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Chapter 10: Problem 25

You have a gas at \(25^{\circ} \mathrm{C}\) confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? \((\mathbf{a})\) Lifting up on the piston to double the volume while keeping the temperature constant; \((\mathbf{b})\) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) , while keeping the volume constant; \((\mathbf{c})\) Pushing down on the piston to halve the volume while keeping the temperature constant.

Short Answer

Expert verified
The correct action that doubles the gas pressure is: \(\boxed{\mathrm{(c)}}\) Pushing down on the piston to halve the volume while keeping the temperature constant.

Step by step solution

01

a) Lifting up the piston to double the volume while keeping the temperature constant

In this action, since temperature (T) remains constant and the number of moles (n) and the gas constant (R) are unchanged, the Ideal Gas Law becomes: \(PV = constant\) Now, let's double the volume (V): \(P(2V) = constant\) \(2PV = constant\) Comparing the initial and final states: \(2PV = PV\) This shows that the pressure (P) remains constant in this action. Hence, lifting up the piston to double the volume does not double the pressure.
02

b) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\), while keeping the volume constant

In this action, the volume remains constant, so we rewrite the Ideal Gas Law as: \(P \propto T\) We need to convert the temperatures from Celsius to Kelvin: Initial temperature: \(T_i = 25^{\circ}C + 273.15 = 298.15K\) Final temperature: \(T_f = 50^{\circ}C + 273.15 = 323.15K\) Now, let's find the ratio of the final pressure to the initial pressure: \(\frac{P_f}{P_i} = \frac{T_f}{T_i}\) \(\frac{P_f}{P_i} = \frac{323.15K}{298.15K} = 1.084\) This shows that the pressure increases by a factor of 1.084, which is not doubling the pressure. Hence, heating the gas with this temperature change does not double the pressure.
03

c) Pushing down on the piston to halve the volume while keeping the temperature constant

In this action, the temperature remains constant, so the Ideal Gas Law becomes: \(PV = constant\) Now, let's halve the volume (V): \(P(\frac{1}{2}V) = constant\) \(\frac{P}{2}V = constant\) Comparing the initial and final states: \(\frac{P}{2}V = PV\) This shows that the pressure (P) doubles in this action. Therefore, pushing down the piston to halve the volume while keeping the temperature constant doubles the pressure. In conclusion, the correct action that doubles the gas pressure is: \(\boxed{\mathrm{(c)}}\) Pushing down on the piston to halve the volume while keeping the temperature constant.

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Key Concepts

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

Gas Pressure
Gas pressure is the amount of force that the gas exerts on the walls of its container. It occurs because gas molecules move in continuous, rapid motion, colliding with each other and the container walls. When you increase the number of collisions, you increase the pressure.
In our context, controlling pressure is essential to understanding how gases behave in different conditions.
  • Doubling the volume, with temperature constant, keeps the pressure the same.
  • Halving the volume of the gas doubles the pressure, this is because the same number of molecules are confined in a smaller space, leading to more frequent collisions.
  • Increasing temperature generally increases gas pressure, but not necessarily doubling it, unless specific conditions are met.
Temperature Conversion
Converting temperature from Celsius to Kelvin is a basic yet crucial step when working with gas laws. Kelvin is the standard unit of temperature in gas law calculations, because it is an absolute temperature scale.
To convert Celsius to Kelvin:
  • Add 273.15 to the Celsius temperature.
  • This removes negative values, which makes the math manageable, as Kelvin starts from absolute zero.
Let's see this applied:
If the gas is at \(25^{\circ}\text{C}\), converted to Kelvin it is \(298.15 \text{K}\). If you heat it to \(50^{\circ}\text{C}\), that becomes \(323.15 \text{K}\).
Kelvin simplifies the relationship between temperature and pressure, as doubling temperature in Kelvin doesn't necessarily mean doubling the pressure.
Volume Change
Volume change, especially in gases, refers to the contraction or expansion of the space that the gas occupies. This concept is directly related to gas pressure and temperature. When you increase the volume of gas at constant pressure, each gas molecule has more space to move around.
Consider these effects:
  • Increasing volume with constant temperature leads to lower pressure.
  • Decreasing volume with same temperature results in more pressure, as molecules collide more often.
In the exercise, when the volume is doubled while keeping temperature constant, pressure remains unchanged. On the other hand, halving the volume while maintaining the same temperature leads to a doubling of the pressure. It's essential to understand that volume changes do not affect temperature, but they do affect pressure when other variables remain constant.
Pressure-Volume Relationship
The pressure-volume relationship is central to understanding gas behavior under different conditions, described by Boyle's Law. This law-specific case of the Ideal Gas Law gives insight into how pressure and volume interact when temperature is constant.
Here are key points:
  • The relationship is inversely proportional, meaning \(P \times V = \,\text{constant}\).
  • If pressure increases, volume must decrease if temperature is constant, and vice versa.
From the exercise, when the piston is used to halve the volume, the doubled pressure aligns with Boyle's Law, showing how decreased volume forces the gas molecules closer together, increasing the frequency of collisions and thus the pressure. Understanding this relationship helps explain real-world applications, from engines to weather systems.

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Most popular questions from this chapter

Calculate the pressure that \(\mathrm{CCl}_{4}\) will exert at \(80^{\circ} \mathrm{C}\) if 1.00 mol occupies \(33.3 \mathrm{L},\) 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.

(a) Place the following gases in order of increasing average molecular speed at \(300 \mathrm{K} : \mathrm{CO}, \mathrm{SF}_{6}, \mathrm{H}_{2} \mathrm{S}, \mathrm{Cl}_{2}, \mathrm{HBr}\) . (b) Calculate the rms speeds of CO and \(\mathrm{Cl}_{2}\) molecules at 300 \(\mathrm{K}\) . (c) Calculate the most probable speeds of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) molecules at 300 \(\mathrm{K}\) .

A quantity of \(\mathrm{N}_{2}\) gas originally held at 5.25 atm pressure in a 1.00 -L container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5-\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\) . A quantity of \(\mathrm{O}_{2}\) gas originally at 5.25 atm and \(26^{\circ} \mathrm{C}\) in a \(5.00-\mathrm{L}\) container is transferred to this same container. What is the total pressure in the new container?

To minimize the rate of evaporation of the tungsten filament, \(1.4 \times 10^{-5}\) mol of argon is placed in a \(600-\mathrm{cm}^{3}\) light-bulb. What is the pressure of argon in the lightbulb at \(23^{\circ} \mathrm{C} ?\)

Suppose you are given two \(1-\) flasks and told that one contains a gas of molar mass 30 , the other a gas of molar mass 60 , both at the same temperature. The pressure in flask \(A\) is \(x\) atm, and the mass of gas in the flask is 1.2 \(\mathrm{g}\) . The pressure in flask \(\mathrm{B}\) is 0.5\(x\) atm, and the mass of gas in that flask is 1.2 \(\mathrm{g}\) . Which flask contains the gas of molar mass \(30,\) and which contains the gas of molar mass 60\(?\)
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