Calculating Original Temperature Using Charles Law Chemistry Problem

Jul 22, 2025 by Sam Evans 69 views

Calculating Original Temperature Using Charles's Law

Hey there, science enthusiasts! Today, we're diving into a fascinating concept in chemistry: Charles's Law. This principle helps us understand the relationship between the volume and temperature of a gas when the pressure and the amount of gas remain constant. Let's tackle a problem where we need to find the original temperature of a gas given some changes in its volume and final temperature.

Understanding Charles's Law

Before we jump into the problem, let's quickly recap Charles's Law. This law states that the volume of a gas is directly proportional to its absolute temperature when the pressure and the amount of gas are kept constant. In simpler terms, if you increase the temperature of a gas, its volume will increase proportionally, and vice versa. This relationship can be mathematically expressed as:

$V_1/T_1 = V_2/T_2$

Where:

$V_1$ is the initial volume
$T_1$ is the initial temperature
$V_2$ is the final volume
$T_2$ is the final temperature

It's crucial to remember that temperature must be in Kelvin (K) for this law to work correctly. Kelvin is the absolute temperature scale, where 0 K is absolute zero (theoretically the coldest possible temperature). If you're given temperatures in Celsius (°C), you'll need to convert them to Kelvin using the following formula:

T (K) = T (°C) + 273.15

Problem Breakdown: Finding the Original Temperature

Now, let's get to the heart of the matter. Our problem states:

The volume of a gas changed from 1145 mL to 765 mL. The final temperature of the gas is 640 K. What was the original temperature of the gas?

$T = [?] K$

Assume that the number of moles and the pressure remain constant.

To solve this, we'll use Charles's Law. Here's how we can break down the problem:

Identify the knowns and unknowns:
- Initial volume ( $V_1$ ) = 1145 mL
- Final volume ( $V_2$ ) = 765 mL
- Final temperature ( $T_2$ ) = 640 K
- Initial temperature ( $T_1$ ) = ? (This is what we need to find)
Rearrange Charles's Law to solve for $T_1$ $T_{1}$ :
- We have the equation $V_1/T_1 = V_2/T_2$ $V_{1} / T_{1} = V_{2} / T_{2}$ . To isolate $T_1$ $T_{1}$ , we can rearrange the equation as follows:
  - $T_1 = (V_1 * T_2) / V_2$
Plug in the values and calculate:
- Now, we simply substitute the known values into the rearranged equation:
  - $T_1 = (1145 mL * 640 K) / 765 mL$
  - $T_1 = 732800 K ext{ mL} / 765 mL$
  - $T_1 ≈ 958 K$

Therefore, the original temperature of the gas was approximately 958 K. It's always a good practice to double-check your work and units to ensure accuracy.

Step-by-Step Solution with Detailed Explanations

Let's walk through the solution again, but this time, we'll add some extra explanations to make sure everything is crystal clear. We all know how tricky these problems can be, so let's break it down nice and slow.

1. Identify the Givens and the Unknown

First, let's list out what we know from the problem. This is like gathering our ingredients before we start cooking – it's crucial to a successful outcome! We have:

$V_1$ (Initial volume) = 1145 mL
$V_2$ (Final volume) = 765 mL
$T_2$ (Final temperature) = 640 K
$T_1$ (Initial temperature) = ? (This is what we're trying to find – the mystery ingredient!)

Notice how the problem explicitly states that the number of moles and the pressure remain constant. This is a BIG clue that we can use Charles's Law. If those factors weren't constant, we'd need a different approach.

2. Write Down Charles's Law

Okay, now that we've identified our ingredients, let's get out our recipe – Charles's Law! As we discussed earlier, it's:

$V_1/T_1 = V_2/T_2$

This equation is the key to solving our problem. It tells us exactly how volume and temperature are related when pressure and the amount of gas are constant.

3. Rearrange the Equation to Solve for $T_1$

Here's where things might get a little tricky, but don't worry, we'll take it step by step. Our goal is to get $T_1$ by itself on one side of the equation. To do this, we'll use some algebraic magic. It's like untangling a knot – you just have to be patient and follow the steps.

Cross-multiply: This gives us $V_1 * T_2 = V_2 * T_1$
Divide both sides by $V_2$ : This isolates $T_1$ on the right side: $(V_1 * T_2) / V_2 = T_1$
Rewrite the equation: For clarity, let's flip the equation so that $T_1$ is on the left side: $T_1 = (V_1 * T_2) / V_2$

Voila! We've successfully rearranged Charles's Law to solve for $T_1$ . This is a crucial step, so make sure you understand the algebra behind it.

4. Plug in the Values

Now comes the fun part – plugging in the numbers! This is like adding the ingredients to the bowl. We know:

$V_1 = 1145 mL$
$T_2 = 640 K$
$V_2 = 765 mL$

So, let's substitute these values into our rearranged equation:

$T_1 = (1145 mL * 640 K) / 765 mL$

5. Calculate the Result

Time to crunch the numbers! Grab your calculator (or your mental math skills, if you're feeling brave) and let's do the arithmetic:

Multiply $1145 mL * 640 K = 732800 K ext{ mL}$
Divide $732800 K ext{ mL} / 765 mL ≈ 958 K$

So, we get $T_1 ≈ 958 K$ . This is our answer!

6. Double-Check the Units and the Logic

Before we celebrate, let's make sure our answer makes sense. It's like tasting the dish before you serve it – you want to be sure it's just right!

Units: Notice that the mL units cancel out in our calculation, leaving us with Kelvin (K), which is the correct unit for temperature. Phew! That's a good sign.
Logic: The volume of the gas decreased from 1145 mL to 765 mL. According to Charles's Law, if the volume decreases, the temperature should also decrease. Our calculated initial temperature (958 K) is higher than the final temperature (640 K), which makes sense. Awesome!

Final Answer

So, after all that careful calculation, we've arrived at our final answer:

The original temperature of the gas was approximately 958 K.

$T_1 ≈ 958 K$

Key Takeaways and Practical Applications

Let's wrap up by highlighting the key takeaways from this problem and discussing some real-world applications of Charles's Law. Understanding these concepts isn't just about solving problems; it's about understanding the world around us.

Charles's Law in Action:

Hot Air Balloons: One of the most visible applications of Charles's Law is in hot air balloons. Heating the air inside the balloon increases its volume, making it less dense than the surrounding air. This difference in density creates buoyancy, allowing the balloon to float. The hotter the air, the greater the volume, and the higher the balloon can rise.
Internal Combustion Engines: Car engines rely on the principles of gas laws, including Charles's Law. As the air-fuel mixture heats up during combustion, its volume expands, pushing the piston and generating power. Understanding these relationships is crucial for engine design and efficiency.
Weather Forecasting: Meteorologists use gas laws to predict weather patterns. Changes in temperature and pressure affect the volume of air masses, which can lead to various weather phenomena. For example, warm air rises and expands (Charles's Law), potentially leading to cloud formation and precipitation.
Everyday Life: Even in everyday situations, Charles's Law plays a role. Think about a slightly deflated basketball left in a hot car. The increase in temperature will cause the air inside the ball to expand, potentially increasing its pressure and even inflating it slightly.

Common Mistakes to Avoid

When working with Charles's Law and gas law problems in general, there are a few common pitfalls that students often encounter. Being aware of these mistakes can help you avoid them and ensure you get the correct answer. Let's shine a spotlight on these potential stumbling blocks:

Forgetting to Convert to Kelvin: This is the most common mistake. Charles's Law (and most other gas laws) requires temperature to be in Kelvin (K), the absolute temperature scale. If you use Celsius (°C) or Fahrenheit (°F), your calculations will be incorrect. Always convert to Kelvin using the formula: T (K) = T (°C) + 273.15. Don't let this simple step trip you up!
Mixing Up Variables: It's easy to get the initial and final volumes or temperatures mixed up. Carefully read the problem and clearly label your variables ( $V_1$ , $V_2$ , $T_1$ , $T_2$ ) before plugging them into the equation. A little organization goes a long way!
Incorrectly Rearranging the Equation: Algebra can be tricky! Make sure you correctly rearrange Charles's Law to solve for the unknown variable. If you're unsure, take it step by step and double-check your work. Remember our detailed explanation of the rearrangement process earlier in this article.
Ignoring Constant Conditions: Charles's Law only applies when the pressure and the number of moles of gas are constant. If the problem states that these conditions change, you'll need to use a different gas law (like the Ideal Gas Law). Pay close attention to the problem statement and identify which gas law is appropriate.
Not Checking Units: Always double-check that your units are consistent throughout the problem. If the volume is given in milliliters (mL), make sure you use mL for both $V_1$ and $V_2$ . If you need to convert units, do it carefully and correctly. As we saw in our example, canceling units is a great way to verify your calculations.

By being mindful of these common mistakes, you can approach Charles's Law problems with confidence and increase your chances of success. Remember, practice makes perfect!

Practice Problems to Sharpen Your Skills

Now that we've thoroughly covered Charles's Law and worked through a detailed example, it's time to put your knowledge to the test! The best way to master any scientific concept is through practice. So, grab a pen and paper, and let's tackle a few more problems together. Don't worry, we'll provide hints and explanations along the way.

By working through these practice problems, you'll not only solidify your understanding of Charles's Law but also develop valuable problem-solving skills that will benefit you in future chemistry endeavors. Remember, the key to success is practice, practice, practice!

Let's recap, guys. In this article, we've explored Charles's Law, learned how to apply it to solve problems, discussed real-world applications, and highlighted common mistakes to avoid. With this knowledge, you're well-equipped to tackle any Charles's Law challenge that comes your way! Keep practicing, and you'll become a gas law pro in no time!