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A 3.00-L tank contains air at 3.00 atm and 20.0\(^\circ\)C. The tank is sealed and cooled until the pressure is 1.00 atm. (a) What is the temperature then in degrees Celsius? Assume that the volume of the tank is constant. (b) If the temperature is kept at the value found in part (a) and the gas is compressed, what is the volume when the pressure again becomes 3.00 atm?

Short Answer

Expert verified
(a) \(-175.43^\circ C\); (b) Volume is 1.00 L.

Step by step solution

01

Identify Known Values and Equations

We know that the initial pressure \( P_1 = 3.00 \) atm, initial temperature \( T_1 = 20.0^\circ C \) which is 293.15 K, and final pressure \( P_2 = 1.00 \) atm. We are given that volume \( V \) remains constant, so we can use the ideal gas law relation \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \).
02

Solve for Final Temperature

Using the relation \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), rearrange to solve for \( T_2 \): \( T_2 = \frac{P_2 \times T_1}{P_1} \). Substitute the values: \( T_2 = \frac{1.00 \times 293.15}{3.00} = 97.72 \) K. Convert back to Celsius: \( T_2 = 97.72 - 273.15 = -175.43^\circ C \).
03

Interpret Temperature in Celsius

From the calculation, the final temperature when the pressure is cooled to 1.00 atm is \( -175.43^\circ C \).
04

Apply Ideal Gas Law for Compression

With the gas at \( T = -175.43^\circ C \) or 97.72 K, we now compress it back to the original pressure of 3.00 atm. Utilizing \( PV = nRT \), for constant \( n \) and \( R \), \( \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \). Given that \( P_1 = 1.00 \) atm, \( V_1 = 3.00 \) L, and \( P_2 = 3.00 \) atm, solve for \( V_2 \): \( V_2 = V_1 \times \frac{P_1}{P_2} = 3.00 \times \frac{1.00}{3.00} = 1.00 \) L.

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Key Concepts

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

Gas laws
Gas laws describe the behavior of gases in various conditions by explaining the relationships between pressure, volume, and temperature. One of the most important gas laws is the Ideal Gas Law, which combines several simpler laws to provide insights into how gases behave. The Ideal Gas Law is expressed with the equation \( PV = nRT \). This equation outlines the relationship where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
The Ideal Gas Law helps predict how one property of a gas will change when another is adjusted, assuming all else is constant. It is particularly useful in chemical and physical processes when estimating gas behavior under different conditions. For example:
  • If temperature increases, pressure also increases if the volume stays constant.
  • Similarly, if you compress a gas (decrease its volume), its pressure goes up, assuming constant temperature.
  • Reducing the temperature often leads to a decrease in pressure if the volume remains constant.
The Ideal Gas Law works best for ideal gases, which are theoretical gases composed of many randomly moving particles in straight paths without interactions.
Pressure-Temperature relationship
The Pressure-Temperature relationship in gases is described by Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its absolute temperature when volume is constant. Mathematically, this relation can be written as \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), where \( P \) is pressure and \( T \) is temperature in Kelvin.
In the context of the given exercise, as the gas in the tank was cooled, the pressure decreased because temperature also dropped, keeping the volume fixed. The relationship shows:
  • As the gas cooled to a lower temperature from 293.15 K to 97.72 K (equivalent to starting at 20.0°C and cooling to -175.43°C), the pressure fell from 3.00 atm to 1.00 atm.
  • This relationship reflects how cooling a gas will lower its pressure when confined to a constant volume, simply due to the reduction in the energy and speed of gas particles.
Gay-Lussac's Law is a practical tool in scenarios such as pressure cookers, where temperature changes at a constant volume directly affect pressure.
Volume-Pressure relationship
The connection between pressure and volume in a gas is delineated by Boyle's Law, which states that pressure is inversely proportional to volume when the temperature is constant. In equation form, this is \( P_1V_1 = P_2V_2 \).
This inverse relationship signifies that increasing the volume of a gas results in lower pressure, assuming there is no temperature change, and vice versa. This concept was pivotal in the exercise when the temperature (97.72 K from -175.43°C) was held constant as the gas was compressed back to its initial pressure.
  • In the exercise scenario, when the pressure was restored to 3.00 atm, the volume decreased from 3.00 L to 1.00 L.
  • This simple relationship helps in understanding phenomena such as how air pressure inflates balloons or how syringes work.
Boyle's Law is vital in any field involving gaseous behavior, ensuring clarity about how gases will react under different volume and pressure adjustments.

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