Problem 44
Question
A sample of gas has a volume of \(2.75 \mathrm{~L}\) at a temperature of \(100 \mathrm{~K}\). What is the volume of the gas when the temperature increases to \(200 \mathrm{~K}\) ? Assume constant pressure and moles.
Step-by-Step Solution
Verified Answer
The final volume of the gas is 5.5 L.
1Step 1: Identify the Relationship
In this problem, we use Charles's Law because the pressure and amount of gas are constant. Charles's Law states that for a given amount of gas at constant pressure, the volume is directly proportional to its temperature (in Kelvin), expressed as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).
2Step 2: List the Known Values
We are given the initial volume \( V_1 = 2.75 \) L and initial temperature \( T_1 = 100 \) K. The final temperature \( T_2 \) is 200 K, and we need to find the final volume \( V_2 \).
3Step 3: Apply Charles's Law
Substitute the known values into Charles's Law: \( \frac{2.75}{100} = \frac{V_2}{200} \).
4Step 4: Solve for Final Volume
Cross-multiply to solve for \( V_2 \): \( V_2 = \frac{2.75 \times 200}{100} \).
5Step 5: Calculate the Answer
Perform the calculation: \( V_2 = \frac{550}{100} = 5.5 \) L.
Key Concepts
Gas Volume-Temperature RelationshipConstant Pressure in GasesVolume Calculation Using Charles's Law
Gas Volume-Temperature Relationship
When dealing with gases, understanding the volume-temperature relationship is crucial. As a foundational concept in chemistry, it states that the volume of a gas is directly proportional to its temperature when measured in Kelvin. This means that if you increase the temperature of a gas, its volume expands, provided the pressure remains the same.
Think of how hot air balloons work; as the air inside is heated, the balloon inflates. This is a perfect real-life application of the gas volume-temperature relationship. This relationship is linear, so doubling the temperature will double the volume, and halving the temperature will halve the volume. It's an elegant and straightforward concept when you remember that temperature changes should be measured using Kelvin rather than degrees Celsius or Fahrenheit.
When you're working with gases, keep this principle in mind. It's fundamental to understanding and predicting the behavior of gases in different temperature scenarios.
Think of how hot air balloons work; as the air inside is heated, the balloon inflates. This is a perfect real-life application of the gas volume-temperature relationship. This relationship is linear, so doubling the temperature will double the volume, and halving the temperature will halve the volume. It's an elegant and straightforward concept when you remember that temperature changes should be measured using Kelvin rather than degrees Celsius or Fahrenheit.
When you're working with gases, keep this principle in mind. It's fundamental to understanding and predicting the behavior of gases in different temperature scenarios.
Constant Pressure in Gases
For Charles's Law to be applicable, the pressure of the gas must remain constant. But what does constant pressure mean? In simple terms, it indicates that the force exerted by the gas molecules on the walls of its container doesn't change as the temperature varies.
Why is this important? The pressure in a gas can control how easily its volume changes when heated or cooled. If pressure changes, it will affect the gas's volume independently of what the temperature does. Therefore, by maintaining steady pressure, we isolate temperature as the only variable impacting volume changes.
In practical terms, maintaining constant pressure means using a flexible container, like a balloon, which expands or contracts without altering the internal pressure significantly. Keeping pressure constant is a key condition for applying Charles's Law correctly.
Why is this important? The pressure in a gas can control how easily its volume changes when heated or cooled. If pressure changes, it will affect the gas's volume independently of what the temperature does. Therefore, by maintaining steady pressure, we isolate temperature as the only variable impacting volume changes.
In practical terms, maintaining constant pressure means using a flexible container, like a balloon, which expands or contracts without altering the internal pressure significantly. Keeping pressure constant is a key condition for applying Charles's Law correctly.
Volume Calculation Using Charles's Law
Charles's Law provides a handy mathematical formula to calculate gas volume changes:
With this simple equation, you can determine the new volume of a gas when its temperature changes, as long as you know two initial conditions:
You also need the final temperature \( T_2 \) in Kelvin.
\( V_2 \), the new volume, can be found by rearranging the formula to \( V_2 = \frac{V_1 \times T_2}{T_1} \). Plug in your known values, and you can compute \( V_2 \).
For example, if a gas starts at a volume of 2.75 L at 100 K, increasing the temperature to 200 K under constant pressure would result in a volume of 5.5 L. It's a clear demonstration of how useful Charles's Law is in predicting the behavior of gases when temperatures shift.
- \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
With this simple equation, you can determine the new volume of a gas when its temperature changes, as long as you know two initial conditions:
- The starting volume \( V_1 \)
- The starting temperature \( T_1 \) in Kelvin
You also need the final temperature \( T_2 \) in Kelvin.
\( V_2 \), the new volume, can be found by rearranging the formula to \( V_2 = \frac{V_1 \times T_2}{T_1} \). Plug in your known values, and you can compute \( V_2 \).
For example, if a gas starts at a volume of 2.75 L at 100 K, increasing the temperature to 200 K under constant pressure would result in a volume of 5.5 L. It's a clear demonstration of how useful Charles's Law is in predicting the behavior of gases when temperatures shift.
Other exercises in this chapter
Problem 42
A \(0.50 \mathrm{~L}\) container of helium expands to \(1.50 \mathrm{~L}\). By what factor does the pressure change? Assume constant moles and temperature.
View solution Problem 43
A sample of oxygen gas has an initial pressure and volume of \(1.0 \mathrm{~L}\) and \(1.0\) atm. What is the final pressure if the volume is compressed to \(0.
View solution Problem 45
What is the final volume of a gas that was originally at \(0.75 \mathrm{~L}\) at \(25^{\circ} \mathrm{C}\) and a final temperature of \(50^{\circ} \mathrm{C}\)
View solution Problem 46
A sample of nitrogen is at \(45^{\circ} \mathrm{C}\) with a volume of \(2.5 \mathrm{~L}\). What is the final temperature in \({ }^{\circ} \mathrm{C}\) if the vo
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