Calculating Entropy Change: Argon Expansion And Heating

by TheNnagam 56 views

Hey there, fellow science enthusiasts! Today, we're diving into the fascinating world of thermodynamics, specifically focusing on entropy and how it changes when a gas undergoes expansion and heating. We'll use Argon as our model gas and walk through the calculations step-by-step. Buckle up, because we're about to have some fun exploring the changes in entropy! Let's get started with this entropy calculation adventure.

Understanding Entropy and Its Significance

Before we jump into the calculations, let's get our heads around the concept of entropy. In simple terms, entropy is a measure of the disorder or randomness within a system. The higher the entropy, the more disordered the system is. Think of it like this: a perfectly organized room has low entropy, while a messy room has high entropy. In thermodynamics, entropy helps us understand the direction of natural processes and how energy flows. It's a fundamental concept that helps us understand whether a process is spontaneous or not. Understanding entropy is key to understanding the second law of thermodynamics, which states that the entropy of an isolated system always increases over time. This means that, in a closed system, things tend to become more disordered. So, why is entropy so important? Well, it helps us predict the feasibility of a process. For example, a gas expanding into a larger volume, or heat flowing from a hot object to a cold object, are both processes that increase entropy and are therefore spontaneous. Processes that decrease entropy are non-spontaneous and require energy input. Now, let's look at the formula we are going to use. We have to consider both the expansion and the heating of the gas, the equation becomes a little more complex.


Step-by-Step Calculation of Entropy Change for Argon

Alright, guys, let's get down to the nitty-gritty. Our goal is to calculate the change in entropy (Ī”S) when Argon gas undergoes two changes: expansion and heating. First off, let's recall the problem. Argon at 298K and 1 atm pressure in a 500 ml container is allowed to expand to 1000 ml and heated to 373K. To make things easy, we'll break the process down into two parts and calculate the entropy change for each part separately, then we will add them up. Then we calculate the total entropy change, Ī”S, is the sum of the entropy changes for each step. The formula for the entropy change due to volume change (isothermal process) is: Ī”S₁ = nRln(Vā‚‚/V₁). Here, n is the number of moles of gas, R is the ideal gas constant (8.314 J/molĀ·K), Vā‚‚ is the final volume, and V₁ is the initial volume. Also, to calculate the entropy change due to temperature change (isochoric process), we use the formula: Ī”Sā‚‚ = nCvln(Tā‚‚/T₁). Here, Cv is the molar heat capacity at constant volume, Tā‚‚ is the final temperature, and T₁ is the initial temperature. We can also make another assumption. Argon is a monoatomic gas. Therefore, its heat capacity at constant volume (Cv) is equal to 3/2R, where R is the ideal gas constant. In the following steps we will perform calculations, so that we can clearly get the value of entropy change. Now, let's walk through the calculations one by one!


Step 1: Calculate the Number of Moles of Argon (n)

First things first, we need to determine the number of moles of Argon present. We can use the ideal gas law, PV = nRT, to do this. But, since our volume is in milliliters, let's convert the volumes to liters (1 L = 1000 mL). The initial volume V₁ = 500 mL = 0.5 L and the final volume Vā‚‚ = 1000 mL = 1.0 L. We are also going to use the gas constant R = 0.0821 LĀ·atm/molĀ·K or R = 8.314 J/molĀ·K. The ideal gas law is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. First, we need to use the ideal gas law to find the number of moles (n) of Argon at the initial conditions, since the problem gave us the pressure, temperature, and volume. So we will rearrange the formula to n = PV/RT. We know P = 1 atm, V₁ = 0.5 L, R = 0.0821 LĀ·atm/molĀ·K and T = 298 K. So then we can substitute the values and get n = (1 atm * 0.5 L) / (0.0821 LĀ·atm/molĀ·K * 298 K), and after performing the calculation, we get n ā‰ˆ 0.0204 mol. Let’s keep this value ready for the next step. Let’s make sure we have all the values ready before starting the calculation, shall we?


Step 2: Calculate Entropy Change Due to Volume Change (Ī”S₁)

Now, let's calculate the entropy change due to the expansion of Argon, Ī”S₁. We will use the formula Ī”S₁ = nRln(Vā‚‚/V₁). We already calculated n ā‰ˆ 0.0204 mol, R = 8.314 J/molĀ·K, Vā‚‚ = 1.0 L, and V₁ = 0.5 L. By substituting the values, we get Ī”S₁ = 0.0204 mol * 8.314 J/molĀ·K * ln(1.0 L / 0.5 L). Doing the math, we find that Ī”S₁ ā‰ˆ 0.117 J/K. Remember, this value tells us the increase in entropy just from the gas expanding to twice its original volume. Pretty cool, huh? This is why the entropy change is positive, indicating an increase in disorder.


Step 3: Calculate Entropy Change Due to Temperature Change (ΔS₂)

Next up, we need to calculate the entropy change due to the temperature increase, Ī”Sā‚‚. The formula we will use is Ī”Sā‚‚ = nCvln(Tā‚‚/T₁). We have n ā‰ˆ 0.0204 mol, Tā‚‚ = 373 K, and T₁ = 298 K. As mentioned earlier, for a monoatomic gas like Argon, Cv = (3/2)R. Therefore, Cv = (3/2) * 8.314 J/molĀ·K ā‰ˆ 12.471 J/molĀ·K. Now we can substitute all the values and get Ī”Sā‚‚ = 0.0204 mol * 12.471 J/molĀ·K * ln(373 K / 298 K). Calculating this, we find Ī”Sā‚‚ ā‰ˆ 0.022 J/K. This tells us the entropy change related to heating the gas from 298 K to 373 K. Now that we have the values of the entropy change, let's move on to the next step.


Step 4: Calculate the Total Entropy Change (ΔS)

Finally, to get the total entropy change (Ī”S), we simply add the entropy changes from both steps: Ī”S = Ī”S₁ + Ī”Sā‚‚. So, Ī”S = 0.117 J/K + 0.022 J/K, which gives us Ī”S ā‰ˆ 0.139 J/K. And there you have it! The total entropy change for the process is approximately 0.139 J/K. This positive value confirms that the overall process leads to an increase in the disorder or randomness of the Argon gas system. We have successfully determined the total entropy change, reflecting the combined effect of both the expansion and heating of the gas. Congratulations, we are done with the entropy calculation!


Conclusion: Understanding the Significance

So, what have we learned, guys? We've successfully calculated the total entropy change for Argon undergoing expansion and heating. We found that the entropy increased, which is consistent with the second law of thermodynamics, as the system becomes more disordered. This exercise helps us understand how entropy quantifies the direction of spontaneous processes. As a reminder, the positive value of the entropy change (ΔS) confirms that the system became more disordered due to both the expansion and the heating. It all goes back to the second law of thermodynamics. Keep in mind that entropy is a state function. It only depends on the initial and final states of the system and not on the path taken. This makes entropy a very useful tool in predicting the behavior of various processes. Understanding this is key to many real-world applications, from designing more efficient engines to understanding climate change. Now that you've got this down, you can tackle similar problems. So keep practicing and you'll become an entropy expert in no time!