1\. A container with a piston contains a sample of gas. Initially, the pressure in the container is exactly 1 atm, but the volume is unknown. The piston is adjusted so that the volume is \(0.155 \mathrm{~L}\) and the pressure is \(956 \mathrm{~mm}\) Hg; what was the initial volume? 2\. The pressure of \(12.5 \mathrm{~L}\) of a gas is \(0.82 \mathrm{~atm}\). If the pressure changes to \(1.32 \mathrm{~atm}\), what will the final volume be? A sample of helium gas has a pressure of \(860.0 \mathrm{~mm}\) Hg. This gas is transferred to a different container having a volume of \(25.0 \mathrm{~L}\); in this new container, the pressure is determined to be \(770.0 \mathrm{~mm}\) Hg. What was the initial volume of the gas?

Short Answer

Expert verified
The initial volumes for the gases are: 1. Exercise: 0.195 L, 2. Exercise: 7.765 L, 3. Exercise: 22.368 L.

Step by step solution

01

- Converting pressures to the same units

Convert the pressures to the same unit to make calculations consistent. Since the given pressures are in atm and mm Hg, we can use the conversion 1 atm = 760 mm Hg. Therefore, the final pressure in the first case is \(956 \text{mm Hg} \times \frac{1 \text{atm}}{760 \text{mm Hg}} = 1.2589 \text{atm}\). For the helium gas, the initial pressure is \(860.0 \text{mm Hg} \times \frac{1 \text{atm}}{760 \text{mm Hg}} = 1.1316 \text{atm}\) and the final pressure is \(770.0 \text{mm Hg} \times \frac{1 \text{atm}}{760 \text{mm Hg}} = 1.0132 \text{atm}\).
02

- Applying Boyle's Law for the first scenario

Use Boyle's Law \(P_1V_1 = P_2V_2\) to calculate the initial volume of the gas. Here \(P_1 = 1 \text{atm}\), \(P_2 = 1.2589 \text{atm}\), and \(V_2 = 0.155 \text{L}\). Solving for \(V_1\), we get \(V_1 = \frac{P_2V_2}{P_1} = \frac{1.2589 \text{atm} \times 0.155 \text{L}}{1 \text{atm}} = 0.195 \text{L}\).
03

- Applying Boyle's Law for the second scenario

Again using Boyle's Law \(P_1V_1 = P_2V_2\), but now for the second scenario where \(P_1 = 0.82 \text{atm}\), \(P_2 = 1.32 \text{atm}\), and \(V_1 = 12.5 \text{L}\). We solve for \(V_2\), giving us \(V_2 = \frac{P_1V_1}{P_2} = \frac{0.82 \text{atm} \times 12.5 \text{L}}{1.32 \text{atm}} = 7.765 \text{L}\).
04

- Applying Boyle's Law for the helium gas

Use Boyle's Law one more time for the helium gas situation with the converted pressures. With \(P_1 = 1.1316 \text{atm}\), \(V_2 = 25.0 \text{L}\), and \(P_2 = 1.0132 \text{atm}\), we find \(V_1\) as \(V_1 = \frac{P_2V_2}{P_1} = \frac{1.0132 \text{atm} \times 25.0 \text{L}}{1.1316 \text{atm}} = 22.368 \text{L}\).

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Key Concepts

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

Gas Laws
To better understand the behavior of gases under various conditions, scientists have formulated principles known as gas laws. These include Boyle's Law, Charles's Law, Gay-Lussac's Law, and the Ideal Gas Law, among others.

Boyle's Law, in particular, focuses on the pressure-volume relationship of a gas held at a constant temperature. The law states that the pressure of a fixed amount of gas is inversely proportional to its volume when temperature is held constant. The mathematical representation of Boyle's Law is often written 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 final pressure and volume, respectively.

Understanding Boyle's Law is crucial because it applies to several real-world situations, such as breathing, where the lungs expand and contract, altering the volume and consequently the pressure of the air inside.
Pressure Volume Relationship
The pressure volume relationship highlighted in Boyle's Law is a core concept in understanding how gases behave under pressure changes. When the volume of a gas is decreased, its particles are squeezed closer together, causing more frequent collisions and thus increasing the pressure. Conversely, when the volume is increased, the particles are further apart, and the pressure decreases, as the collisions become less frequent.

Practical Application

Take a syringe for example: when you pull the plunger back, the volume inside increases, leading to a decrease in pressure which draws liquid into the syringe. Pushing the plunger in the opposite direction decreases the volume and the liquid is expelled due to increased pressure. These concepts are directly applicable to many scientific fields, including respiratory physiology, engineering, and more.

To solve problems involving this relationship, it's essential to understand both the direct and inverse proportionality, and ensure consistent units of measurement for accurate calculations. For instance, when converting from mmHg to atm or vice versa, using a conversion factor is necessary to maintain the integrity of the calculations.
Chemistry Problems Solving
Solving chemistry problems, especially those related to gas laws, requires a methodical approach. It's important to follow a series of steps to ensure that the problems are approached in a consistent and effective manner. Here's a simplified method often used by chemists and students alike:
  • Identify which gas law applies to the problem at hand.
  • Write down the known variables and what you are solving for.
  • Convert units where necessary to maintain consistency throughout the problem.
  • Rearrange the gas law formula to solve for the unknown variable.
  • Perform the calculations carefully and check that the units cancel out as expected.

By breaking down the problem, converting units (such as atm to mmHg), rearranging equations to isolate the unknown variable, and methodically plugging in the given information, students can competently tackle and solve complex problems. Even the most challenging chemistry exercises become more approachable when each step is followed with care and precision, as illustrated in the given Boyle's Law problems.

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Most popular questions from this chapter

1\. An automobile air bag requires about \(62 \mathrm{~L}\) of nitrogen gas in order to inflate. The nitrogen gas is produced by the decomposition of sodium azide, according to the equation shown below $$ 2 \mathrm{NaN}_{3}(s) \rightarrow 2 \mathrm{Na}(s)+3 \mathrm{~N} 2(g) $$ What mass of sodium azide is necessary to produce the required volume of nitrogen at \(25^{\circ} \mathrm{C}\) and 1 atm? 2\. When \(\mathrm{Fe} 2 \mathrm{O}_{3}\) is heated in the presence of carbon, CO 2 gas is produced, according to the equation shown below. A sample of \(96.9\) grams of \(\mathrm{Fe} 2 \mathrm{O}_{3}\) is heated in the presence of excess carbon and the CO2 produced is collected and measured at 1 atm and \(453 \mathrm{~K}\). What volume of \(\mathrm{CO}_{2}\) will be observed? $$ 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{C}(\mathrm{s}) \rightarrow 4 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{~g}) $$ 3\. The reaction of zinc and hydrochloric acid generates hydrogen gas, according to the equation shown below. An unknown quantity of zinc in a sample is observed to produce \(7.50 \mathrm{~L}\) of hydrogen gas at a temperature of \(404 \mathrm{~K}\) and a pressure of \(1.75 \mathrm{~atm} .\) How many moles of zinc were in the sample? $$ \mathrm{Zn}(\mathrm{s})+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl} 2(a q)+\mathrm{H} 2(g) $$

1\. A sample of methane has a volume of \(17.5 \mathrm{~L}\) at \(100.0^{\circ} \mathrm{C}\) and \(1.72 \mathrm{~atm}\). How many moles of methane are in the sample? 2\. A \(0.0500 \mathrm{~L}\) sample of a gas has a pressure of \(745 \mathrm{~mm} \mathrm{Hg}\) at \(26.4^{\circ} \mathrm{C}\). The temperatureis now raised to \(404.4 \mathrm{~K}\) and the volume is allowed to expand until a final pressure of \(1.06\) atm is reached. What is the final volume of the gas? 3\. When \(128.9\) grams of cyclopropane \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right)\) are placed into an \(8.00 \mathrm{~L}\) cylinder at \(298 \mathrm{~K}\), the pressure is observed to be \(1.24\) atm. A piston in the cylinder is now adjusted so that the volume is now \(12.00 \mathrm{~L}\) and the pressure is \(0.88 \mathrm{~atm} .\) What is the final temperature of the gas?

The value of the proportionality constant \(R\), can be calculated from the fact that exactly one mole of a gas at exactly 1 atm and at 0 "C \((273 \mathrm{~K})\) has a volume of \(22.414 \mathrm{~L}\). Solution Substituting in the equation: $$ \begin{array}{c} P V=n R T \text { or } R=\frac{P V}{n T} \\ R=\frac{(1 a t m)(22.414 L)}{(1 m o l e)(273 K)}=0.082057 L \text { atm } m o l^{-1} K^{-1} \end{array} $$

A sample of oxygen occupies \(17.5 \mathrm{~L}\) at \(0.75 \mathrm{~atm}\) and \(298 \mathrm{~K}\). The temperature is raised to \(303 \mathrm{~K}\) and the pressure is increased to \(0.987\) atm. What is the final volume of the sample?

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