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A fixed quantity of gas at \(21^{\circ} \mathrm{C}\) exhibits a pressure of 752 torr and occupies a volume of \(4.38 \mathrm{~L}\). (a) Use Boyle's law to calculate the volume the gas will occupy if the pressure is increased to \(1.88\) atm while the temperature is held constant. (b) Use Charles's law to calculate the volume the gas will occupy if the temperature is increased to \(175^{\circ} \mathrm{C}\) while the pressure is held constant.

Short Answer

Expert verified
In conclusion, using Boyle's Law, if the pressure is increased to 1.88 atm while the temperature is held constant, the gas will occupy a volume of approximately 2.29 L. Using Charles's Law, if the temperature is increased to 175°C while the pressure is held constant, the gas will occupy a volume of approximately 6.60 L.

Step by step solution

01

Part (a)

To solve part (a) using Boyle's Law, we need to first convert the given pressure from torr to atm. We can use the conversion factor: 1 atm = 760 torr. Perform the conversion: \(P_1 = 752 \textrm{ torr} \times \frac{1\textrm{ atm}}{760 \textrm{ torr}} \approx 0.99 \textrm{ atm}\) Now, we can use Boyle's Law to find the final volume \(V_2\): \(P_1V_1 = P_2V_2\) Plug in the known values: \((0.99 \textrm{ atm})(4.38 \textrm{ L}) = (1.88 \textrm{ atm})V_2\) Now, we can solve for the final volume \(V_2\): \(V_2 = \frac{(0.99 \textrm{ atm})(4.38 \textrm{ L})}{(1.88 \textrm{ atm})} \approx 2.29 \textrm{ L}\)
02

Part (b)

To solve part (b) using Charles's Law, we first need to convert the given temperatures from Celsius to Kelvin. Remember that to convert Celsius to Kelvin, you simply add 273.15 to the Celsius temperature: \(T_1 = 21^\circ \mathrm{C} + 273.15 = 294.15\mathrm{K}\) \(T_2 = 175^\circ \mathrm{C} + 273.15 = 448.15\mathrm{K}\) Now, we can use Charles's Law to find the final volume \(V_2\): \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) Plug in the known values: \(\frac{4.38 \textrm{ L}}{294.15 \textrm{ K}} = \frac{V_2}{448.15 \textrm{ K}}\) Now, we can solve for the final volume \(V_2\): \(V_2 = \frac{4.38 \textrm{ L} \times 448.15 \textrm{ K}}{294.15 \textrm{ K}} \approx 6.60 \textrm{ L}\) In conclusion, using Boyle's Law, if the pressure is increased to 1.88 atm while the temperature is held constant, the gas will occupy a volume of approximately 2.29 L. Using Charles's Law, if the temperature is increased to 175°C while the pressure is held constant, the gas will occupy a volume of approximately 6.60 L.

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Key Concepts

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

Boyle's Law
Boyle's Law is one of the fundamental principles that describe how gases behave. It dictates the relationship between pressure and volume in a closed system, where the temperature is kept constant. According to Boyle's Law, the pressure of a fixed quantity of gas is inversely proportional to its volume. This means:
  • When you increase the pressure exerted on a gas, its volume decreases.
  • Conversely, if the pressure decreases, the volume increases.
The mathematical expression for Boyle's Law is given by:\( P_1 V_1 = P_2 V_2 \)where \( P_1 \) and \( P_2 \) are the initial and final pressures, and \( V_1 \) and \( V_2 \) are the initial and final volumes, respectively.
In practical terms, this law is used to predict how a change in pressure affects the volume of a gas. For example, when a balloon is compressed, the pressure inside it goes up, causing the balloon to shrink according to Boyle's Law.
Charles's Law
Charles's Law describes the relationship between the temperature and volume of a gas, keeping the pressure constant. This law states that the volume of a gas is directly proportional to its temperature when measured in Kelvin. Simply put:
  • If you heat a gas, it expands and takes up more volume.
  • If you cool a gas, it contracts, decreasing its volume.
Charles's Law can be expressed as:\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)where \( V_1 \) and \( V_2 \) are the initial and final volumes, and \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.
This concept is crucial in understanding how gases react to temperature changes. For example, a hot air balloon rises because the air inside is heated, which increases the volume and reduces the density compared to the cooler air outside.
Pressure-Volume Relationship
The pressure-volume relationship is a fundamental concept in gas laws that can be better understood through Boyle’s Law. This relationship indicates that the volume of a gas is inversely related to its pressure when temperature remains constant. So, as pressure builds up, the volume diminishes and vice versa.
This relationship is a key aspect of many natural and man-made systems. For instance:
  • In our lungs, when the diaphragm contracts and the chest cavity expands, the pressure inside the lungs decreases, allowing air to fill the space.
  • In a syringe, when the plunger is pulled back, the pressure inside decreases, causing fluid to flow in.
The pressure-volume relationship is a simple yet powerful tool for predicting how changes in pressure can affect the volume of gases, which is instrumental in fields ranging from medicine to engineering and meteorology.
Temperature-Volume Relationship
The temperature-volume relationship described by Charles's Law is essential to understanding how gases behave when heated or cooled. In a constant pressure setting, the volume of a gas changes directly with its temperature. This means:
  • Increasing the temperature results in an increase in volume.
  • Decreasing the temperature leads to a decrease in volume.
This relationship is expressed mathematically through:\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)where temperature is always considered in Kelvin to ensure proportionality.
This fundamental concept explains why heating a gas causes it to inflate. It's applicable in everyday situations, such as why car tire pressure varies with temperature and why balloons shrink when exposed to cold environments. Understanding this relationship helps us predict and control how gases will respond to temperature changes in various applications.

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Most popular questions from this chapter

An open-end manometer containing mercury is connected to a container of gas, as depicted in Sample Exercise \(10.2 .\) What is the pressure of the enclosed gas in torr in each of the following situations? (a) The mercury in the arm attached to the gas is \(15.4 \mathrm{~mm}\) higher than in the one open to the atmosphere; atmospheric pressure is \(0.966\) atm. (b) The mercury in the arm attached to the gas is \(8.7 \mathrm{~mm}\) lower than in the one open to the atmosphere; atmospheric pressure is \(0.99\) atm.

Arsenic(III) sulfide sublimes readily, even below its melting point of \(320^{\circ} \mathrm{C}\). The molecules of the vapor phase are found to effuse through a tiny hole at \(0.28\) times the rate of effusion of Ar atoms under the same conditions of temperature and pressure. What is the molecular formula of arsenic(III) sulfide in the gas phase?

Suppose you havetwo 1-L flasks, one containing \(\mathrm{N}_{2}\) at STP, the other containing \(\mathrm{CH}_{4}\) at STP. How do these systems compare with respect to (a) number of molecules, (b) density, (c) average kinetic energy of the molecules, (d) rate of effusion through a pinhole leak?

Complete the following table for an ideal gas: $$ \begin{array}{llll} \hline P & V & n & T \\ \hline 2.00 \mathrm{~atm} & 1.00 \mathrm{~L} & 0.500 \mathrm{~mol} & ? \mathrm{~K} \\ 0.300 \mathrm{~atm} & 0.250 \mathrm{~L} & ? \mathrm{~mol} & 27^{\circ} \mathrm{C} \\ 650 \text { torr } & ? \mathrm{~L} & 0.333 \mathrm{~mol} & 350 \mathrm{~K} \\ ? \mathrm{~atm} & 585 \mathrm{~mL} & 0.250 \mathrm{~mol} & 295 \mathrm{~K} \\ \hline \end{array} $$

Calculate the pressure that \(\mathrm{CCl}_{4}\) will exert at \(40^{\circ} \mathrm{C}\) if \(1.00\) mol occupies \(28.0 \mathrm{~L}\), assuming that (a) \(\mathrm{CCl}_{4}\) obeys the ideal-gas equation; (b) \(\mathrm{CCl}_{4}\) obeys the van der Waals equation. (Values for the van der Waals constants are given in Table 10.3.) (c) Which would you expect to deviate more from ideal behavior under these conditions, \(\mathrm{Cl}_{2}\) or \(\mathrm{CCl}_{4} ?\) Explain.

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