Adiabatic and Isothermal work
Constant Pressure Work vs. Variable Pressure Work in Thermodynamics
Discussion Objectives
Differentiate between constant pressure work and variable pressure work.
Understand the implications of using constant pressure work equations, leading to overestimation or underestimation of results.
Constant Pressure Work
Definition: Constant pressure work refers to the work done by an ideal gas when it expands or contracts against a constant external pressure. This is an idealized scenario often used for approximation in systems where the external resistance to expansion or compression remains largely unchanged, such as a piston moving against constant atmospheric pressure or a fixed weight.
Mathematical Expression:
The work done (W) is given by the equation:
W = -P\Delta V
where:
P = constant external pressure
\Delta V = change in volume, calculated as (V{final} - V{initial})
The negative sign convention indicates that work done by the system (expansion) is negative, and work done on the system (compression) is positive.
Example: If a gas expands against a constant external pressure of 2\,atm and its volume changes from 5\,L to 15\,L, the work done by the gas isW = -(2\,atm)(15\,L - 5\,L) = - (2\,atm)(10\,L) = -20\,L\cdot atm . To convert to Joules, we can use the conversion factor 1\,L\cdot atm = 101.325\,J, so W = -20 \cdot 101.325\,J = -2026.5\,J.
Visual Representation:
When plotting pressure against volume (a P-V diagram), the work done by the system corresponds to the area under the curve. For constant pressure work, this area is a simple rectangle, reflecting the constant height (pressure) over the width (volume change).
Expansion Scenario:
When a gas expands, its internal pressure typically decreases. If work is calculated using an initial (higher) constant pressure assumption for the entire expansion, it will overestimate the actual work done by the system. This occurs because the constant pressure assumption does not account for the natural drop in the system's pressure as its volume increases, meaning the work is calculated based on a pressure higher than what is truly exerted during the later stages of expansion. A shaded area on the graph would represent this excess, overestimated work.
Compression Scenario:
Conversely, during compression, the system's pressure typically rises. If work is estimated using an initial (lower) constant pressure, the calculation will underestimate the actual work done on the system. The constant pressure assumption fails to capture the increasing pressure resistance as the volume shrinks, leading to a calculated work value that is less than the true energy input required. Again, shaded areas on the graph can illustrate where these estimations fall short of the real process.
Real-world Example:
An example of constant pressure work can be described as a gas performing work against a piston with constant mass on top, meaning the external force (and thus external pressure) remains constant during expansion or compression, for instance, against atmospheric pressure.
Variable Pressure Work
Definition: Variable pressure work encompasses situations where pressure changes dynamically and continuously during gas expansion or compression, requiring a more sophisticated mathematical approach for accuracy. This method is crucial when the system's internal pressure significantly impacts the work done throughout the process.
Mathematical Expression:
For variable pressure systems, we must sum up an infinite number of infinitesimal work contributions to accurately capture the work done over continuously changing pressure intervals. Thus, work is expressed through integration:
W = -\int_{V1}^{V2} P\,dV
For ideal gases undergoing a reversible process, the pressure P can be defined by the ideal gas law: P = \frac{nRT}{V}. Substituting this into the integral leads to:
W = -\int_{V1}^{V2} \frac{nRT}{V}\,dV
Adjustment of Integral:
In an isothermal process (constant temperature, T), n (moles of gas) and R (ideal gas constant) are also constant. Therefore, nRT can be pulled out of the integral:
W = -nRT\int_{V1}^{V2} \frac{1}{V}\,dV
Logarithmic Solution:
The integral \int \frac{1}{V}\,dV evaluates to \log(V), which allows us to express work for an isothermal, reversible expansion/compression of an ideal gas as:
W = -nRT[\log(V2) - \log(V1)]
Using logarithmic properties, this can be simplified to:
W = -nRT\log\left(\frac{V2}{V1}\right)
Example: For 1 mole of ideal gas expanding isothermally at 300\,K from an initial volume of 10\,L to a final volume of 20\,L, the work done is W = -(1\,mol)(8.314\,J\cdot mol^{-1}\cdot K^{-1})(300\,K)\log\left(\frac{20\,L}{10\,L}\right) = -2494.2 \log(2)\,J \approx -1728.8\,J.
Key Distinction: This variable pressure equation, specifically derived for reversible isothermal processes, results in neither overestimation nor underestimation. By integrating over the actual path taken by the system on the P-V diagram, it accurately reflects the continuous changes in pressure and volume, providing a true measure of the work done.
Comparison of Isothermal and Adiabatic Processes
Definitions:
Isothermal Process: A thermodynamic process where the temperature of the system remains constant throughout (\Delta T = 0). This typically requires the system to be in thermal contact with a heat reservoir to allow for heat exchange (Q \neq 0) to maintain a constant temperature. For an ideal gas undergoing an isothermal process, the internal energy change is zero (\Delta U = 0), which means that any heat added to the system is entirely converted into work done by the system (Q = W_{rev}).
Adiabatic Process: A thermodynamic process where no heat exchange occurs between the system and its surroundings (Q = 0). The system is thermally isolated. Consequently, any work done by or on the system directly results in a change in its internal energy (\Delta U = W). This change in internal energy leads to a change in the system's temperature.
Slope Analysis:
In isothermal expansion on a P-V diagram, the pressure decreases as volume increases along a gentle curve (adhering to Boyle's law, PV = constant). In contrast, adiabatic expansion exhibits a much steeper slope. This is because, during adiabatic expansion, the gas does work and, having no heat input to compensate, its internal energy decreases, leading to a significant drop in temperature. This temperature drop further reduces the pressure for a given volume change compared to an isothermal process, causing the adiabatic curve to fall more sharply.
First Law of Thermodynamics:
The First Law of Thermodynamics states that \Delta U = Q - W. For an adiabatic process, since Q = 0, the change in internal energy (\Delta U) is solely equal to the work done (W). For an ideal gas, the change in internal energy can also be expressed as:
\Delta U = nC_V\Delta T
where CV is the molar heat capacity at constant volume. Therefore, for an adiabatic process, W = nCV\Delta T. All work comes from or goes into the internal energy, directly affecting the temperature.
Example: If 2 moles of a monatomic ideal gas (C_V = \frac{3}{2}R) undergo an adiabatic process where its temperature decreases by 50\,K, the change in internal energy is \Delta U = (2\,mol)\left(\frac{3}{2} \cdot 8.314\,J\cdot mol^{-1}\cdot K^{-1}\right)(-50\,K) = -1247.1\,J. Since it's adiabatic, W = \Delta U = -1247.1\,J.
Key Relationships:
The relationship between work done during an adiabatic process encapsulates changes in internal energy. Utilizing the ideal gas law and the first law, we can derive key adiabatic process equations. For reversible adiabatic processes, the relationship between temperature and volume is:
T2V2^{(\gamma-1)} = T1V1^{(\gamma-1)}
where \gamma (gamma) is the heat capacity ratio, defined as \gamma = \frac{CP}{CV}, with C_P being the molar heat capacity at constant pressure. (See 'Example Calculation' for an application of this equation.)
Combined Gas Law Application:
By combining the ideal gas law with the adiabatic temperature-volume relationship, we arrive at the adiabatic variations of the gas law, which relates pressure and volume:
P1V1^{\gamma} = P2V2^{\gamma}
This equation is fundamental for calculating changes in pressure or volume during an adiabatic process. (See 'Example Calculation' for an application of this equation.)
Example Calculation
Problem Statement:
Given: 5 moles of an ideal monatomic gas at an initial temperature of 298 K and a pressure of 10 atmospheres expands adiabatically and reversibly until the pressure decreases to 1 atmosphere. Calculate the final volume (V2) and temperature (T2), energy and enthalpy changes (\Delta U and \Delta H), and work done (W).
Initial Volume (V1):
Using the ideal gas law (PV = nRT) to find the initial volume:
V1 = \frac{nRT}{P1} = \frac{(5\,mol)(0.08206\,L\cdot atm\cdot mol^{-1}\cdot K^{-1})(298\,K)}{10\,atm} = 12.226\,L
Calculating Gamma (\gamma):
For monatomic ideal gases, the molar heat capacities are known:
CP = \frac{5}{2}R\qquad CV = \frac{3}{2}R
Therefore, the heat capacity ratio is:
\gamma = \frac{CP}{CV} = \frac{5/2 R}{3/2 R} = \frac{5}{3}
Final Volume (V2):
Using the adiabatic relationship P1V1^{\gamma} = P2V2^{\gamma}:
V2^{\gamma} = \left(\frac{P1}{P2}\right)V1^{\gamma}
Substituting values, we find:
V2 = \left(\frac{P1}{P2}V1^{\gamma}\right)^{1/\gamma} = \left(\frac{10\,atm}{1\,atm}(12.226\,L)^{5/3}\right)^{3/5} \approx 48.7\,L
Final Temperature (T2):
Now, use the ideal gas law again with the final pressure (P2) and the calculated final volume (V2) to find the final temperature:
T2 = \frac{P2V2}{nR} = \frac{(1\,atm)(48.7\,L)}{(5\,mol)(0.08206\,L\cdot atm\cdot mol^{-1}\cdot K^{-1})} = 118.62\,K
Energy Change (\Delta U) and Work Done (W):
For an adiabatic process, \Delta U = W. We can calculate \Delta U using the temperature change:
\Delta U = nC_V\Delta T = (5\,mol)\left(\frac{3}{2}R\right)(T2 - T1)
\Delta U = (5\,mol)\left(\frac{3}{2})(8.314\,J\cdot mol^{-1}\cdot K^{-1})\right)(118.62\,K - 298\,K)
\Delta U \approx -11,162\,J
Thus, the work done is \mathbf{W \approx -11,162\,J}. The negative sign indicates work done by the gas.
Enthalpy Change (\Delta H):
The enthalpy change can be calculated using the constant pressure molar heat capacity:
\Delta H = nC_P\Delta T = (5\,mol)\left(\frac{5}{2}R\right)(T2 - T1)
\Delta H = (5\,mol)\left(\frac{5}{2})(8.314\,J\cdot mol^{-1}\cdot K^{-1})\right)(118.62\,K - 298\,K)
\Delta H \approx -18,604\,J
Conclusion
Understanding the fundamental differences between constant pressure and variable pressure work is crucial for accurate thermodynamic calculations. Relying on erroneous assumptions, such as treating a variable pressure process as constant, can significantly skew results through overestimation or underestimation of work.
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