Ideal Gas Expansion: Work And Heat Calculation

Hey guys! Ever wondered what happens when an ideal gas decides to stretch its legs? Let's dive into a scenario where we have one mole of an ideal gas chilling at 300K, and it expands from a cozy 2 liters to a more spacious 6 liters. The big question is: how much work does this gas do, and how much heat is involved in the process?

Understanding the Basics

Before we crunch any numbers, let's quickly recap some fundamental concepts. An ideal gas is a theoretical gas composed of randomly moving point particles that don't interact except when they collide elastically. Real gases behave like ideal gases under certain conditions, like high temperature and low pressure. For ideal gases, we often use the ideal gas law:

PV = nRT

Where:

• P = Pressure
• V = Volume
• n = Number of moles
• R = Ideal gas constant (8.314 J/mol·K)
• T = Temperature

Work, in thermodynamics, is the energy transferred when a force causes displacement. For a gas expanding against external pressure, the work done by the gas is:

W = -∫PdV

Where the integral is evaluated from the initial volume to the final volume.

Heat, on the other hand, is the energy transferred due to temperature differences. The amount of heat transferred depends on the process (e.g., constant pressure, constant volume, isothermal, adiabatic).

Types of Expansion

The amount of work done and heat transferred drastically depends on how the expansion occurs. Let's consider a few common types:

1. Isothermal Expansion: This happens at a constant temperature. Think of it like the gas expanding in a perfectly controlled environment where any change in internal energy due to expansion is compensated by heat entering or leaving the system to keep the temperature steady.
2. Adiabatic Expansion: This occurs without any heat exchange with the surroundings. Imagine the gas expanding so rapidly that there's no time for heat to enter or leave the system. This often results in a temperature drop.
3. Isobaric Expansion: This takes place at constant pressure. Envision the gas expanding inside a container with a movable piston, where the external pressure on the piston remains constant.
4. Isochoric Expansion (or Constant Volume): Although not technically an expansion, it's worth mentioning. In this case, the volume remains constant, so no work is done (W = 0).

Calculating Work and Heat for Isothermal Expansion

Let's assume the expansion is isothermal. This means the temperature remains constant at 300K. For an isothermal process, the work done by the gas is:

W = -nRT ln(V₂/V₁)

Where:

• n = 1 mole
• R = 8.314 J/mol·K
• T = 300 K
• V₁ = 2 L
• V₂ = 6 L

Plugging in the values:

W = -(1 mol) * (8.314 J/mol·K) * (300 K) * ln(6 L / 2 L) W = -8.314 * 300 * ln(3) W ≈ -8.314 * 300 * 1.0986 W ≈ -2741.57 J

So, the work done by the gas during isothermal expansion is approximately -2741.57 Joules. The negative sign indicates that the gas is doing work on the surroundings.

For an isothermal process, the change in internal energy (ΔU) is zero because the temperature is constant. Therefore, according to the first law of thermodynamics:

ΔU = Q + W 0 = Q + W Q = -W

This means the heat absorbed by the gas (Q) is equal to the negative of the work done by the gas:

Q ≈ 2741.57 J

So, the gas absorbs approximately 2741.57 Joules of heat from the surroundings to maintain a constant temperature during the expansion.

Calculating Work and Heat for Isobaric Expansion

Now, let's consider the expansion is isobaric, meaning the pressure remains constant. First, we need to find the pressure. Using the ideal gas law at the initial state:

P₁V₁ = nRT₁ P = (nRT₁) / V₁ P = (1 mol * 8.314 J/mol·K * 300 K) / (0.002 m³) P = (1 * 8.314 * 300) / 0.002 P = 2494.2 / 0.002 P = 1247100 Pa

Where we converted 2L to m³ (2L = 0.002 m³).

For isobaric expansion, the work done is:

W = -P(V₂ - V₁) W = -1247100 Pa * (0.006 m³ - 0.002 m³) W = -1247100 * 0.004 W = -4988.4 J

The work done by the gas during isobaric expansion is approximately -4988.4 Joules.

To find the heat (Q), we use the equation:

Q = nCpΔT

Where Cp is the molar heat capacity at constant pressure. For an ideal gas, Cp = (5/2)R.

First, we need to find the final temperature (T₂) using the ideal gas law:

P₂V₂ = nRT₂ T₂ = (P₂V₂) / (nR) T₂ = (1247100 Pa * 0.006 m³) / (1 mol * 8.314 J/mol·K) T₂ = (1247100 * 0.006) / (1 * 8.314) T₂ = 7482.6 / 8.314 T₂ ≈ 900 K

Now we can calculate the heat:

Q = nCpΔT Q = (1 mol) * (5/2 * 8.314 J/mol·K) * (900 K - 300 K) Q = 1 * (5/2 * 8.314) * 600 Q = 1.0 * 20.785 * 600 Q = 12471 J

So, approximately 12471 Joules of heat are absorbed by the gas during isobaric expansion.

Calculating Work and Heat for Adiabatic Expansion

Finally, let's examine adiabatic expansion, where no heat is exchanged with the surroundings (Q = 0). In this case, the relationship between pressure and volume is given by:

P₁V₁^γ = P₂V₂^γ

Where γ (gamma) is the adiabatic index, defined as Cp/Cv. For an ideal monatomic gas, γ = 5/3 = 1.667. For a diatomic gas, γ = 7/5 = 1.4. Let's assume we're dealing with a monatomic ideal gas, so γ = 5/3.

First, we need to find the final pressure (P₂) after expansion. Using the adiabatic relation:

P₁V₁^γ = P₂V₂^γ

We already know P₁ = 1247100 Pa, V₁ = 2L, and V₂ = 6L. So,

1247100 * (2)^(5/3) = P₂ * (6)^(5/3)

P₂ = 1247100 * (2/6)^(5/3) P₂ = 1247100 * (1/3)^(5/3) P₂ = 1247100 * (0.3333)^(1.667) P₂ = 1247100 * 0.19245 P₂ ≈ 239995 Pa

Now that we have the final pressure, we can calculate the work done during adiabatic expansion:

W = -(P₂V₂ - P₁V₁) / (1 - γ) W = -((239995 * 0.006) - (1247100 * 0.002)) / (1 - 5/3) W = -((1439.97) - (2494.2)) / (-2/3) W = -(-1054.23) / (-0.6667) W = -1581.34 J

So, the work done by the gas during adiabatic expansion is approximately -1581.34 Joules.

Since this is an adiabatic process, no heat is exchanged with the surroundings:

Q = 0

Summary Table

To make it easier, let's summarize our findings in a table:

Conclusion

Alright, folks! We've calculated the work done and heat transferred for one mole of an ideal gas expanding from 2L to 6L under different conditions: isothermal, isobaric, and adiabatic. Remember that the type of expansion significantly affects the amount of work done and heat exchanged. Whether it's keeping the temperature constant, maintaining constant pressure, or preventing any heat exchange, each scenario gives us unique values for work and heat. Keep these principles in mind, and you'll be well-equipped to tackle similar thermodynamic problems! Understanding these concepts is super useful in many fields, including engineering and even weather forecasting. Keep exploring and happy calculating!