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Chapter 13: Problem 72

Given below are pressure-volume data for a sample of \(0.28\) grams of \(\mathrm{N}_{2}\left(g\right.\) ) at \(25^{\circ} \mathrm{C}\). Verify Boyle's law for these data. Plot the data so that a straight line is obtained. $$ \begin{array}{llllllll} \hline P / \mathrm{atm} & 0.26 & 0.41 & 0.83 & 1.20 & 2.10 & 2.63 & 3.14 \\ V / \mathrm{L} & 0.938 & 0.595 & 0.294 & 0.203 & 0.116 & 0.093 & 0.078 \\ \hline \end{array} $$

Short Answer

Expert verified
PV products are nearly constant, and a plot of P vs 1/V is a straight line, verifying Boyle's Law.

Step by step solution

01

Recall Boyle's Law

Boyle's Law states that for a given mass of gas at constant temperature, the product of pressure (P) and volume (V) is constant. Mathematically, it is expressed as: \( PV = k \), where \( k \) is a constant.
02

Calculate PV for Each Pair

For each pressure-volume pair given, calculate the product \( PV \) to see if it is approximately the same constant value. This will verify Boyle's Law.
03

Compute PV Products

- For \( P = 0.26 \, \text{atm} \) and \( V = 0.938 \, \text{L} \), \( PV = 0.26 \times 0.938 = 0.24388 \, \text{atm}\cdot\text{L} \).- For \( P = 0.41 \, \text{atm} \) and \( V = 0.595 \, \text{L} \), \( PV = 0.41 \times 0.595 = 0.24395 \, \text{atm}\cdot\text{L} \).- For \( P = 0.83 \, \text{atm} \) and \( V = 0.294 \, \text{L} \), \( PV = 0.83 \times 0.294 = 0.24342 \, \text{atm}\cdot\text{L} \).- For \( P = 1.20 \, \text{atm} \) and \( V = 0.203 \, \text{L} \), \( PV = 1.20 \times 0.203 = 0.2436 \, \text{atm}\cdot\text{L} \).- For \( P = 2.10 \, \text{atm} \) and \( V = 0.116 \, \text{L} \), \( PV = 2.10 \times 0.116 = 0.2436 \, \text{atm}\cdot\text{L} \).- For \( P = 2.63 \, \text{atm} \) and \( V = 0.093 \, \text{L} \), \( PV = 2.63 \times 0.093 = 0.24459 \, \text{atm}\cdot\text{L} \).- For \( P = 3.14 \, \text{atm} \) and \( V = 0.078 \, \text{L} \), \( PV = 3.14 \times 0.078 = 0.24492 \, \text{atm}\cdot\text{L} \).
04

Analyze the PV Products

The calculated \( PV \) products are all approximately equal, suggesting that \( PV \) is indeed a constant, as expected for a relation following Boyle's Law.
05

Prepare Data for Plotting

To verify with a straight line, plot the pressure \( P \) on the x-axis and the reciprocal of volume \( 1/V \) on the y-axis. If Boyle's Law holds, \( P = k/V \), which means a plot of \( P \) vs. \( 1/V \) should be a straight line.
06

Compute 1/V Values

Calculate the reciprocal of volume \( 1/V \) for each given volume:- \( V = 0.938 \), \( 1/V = 1.066 \, \text{L}^{-1} \).- \( V = 0.595 \), \( 1/V = 1.681 \, \text{L}^{-1} \).- \( V = 0.294 \), \( 1/V = 3.401 \, \text{L}^{-1} \).- \( V = 0.203 \), \( 1/V = 4.926 \, \text{L}^{-1} \).- \( V = 0.116 \), \( 1/V = 8.621 \, \text{L}^{-1} \).- \( V = 0.093 \), \( 1/V = 10.753 \, \text{L}^{-1} \).- \( V = 0.078 \), \( 1/V = 12.821 \, \text{L}^{-1} \).
07

Plot P vs 1/V

Create a plot with pressure \( P \) on the x-axis and \( 1/V \) on the y-axis. The points should form a straight line, confirming the inverse relationship between \( P \) and \( V \), which is consistent with Boyle's Law.
08

Interpret the Plot

Examine the plotted line. A straight line through the data points confirms that the relation between pressure and inverse volume is linear, verifying Boyle's Law for the given data.

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Key Concepts

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

Pressure-Volume Relationship
Boyle's Law explores the fascinating relationship between pressure and volume in a gas. When you hold the temperature constant, the pressure and volume of a gas become inversely proportional. This means that as the pressure increases, the volume decreases, and vice versa. Hence the product of pressure (P) and volume (V) remains constant:
  • For the equation, we use: \( PV = k \), where \( k \) represents a constant value.
  • This equation tells us that if you double the pressure of a gas, its volume is halved, provided the temperature doesn't change.
So when you measure these variables for a gas under constant temperature, as we did in our analysis, you should see that the multiplication of pressure and volume gives nearly the same result, verifying this beautiful principle of physics.
Ideal Gas Law
The ideal gas law interconnects various properties of gases, bringing together pressure, volume, and temperature into one elegant equation. It is expressed as:
  • \( PV = nRT \), where \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) represents temperature in Kelvin.
  • This formula helps in understanding how a change in one of the gas parameters affects the others, while considering temperature changes.
Boyle's Law is, in fact, a part of the broader ideal gas law, isolating the pressure and volume relationship while keeping temperature constant. This interconnected nature makes the ideal gas law exceedingly useful for scientists and engineers working with gas calculations in different conditions.
Graphical Analysis of Gas Laws
Graphical analysis offers a powerful visual tool for understanding gas behaviors. By plotting pressure \( P \) against the reciprocal of volume \( 1/V \), we transform the data into a clear straight line. Here's how:
  • In our exercise, the transformation \( P = k/V \) illustrates that you can plot a linear graph using these variables, confirming the inverse relationship indicated by Boyle's Law.
  • This method gives an immediate visual cue on the direct or inverse relationships within the gas laws. Observing these trends aids in reinforcing conceptual understanding.
  • Straight-line graphs simplify the determination of constants like \( k \), enhancing ease of real-world gas behavior modeling.
By utilizing plots, we not only verify theoretical laws like Boyle's Law but also engage in a meaningful analysis of how gases behave across varying conditions.

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Most popular questions from this chapter

Calculate the number of grams of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(g)\), in a \(50.0\) -liter container at a pressure of \(7.50\) atm and a temperature of \(25^{\circ} \mathrm{C}\).

A gaseous mixture consisting of \(0.513\) grams of \(\mathrm{H}_{2}(g)\) and \(16.1\) grams of \(\mathrm{N}_{2}(g)\) occupies \(10.0\) liters at \(20.0^{\circ} \mathrm{C}\). Calculate the partial pressures of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) in the mixture in units of atmospheres.

Convert the following temperatures to the Kelvin scale: (a) \(-183^{\circ} \mathrm{C}\) (the melting point of oxygen) (b) \(6000^{\circ} \mathrm{C}\) (temperature at the surface of the sun) (c) \(-269^{\circ} \mathrm{C}\) (the boiling point of helium) (d) \(800^{\circ} \mathrm{C}\) (the melting point of sodium chloride)

Hydrogen and oxygen react violently with each other once the reaction is initiated. For example, a spark can set off the reaction and cause the mixture to explode. What volume of oxygen gas will react with \(0.55\) liters of hydrogen gas if both are at \(300^{\circ} \mathrm{C}\) and one bar? What volume of water vapor will be produced at \(300^{\circ} \mathrm{C}\) and one bar?

Nitrogen is often injected into cans and bottles to keep the contents from oxidizing. If the nitrogen injected into a particular bottle occupies a volume of \(10.0\) milliliters at a pressure of \(1.2\) bar, what volume will it occupy at a pressure of \(0.95\) bar at the same temperature?
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