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A 28.4-L volume of methane gas is heated from \(35^{\circ} \mathrm{C}\) to \(72^{\circ} \mathrm{C}\) at constant pressure. What is the final volume of the gas?

Short Answer

Expert verified
The final volume of the gas is approximately 31.84 L.

Step by step solution

01

Identify the Known Variables

First, we need to identify the known variables and the formula to use. We have the initial volume \( V_1 = 28.4 \text{ L} \), the initial temperature \( T_1 = 35^{\circ} \text{C} = 308.15 \text{ K} \) after converting to Kelvin, and the final temperature \( T_2 = 72^{\circ} \text{C} = 345.15 \text{ K} \).
02

Apply Charles's Law

Charles's Law states that \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \( V_2 \) is the final volume we need to find. Rearrange this equation to solve for \( V_2 \): \( V_2 = \frac{V_1 T_2}{T_1} \).
03

Substitute the Known Values

Now substitute the known values into the equation: \( V_2 = \frac{28.4 \text{ L} \times 345.15 \text{ K}}{308.15 \text{ K}} \).
04

Calculate the Final Volume

Perform the calculation: \( V_2 = \frac{28.4 \times 345.15}{308.15} \approx 31.84 \text{ L} \). So, the final volume of the gas is approximately 31.84 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 how gases behave under various conditions of pressure, volume, and temperature. A primary gas law relevant here is Charles's Law. Charles's Law shows how the volume of a gas changes in relation to its temperature when the pressure is constant.

Specifically, it states that the volume of a gas is directly proportional to its temperature in Kelvin, assuming pressure remains constant. This means if you increase the temperature of a gas, its volume will also increase and vice versa, as long as the pressure doesn’t change.
  • The mathematical representation is: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \( V \) is volume and \( T \) is temperature.
  • This relationship is crucial for understanding how gases expand and contract in different conditions.
Temperature Conversions
In gas law calculations, temperatures must be in Kelvin. Kelvin is an absolute temperature scale and necessary for calculations because it starts at absolute zero.

To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, \( 35^{\circ} \text{C} \) becomes \( 308.15 \text{ K} \), as seen in the original exercise.
  • Always convert to Kelvin before using gas law formulas like Charles's Law.
  • Remember, no negative numbers occur in Kelvin, which simplifies calculations and maintains consistency with the laws of thermodynamics.
Volume Calculations
To find the final volume of a gas after a temperature change, apply Charles's Law once you've converted all temperatures to Kelvin. This involves substituting known values into the rearranged formula \( V_2 = \frac{V_1 T_2}{T_1} \).

Volume calculations like this require careful handling of units and temperatures to ensure accuracy:
  • With an initial volume \( V_1 \) of 28.4 L, initial temperature \( T_1 \) of 308.15 K, and final temperature \( T_2 \) of 345.15 K, plug into the formula to isolate \( V_2 \).
  • It's important to follow through with each step completely to avoid errors and attain the correct final volume.
Ideal Gas Behavior
Ideal gas behavior assumes gases consist of small particles in random, swift motion and that they exhibit no intermolecular forces. In real-world contexts, gases approximate ideal behavior at high temperatures and low pressures.

For the purposes of calculations like with Charles's Law, assuming ideal gas behavior means the gas will follow the gas laws accurately without deviation. This assumption simplifies equations and makes calculations more straightforward.
  • Ideal behavior implies gas particles do not attract or repel each other.
  • Most gases, under normal conditions, behave ideally, thus making this assumption valid in many practical scenarios.

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Most popular questions from this chapter

The atmosphere on Mars is composed mainly of carbon dioxide. The surface temperature is \(220 \mathrm{~K},\) and the atmospheric pressure is about \(6.0 \mathrm{mmHg}\). Taking these values as Martian "STP" calculate the molar volume in liters of an ideal gas on Mars.

When ammonium nitrite \(\left(\mathrm{NH}_{4} \mathrm{NO}_{2}\right)\) is heated, it decomposes to give nitrogen gas. This property is used to inflate some tennis balls. (a) Write a balanced equation for the reaction. (b) Calculate the quantity (in grams) of \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) needed to inflate a tennis ball to a volume of \(86.2 \mathrm{~mL}\) at \(1.20 \mathrm{~atm}\) and \(22^{\circ} \mathrm{C}\).

At what temperature will He atoms have the same \(u_{\mathrm{rms}}\) value as \(\mathrm{N}_{2}\) molecules at \(25^{\circ} \mathrm{C} ?\)

The apparatus shown here can be used to measure atomic and molecular speeds. Suppose that a beam of metal atoms is directed at a rotating cylinder in a vacuum. A small opening in the cylinder allows the atoms to strike a target area. Because the cylinder is rotating, atoms traveling at different speeds will strike the target at different positions. In time, a layer of the metal will deposit on the target area, and the variation in its thickness is found to correspond to Maxwell's speed distribution. In one experiment it is found that at \(850^{\circ} \mathrm{C}\) some bismuth (Bi) atoms struck the target at a point \(2.80 \mathrm{~cm}\) from the spot directly opposite the slit. The diameter of the cylinder is \(15.0 \mathrm{~cm},\) and it is rotating at 130 revolutions per second. (a) Calculate the speed (in \(\mathrm{m} / \mathrm{s}\) ) at which the target is moving. (Hint: The circumference of a circle is given by \(2 \pi r\), where \(r\) is the radius.) (b) Calculate the time (in seconds) it takes for the target to travel \(2.80 \mathrm{~cm} .\) (c) Determine the speed of the \(\mathrm{Bi}\) atoms. Compare your result in part (c) with the \(u_{\mathrm{rms}}\) of \(\mathrm{Bi}\) at \(850^{\circ} \mathrm{C}\). Comment on the difference.

Some ballpoint pens have a small hole in the main body of the pen. What is the purpose of this hole?

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