Calculate Work Done By Ideal Gas Expansion Step-by-Step

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Hey guys! Let's dive into a fascinating problem in thermodynamics: calculating the work done by an ideal gas during expansion. This is a classic physics problem that helps us understand how gases behave under changing conditions. We'll break down the problem step by step, so you can follow along easily.

Problem Statement

The problem we're tackling today involves calculating the work done by 10 moles of an ideal gas. This gas undergoes expansion from an initial pressure of 2 atm to a final pressure of 0.1 atm, all while maintaining a constant temperature of 0°C (273.15 K). The key here is that the external pressure is always equal to the pressure of the gas itself. This process is conducted through a series of gradual variations, which makes it an interesting scenario to analyze.

Understanding the Concepts

Before we jump into the calculations, let's make sure we're on the same page with some essential concepts.

Ideal Gas

An ideal gas is a theoretical gas that follows the ideal gas law. This law assumes that gas particles have negligible volume and do not interact with each other. While no real gas is truly ideal, many gases behave closely enough to ideal gas behavior under certain conditions, making this a useful approximation. The ideal gas law is expressed as:

PV = nRT

Where:

  • P is the pressure of the gas.
  • V is the volume of the gas.
  • n is the number of moles of the gas.
  • R is the ideal gas constant (0.0821 L atm / (mol K)).
  • T is the temperature of the gas in Kelvin.

Work Done by a Gas

When a gas expands, it does work on its surroundings. The work done by a gas during a volume change is given by:

W = -∫P_ext dV

Where:

  • W is the work done.
  • P_ext is the external pressure.
  • dV is the change in volume.

The negative sign indicates that the work done by the gas is positive when the volume increases (expansion) and negative when the volume decreases (compression).

Isothermal Process

An isothermal process is a thermodynamic process that occurs at a constant temperature. In our problem, the expansion happens at a constant temperature of 0°C. For an isothermal process, the work done can be calculated as:

W = -nRT ln(V₂/V₁)

Where:

  • V₁ is the initial volume.
  • V₂ is the final volume.

However, there's a catch in our problem. The external pressure is always equal to the pressure of the gas. This means we can't directly use the above formula, which assumes a constant external pressure. Instead, we need to use the integral form of the work equation.

Step-by-Step Solution

Now that we've reviewed the key concepts, let's solve the problem step by step.

1. Calculate Initial Volume (V₁)

We know the initial pressure (P₁ = 2 atm), the number of moles (n = 10 moles), the ideal gas constant (R = 0.0821 L atm / (mol K)), and the temperature (T = 273.15 K). We can use the ideal gas law to find the initial volume (V₁):

P₁V₁ = nRT

V₁ = nRT / P₁

V₁ = (10 moles) * (0.0821 L atm / (mol K)) * (273.15 K) / (2 atm)

V₁ ≈ 112.07 L

So, the initial volume of the gas is approximately 112.07 liters.

2. Calculate Final Volume (V₂)

Similarly, we know the final pressure (P₂ = 0.1 atm), and we can use the ideal gas law again to find the final volume (V₂):

P₂V₂ = nRT

V₂ = nRT / P₂

V₂ = (10 moles) * (0.0821 L atm / (mol K)) * (273.15 K) / (0.1 atm)

V₂ ≈ 2241.47 L

Thus, the final volume of the gas is approximately 2241.47 liters.

3. Calculate Work Done (W)

Since the external pressure is always equal to the pressure of the gas, we can express the work done as:

W = -∫P dV

Where P is the pressure of the gas at each stage of the expansion. From the ideal gas law, we know:

P = nRT / V

So, we can substitute this into the work equation:

W = -∫(nRT / V) dV

Since n, R, and T are constants, we can take them out of the integral:

W = -nRT ∫(1/V) dV

The integral of (1/V) dV is ln(V), so we have:

W = -nRT [ln(V)] (from V₁ to V₂)

W = -nRT (ln(V₂) - ln(V₁))

Using the properties of logarithms, we can rewrite this as:

W = -nRT ln(V₂/V₁)

Now, we can plug in the values:

W = -(10 moles) * (0.0821 L atm / (mol K)) * (273.15 K) * ln(2241.47 L / 112.07 L)

W ≈ -2241.46 J * ln(20.0)

W ≈ -2241.46 J * 2.9957

W ≈ -6715.01 J

Therefore, the work done by the gas during the expansion is approximately -6715.01 Joules. The negative sign indicates that the gas is doing work on its surroundings.

Converting to Calories (Optional)

If we want to express the work in calories, we can use the conversion factor 1 cal = 4.184 J:

W (in calories) = -6715.01 J / (4.184 J/cal)

W ≈ -1604.93 cal

So, the work done is approximately -1604.93 calories.

Conclusion

Alright, guys! We've successfully calculated the work done by 10 moles of an ideal gas expanding from 2 atm to 0.1 atm at a constant temperature of 0°C. The key to solving this problem was understanding the ideal gas law, the concept of work done by a gas, and how to apply the integral form of the work equation when the external pressure varies. Remember, the work done by the gas is negative, indicating that the gas is doing work on its surroundings during expansion.

This type of problem is fundamental in thermodynamics and helps us grasp the behavior of gases in various processes. Keep practicing, and you'll master these concepts in no time!

Keywords and SEO Optimization

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Repaired Input Keywords

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  • Original Keyword: "Calcular o trabalho executado por 10 moles de gĂĄs perfeito, quando submetidos a uma dilatação de 2 atm a 0°C atĂ© 0,1 atm a 0°C mantendo a pressĂŁo externa sempre igual Ă quela do gĂĄs. Neste caso, o processo Ă© conduzido por uma sucessĂŁo de variaçÔes."
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SEO Title

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