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Chapter 11: Problem 25

Calculate the final Celsius temperature when \(125 \mathrm{~mL}\) of chlorine gas at \(25^{\circ} \mathrm{C}\) is heated to give a volume of \(175 \mathrm{~mL}\). Assume that the pressure remains constant.

Short Answer

Expert verified
The final Celsius temperature is \(144.26^{\circ} \text{C}\).

Step by step solution

01

Identify the Variables

The initial volume of chlorine gas is \(V_1 = 125 \text{ mL}\) and the initial temperature is \(T_1 = 25^{\circ} \text{C}\). Convert this temperature to Kelvin by adding 273.15 to get \(T_1 = 298.15 \text{ K}\). The final volume is \(V_2 = 175 \text{ mL}\). We need to find the final temperature \(T_2\).
02

Use Charles' Law

According to Charles' Law, for an ideal gas at constant pressure, \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\). This means that the ratio of volume to temperature remains constant. We will use this formula to solve for \(T_2\).
03

Rearrange Charles' Law for Final Temperature

Rearrange the equation \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) to solve for \(T_2\): \(T_2 = \frac{V_2 \cdot T_1}{V_1}\).
04

Plug in the Known Values

Substitute the known values into the equation: \(T_2 = \frac{175 \text{ mL} \times 298.15 \text{ K}}{125 \text{ mL}}\).
05

Calculate the Final Temperature in Kelvin

Perform the calculation: \(T_2 = \frac{175 \times 298.15}{125} = 417.41 \text{ K}\).
06

Convert the Final Temperature to Celsius

To convert from Kelvin back to Celsius, subtract 273.15 from the Kelvin temperature: \(T_2 = 417.41 - 273.15 = 144.26^{\circ} \text{C}\).
07

Conclusion

The final temperature of the chlorine gas is \(144.26^{\circ} \text{C}\).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry describing the behavior of ideal gases. It combines several gas laws, including Boyle's, Charles', and Avogadro's Law, into a single equation: \[ PV = nRT \] where:
  • \( P \) stands for pressure
  • \( V \) is the volume of the gas
  • \( n \) is the number of moles of the gas
  • \( R \) is the universal gas constant
  • \( T \) is the temperature in Kelvin
While the exercise provided specifically uses Charles' Law, it's important to recognize that the ideal gas law connects all these variables.
Typically, when working with gases, assumptions are made:
  • Gases consist of a large number of particles in constant, random motion.
  • These particles are considered point masses with no volume.
  • Collisions between particles or with container walls are perfectly elastic.
The understanding of this "ideal" behavior simplifies calculations but remember, real gases may deviate under high pressure or low temperatures.
Temperature Conversion
Converting temperature units between Celsius and Kelvin is a common task in thermodynamics. Kelvin is the SI unit for temperature and is preferred in scientific calculations because it starts at absolute zero, simplifying calculations involving gas laws.
**Here's how you do the conversion:**
  • To convert from Celsius to Kelvin, add 273.15 to the Celsius temperature: \[ T( ext{K}) = T(^{ ext{C}}) + 273.15 \]
  • To convert from Kelvin to Celsius, subtract 273.15 from the Kelvin temperature: \[ T(^{ ext{C}}) = T( ext{K}) - 273.15 \]
In Charles' Law problems, such as the one in the exercise, converting Celsius to Kelvin is crucial because the direct relationship between volume and temperature is only linear when the temperature is in Kelvin. This ensures accuracy when using the formula \( \frac{V_i}{T_i} = \frac{V_f}{T_f} \).
Volume-Temperature Relationship
Understanding the volume-temperature relationship as described by Charles' Law is crucial in gas behavior studies. According to Charles' Law, if the pressure is held constant, the volume of a gas is directly proportional to its temperature in Kelvin. The equation form is:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]This means, as the temperature of a gas increases, its volume increases as well, provided the pressure doesn't change.
**Key points to remember:**
  • The formula indicates a linear relationship that allows predictions about how changes in temperature affect volume.
  • The relationship relies on temperature being in Kelvin to ensure that the volume never approaches zero, as this is not possible physically for a real gas.
  • Charles' Law helps in situations like predicting how a gas-filled balloon would expand if the temperature rises.
By rearranging the equation to solve for an unknown variable, you can tackle many practical problems, similar to the one in the exercise. This makes it a valuable tool for chemists and engineers alike.

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Most popular questions from this chapter

Distinguish between a real gas and an ideal gas.

A sample of hydrogen chloride gas occupies \(0.750 \mathrm{~L}\) at STP. What is the pressure in atm if the volume is \(0.100 \mathrm{~L}\) at \(25^{\circ} \mathrm{C} ?\)

What is the relationship between the vapor pressure of \(a\) liquid and temperature?

A sample of unknown gas has a mass of \(2.85 \mathrm{~g}\) and occupies \(750 \mathrm{~mL}\) at \(760 \mathrm{~mm} \mathrm{Hg}\) and \(100{ }^{\circ} \mathrm{C}\). What is the molar mass of the unknown gas?

If \(1.25 \mathrm{~mol}\) of oxygen gas exerts a pressure of \(1200 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C},\) what is the volume in liters?
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