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Starting with Charles's law (stated as an equation), obtain an equation for the final volume of a gas from its initial volume when the temperature is changed at constant pressure.

Short Answer

Expert verified
The final volume equation is \( V_2 = V_1 \times (T_2/T_1) \).

Step by step solution

01

Understanding Charles's Law

Charles's Law states that the volume of a gas is directly proportional to its temperature when the pressure is kept constant. The mathematical form of Charles's Law is \( V_1/T_1 = V_2/T_2 \), where \( V_1 \) and \( T_1 \) are the initial volume and temperature, and \( V_2 \) and \( T_2 \) are the final volume and temperature.
02

Rearranging the Equation

We need to solve for the final volume \( V_2 \). Starting from the equation \( V_1/T_1 = V_2/T_2 \), multiply both sides by \( T_2 \) to isolate \( V_2 \). This gives us \( V_2 = V_1 imes (T_2/T_1) \).
03

Substituting Values

The equation \( V_2 = V_1 imes (T_2/T_1) \) allows us to calculate the final volume if we know the initial volume \( V_1 \) and both temperatures \( T_1 \) and \( T_2 \). Ensure the temperatures are in Kelvins for accuracy.

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Key Concepts

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

Charles's Law
Charles's Law is a fundamental principle in thermodynamics that describes how gases behave under changing temperatures. It tells us that the volume of a gas is directly proportional to its temperature, as long as the pressure remains constant. This means that if you increase the temperature of a gas, its volume will increase accordingly, and if you decrease the temperature, the volume will decrease. We express Charles's Law mathematically as: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]Here, \(V_1\) and \(T_1\) stand for the initial volume and temperature, respectively, while \(V_2\) and \(T_2\) represent the final volume and temperature. Remember, the temperatures in this equation must be in Kelvin, since the Kelvin scale starts from absolute zero and ensures proportional relationships are linear. This law is a key part of understanding gas law calculations in physics and chemistry.
Volume and Temperature Relationship
Understanding the relationship between volume and temperature is crucial to applying Charles's Law. When discussing this relationship, we specifically talk about how one changes when the other changes, without altering the pressure. Key points include:
  • When the temperature of a gas increases, the particles in the gas move more quickly.
  • This increase in particle movement causes the gas to expand, increasing its volume.
  • Conversely, reducing the temperature slows down the particles, resulting in a decrease in volume.
Think of it like a balloon. When heated, the air inside expands, causing the balloon to grow larger. Cooling the same balloon will make it shrink, as the air inside contracts. This dynamic highlights the direct and proportional relationship dictated by Charles's Law.
Constant Pressure Process
In a constant pressure process, the pressure of the gas does not change even as its volume and temperature do. This is an essential condition for applying Charles's Law, making it possible to relate the volume and temperature of a gas directly. Understanding this condition involves several critical points:
  • The container holding the gas allows for volume changes but keeps pressure stable. This might be a flexible vessel, like a balloon.
  • Under constant pressure, any change in temperature will lead to a direct change in volume, as per Charles's Law formula.
  • Real-world applications often adjust variables to maintain this condition, enabling predictable gas behavior.
Maintaining constant pressure is key to ensuring that only temperature impacts the gas's volume. This scenario underpins many applications in science and engineering, such as weather balloons and hot air balloons, where understanding and controlling gas behavior is crucial.

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Most popular questions from this chapter

A rigid 1.0 - \(\mathrm{L}\) container at \(75^{\circ} \mathrm{C}\) is fitted with a gas pressure gauge. A 1.0 -mol sample of ideal gas is introduced into the container. What would the pressure gauge in the container be reading in \(\mathrm{mmHg}\) ? Describe the interactions in the container that are causing the pressure. c Say the temperature in the container were increased to \(150^{\circ} \mathrm{C}\). Describe the effect this would have on the pressure, and, in terms of kinetic theory, explain why this change occurred.

Calculate the rms speed of \(\mathrm{Br}_{2}\) molecules at \(23^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\) What is the rms speed of \(\mathrm{Br}_{2}\) at \(23^{\circ} \mathrm{C}\) and \(2.00 \mathrm{~atm} ?\)

A sample of carbon dioxide gas is placed in a container. The volume of the container is reduced to \(1 / 3\) of its original volume while the pressure is observed to double. In this system did the temperature change? Explain your answer.

A McLeod gauge measures low gas pressures by compressing a known volume of the gas at constant temperature. If \(315 \mathrm{~cm}^{3}\) of gas is compressed to a volume of 0.0457 \(\mathrm{cm}^{3}\) under a pressure of \(2.51 \mathrm{kPa}\), what was the original gas pressure?

Ethanol, the alcohol used in automobile fuels, is produced by the fermentation of sugars present in plants. Corn is often used as the sugar source. The following equation represents the fermentation of glucose, the sugar in corn, by yeast to produce ethanol and carbon dioxide. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ Ethanol is combusted in an automobile engine according to the equation $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) $$ What would be the total volume of \(\mathrm{CO}_{2}\) gas formed at STP when \(3.00 \mathrm{~kg}\) of sugar is fermented and the ethanol is then combusted in an automobile engine?

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