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\- APPLICATION The volume of a gas is \(3.50 \mathrm{~L}\) at \(280 \mathrm{~K}\). The volume of any gas is directly proportional to its temperature on the Kelvin scale \((\mathrm{K})\). a. Find the volume of this gas when the temperature is \(330 \mathrm{~K}\). b. Find the temperature when the volume is \(2.25 \mathrm{~L}\)

Short Answer

Expert verified
a. 4.125 L at 330 K. b. 180 K for volume 2.25 L.

Step by step solution

01

Understand the Relationship

The volume of a gas is proportional to its temperature, meaning \( V \propto T \). This can be expressed as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) when comparing two states of the gas.
02

Find the Constant Ratio for Part (a)

To find the new volume at \( 330 \mathrm{~K} \), use \( \frac{3.50 \mathrm{~L}}{280 \mathrm{~K}} = \frac{V_2}{330 \mathrm{~K}} \). Solve for \( V_2 \):\[ V_2 = \frac{3.50 \times 330}{280} \]
03

Calculate the New Volume

Calculate \( V_2 \):\[ V_2 = \frac{3.50 \times 330}{280} = 4.125 \mathrm{~L} \]This is the new volume when the temperature is \( 330 \mathrm{~K} \).
04

Set Up the Proportion for Part (b)

To find the temperature when the volume is \( 2.25 \mathrm{~L} \), use \( \frac{3.50 \mathrm{~L}}{280 \mathrm{~K}} = \frac{2.25 \mathrm{~L}}{T_2} \). Solve for \( T_2 \):\[ T_2 = \frac{2.25 \times 280}{3.50} \]
05

Calculate the New Temperature

Calculate \( T_2 \):\[ T_2 = \frac{2.25 \times 280}{3.50} \approx 180 \mathrm{~K} \]This is the temperature when the volume is \( 2.25 \mathrm{~L} \).

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Key Concepts

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

Volume
In the realm of gas laws, volume is a critical concept. It refers to the amount of space that a gas occupies. When discussing gas, we typically measure the volume in liters or milliliters.

As seen in the exercise, if we have a gas with an initial volume of \(3.50\,\mathrm{L}\), this is simply referring to how much space the gas fills at a specified state. The relationship between volume and temperature underlines how gases behave physically.

As temperature changes, the volume of a gas can expand or contract. Understanding this notion is key to mastering problems involving gas laws, as shifts in volume are often tied to changes in other conditions, such as temperature or pressure.
Temperature
Temperature plays a pivotal role in determining the behavior of gases. It essentially measures how much thermal energy is available in a substance. In the context of gases, as temperature increases, particles move faster and tend to occupy more space, thus increasing the volume.

In the given exercise, temperatures were provided in Kelvin, which is the standard unit in scientific contexts to prevent negative values that Fahrenheit or Celsius scales might introduce. For example, an initial temperature of \(280\,\mathrm{K}\) was given for a gas volume of \(3.50\,\mathrm{L}\).

Thus, when we computed the new volume at \(330\,\mathrm{K}\), or determined the temperature at \(2.25\,\mathrm{L}\), we fully embraced this dynamic to reveal how temperature governs the expansiveness of gas.
Proportionality
The concept of proportionality is deeply embedded in solving gas law problems. For ideal gases, volume is directly proportional to temperature, meaning if one increases, the other does too, assuming pressure is constant.

The mathematical expression \(V \propto T\) or \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) anchors our calculations. These fractions allow us to find unknowns by setting them equal to each other. By comparing initial and new states of a gas, as exhibited in the exercise, you can determine unknown temperatures or volumes quite efficiently.

This proportionality helps predict how doubling the temperature, for instance, could lead to a doubling of the volume (again, under constant pressure). Understanding this relationship forms the foundation for tackling gas law problems with ease.
Kelvin scale
The Kelvin scale is a thermodynamic temperature scale where absolute zero, the temperature at which all particle motion stops, is \(0\,\mathrm{K}\). It begins at this point and measures degrees incrementally like the Celsius scale. However, one Kelvin degree is equal to one Celsius degree in size.

This scale is crucial when working with gases because it prevents the occurrence of negative temperature values, unlike Celsius or Fahrenheit, which can complicate calculations. All gas law problems should ideally be solved using Kelvin to ensure accuracy and consistency.

In the exercise, both initial and target temperatures were provided in Kelvin, removing ambiguities and allowing a smoother computational process when determining how temperature impacts the volume of the gas.

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