A sample of helium at a pressure of 745 torr and in a volume of \(2.58 \mathrm{~L}\) was heated from 24.0 to \(75.0^{\circ} \mathrm{C}\). The volume of the container expanded to \(2.81 \mathrm{~L}\). What was the final pressure (in torr) of the helium?

Short Answer

Expert verified
The final pressure of the helium was approximately 658 torr.

Step by step solution

01

Convert Initial Pressure to Atmospheres

Convert the initial pressure from torr to atmospheres (atm) using the conversion factor 1 atm = 760 torr. The initial pressure in atm is given by the formula: \( P_{1 \text{ atm}} = \frac{P_{1 \text{ torr}}}{760 \text{ torr/atm}} \).
02

Convert Temperatures to Kelvin

Convert both the initial and final temperatures from Celsius to Kelvin. The conversion is: \( T(K) = T(\circ C) + 273.15 \).
03

Use the Combined Gas Law

Apply the combined gas law \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) and solve for the final pressure \( P_2 \). Make sure to use the correct units for all quantities.
04

Solve for the Final Pressure in Atmospheres

Rearrange the combined gas law to solve for \( P_2 \): \( P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} \) and insert the values converted to the correct units. Then calculate \( P_2 \) in atm.
05

Convert Final Pressure Back to Torr

Convert the calculated final pressure from atmospheres back to torr using the conversion factor 1 atm = 760 torr: \( P_{2 \text{ torr}} = P_{2 \text{ atm}} \times 760 \text{ torr/atm} \).

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Key Concepts

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

Gas Law Calculations
Understanding gas law calculations is crucial to solving problems related to the behavior of gases under various conditions of pressure, volume, and temperature. According to the combined gas law, the relationship between these variables can be represented with the equation \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\), where \(P\) stands for pressure, \(V\) for volume, and \(T\) for temperature. The subscripts 1 and 2 denote the initial and final states of the gas, respectively. To solve these equations, it's important to use consistent units; typically, pressure is in atmospheres (atm), volume in liters (L), and temperature in Kelvin (K).

To find the final pressure \(P_2\) in the provided example, one would rearrange the combined gas law to isolate \(P_2\) and then plug in the known values, making sure to convert all the measurements to appropriate units before doing the calculations. This type of calculation enables us to predict how a gas will behave when it undergoes changes in conditions, which is fundamental in fields like chemistry and physics.
Pressure Conversions
Temperature and pressure are key factors in the study of gases, and converting them correctly is important for gas law calculations. Pressure, the force exerted by a gas per unit area, can be measured in various units, with atmospheres (atm) and torr being common ones used in the laboratory. Conversion between these two units is straightforward: 1 atm is equal to 760 torr. This relationship is especially useful when dealing with gas laws, as many equations require pressure to be in atmospheres.

When solving problems related to gas behavior, it's essential to first convert the given pressure to the units required by the equation in use. For example, in the exercise, the initial pressure measured in torr was converted to atmospheres before applying the combined gas law. This meticulous approach to unit conversion is necessary to prevent errors in the final calculation.
Temperature Conversions
The combined gas law also involves temperature, which must be in Kelvin for the law to be applicable. The Kelvin scale is an absolute temperature scale, meaning it starts from absolute zero, the lowest possible temperature. This scale is key in scientific calculations because it avoids negative numbers for temperature, which can complicate gas law calculations.

To convert Celsius to Kelvin, which is common in gas law problems, the formula used is \( T(K) = T(\circ C) + 273.15 \). This conversion is paramount because Celsius and Kelvin scales have different starting points (0 °C is not the lowest possible temperature, while 0 K is). In our provided exercise, both the initial and final temperatures given in Celsius are converted to Kelvin before inserting them into the combined gas law formula.
Behaviour of Gases
The behavior of gases under different conditions of temperature, pressure, and volume is predictable using the gas laws. These laws were derived from empirical observations and can be combined into the combined gas law, which is a versatile equation that incorporates Boyle's Law, Charles's Law, and Gay-Lussac's Law. The combined gas law shows that if the amount of gas and the amount of moles remain constant, the pressure, volume, and temperature of a gas are interrelated.

For gases, pressure increases as temperature increases, provided the volume is constant. If the volume can expand, the rise in temperature may not cause an increase in pressure. Moreover, increasing the volume of a container while keeping temperature constant leads to reduced pressure. These fundamental principles about the behavior of gases are essential for understanding how to manipulate and predict the state of a gas in different scenarios, such as in the example exercise where the volume expansion and temperature increase led to calculating the new pressure of helium.

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Most popular questions from this chapter

An open-end mercury manometer was connected to a flask containing a gas at an unknown pressure. The mercury in the arm open to the atmosphere was \(65 \mathrm{~mm}\) higher than the mercury in the arm connected to the flask. The atmospheric pressure was 748 torr. What was the pressure of the gas in the flask (in torr)?

What does a small value for the van der Waals constant \(a\) suggest about the molecules of the gas?

Explain in terms of the kinetic theory how raising the temperature of a confined gas makes its pressure increase.

A bubble of air escaping from a diver's mask rises from a depth of \(100 \mathrm{ft}\) to the surface where the pressure is \(1.00 \mathrm{~atm} .\) Initially, the bubble has a volume of \(10.0 \mathrm{~mL}\). Assuming none of the air dissolves in the water, how many times larger is the bubble just as it reaches the surface? Use your answer to explain why scuba divers constantly exhale as they slowly rise from a deep dive. (The density of seawater is approximately \(1.025 \mathrm{~g} \mathrm{~mL}^{-1}\); the density of mercury is \(13.6 \mathrm{~g} \mathrm{~mL}^{-1}\).)

Early one cool \(\left(60.0^{\circ} \mathrm{F}\right)\) morning you start on a bike ride with the atmospheric pressure at \(14.7 \mathrm{lb}\) in. \(^{-2}\) and the tire gauge pressure at \(50.0 \mathrm{lb}\) in. \(^{-2}\). (Gauge pressure is the amount that the pressure exceeds atmospheric pressure.) By late afternoon, the air had warmed up considerably, and this plus the heat generated by tire friction sent the temperature inside the tire to \(104^{\circ} \mathrm{F}\). What will the tire gauge now read, assuming that the volume of the air in the tire and the atmospheric pressure have not changed?

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