathrm{m}^3\) (cubic meters), but often provided in liters, especially in lab settings. Since 1 cubic meter equals 1000 liters, you'll need to adjust units accordingly for calculations.
  • The constant \(k\) combines these units such that \(k = P \cdot V\), leading to units of \(\mathrm{N}\cdot \mathrm{m}\) (newton-meters).
By consistently applying unit conversion principles, complex calculations are simplified and errors are prevented. It's like speaking the same language across different calculations, ensuring everyone is on the same page, which is especially vital in scientific experiments.
Graphing Data
Graphing data, especially when exploring laws like Boyle's, helps in visualizing relationships and patterns. Plotting a graph for pressure versus volume offers a clear depiction of the law's inverse relationship.

Here's how you typically approach graphing Boyle's Law:
  • Assign pressure \(P\) to the y-axis and volume \(V\) to the x-axis. This setup allows you to see how pressure changes with varying volumes.
  • Plot the values from your calculations. The points will likely form a curve that slopes downward, showing the decrease in pressure as volume increases, and vice versa.
  • The curve's shape reinforces the understanding of the inverse relationship described by \( P = \frac{k}{V} \).
Using graphs, students and scientists can better comprehend how one quantity affects another, going beyond abstract equations to intuitive insights. Graphs transform numbers into visuals, aiding in quick analysis and making the science both more accessible and engaging.

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