Question

A gas expands from an initial state where \(p_{1}=500 \mathrm{kPa}\) and \(V_{1}=0.1 \mathrm{~m}^{3}\) to a final state where \(p_{2}=100 \mathrm{kPa}\). The relationship between pressure and volume during the process is \(p V=\) constant. Sketch the process on a \(p-V\) diagram and determine the work, in \(\mathrm{kJ}\).

Short answer

The work done by the gas is 80.45 kJ.

Step-by-step solution

Understand the Relationship

The problem states that the relationship between pressure \(p\) and volume \(V\) during the process is given by \(p \cdot V = \text{constant}\). This implies that the process is an isothermal process for an ideal gas.

Define the Constant

Since \(p \cdot V = \text{constant}\), we can find the constant value using the initial conditions: \(p_{1} \cdot V_{1} = 500 \text{kPa} \cdot 0.1 \text{m}^{3} = 50 \text{kPa} \cdot \text{m}^{3}\).

Determine the Final Volume

Using the final pressure \((p_{2} = 100 \text{kPa})\) and the constant, compute the final volume \((V_{2})\) as follows: \(p_{2} \cdot V_{2} = \text{constant}\) which means \(100 \text{kPa} \cdot V_{2} = 50 \text{kPa} \cdot \text{m}^{3}\), therefore \(V_{2} = \frac{50 \text{kPa} \cdot \text{m}^{3}}{100 \text{kPa}} = 0.5 \text{m}^{3}\).

Plot the Process

On a \(p-V\) diagram, draw a hyperbolic curve starting at \((V_{1}, p_{1}) = (0.1 \text{m}^{3}, 500 \text{kPa})\) and ending at \((V_{2}, p_{2}) = (0.5 \text{m}^{3}, 100 \text{kPa})\), corresponding to the equation \(p \cdot V = 50 \text{kPa} \cdot \text{m}^{3}\).

Calculate the Work Done

For an isothermal process, the work done by the gas is given by: \(\text{Work} = nRT \ln \left( \frac{V_{2}}{V_{1}} \right)\). However, since \(nRT = pV \text{ (constant)} = 50 \text{kPa} \cdot \text{m}^{3} \), we can use: \(\text{Work} = 50 \text{kPa} \ln \left( \frac{V_{2}}{V_{1}} \right)\). Converting to \text{kJ} (1 \text{kPa} \cdot \text{m}^{3} = 1 \text{kJ}), we find: \(\text{Work} = 50 \text{kJ} \ln \left( \frac{0.5 \text{m}^{3}}{0.1 \text{m}^{3}} \right) = 50 \text{kJ} \ln \left(5\right)\).

Compute the Natural Logarithm

Evaluate the natural logarithm: \(\ln \left( 5 \right) \approx 1.609 \). Therefore, the work done by the gas is: \(\text{Work} = 50 \text{kJ} \cdot 1.609 = 80.45 \text{kJ}\).

Key concepts

Pressure-Volume Relationship

In thermodynamics, the pressure-volume (P-V) relationship describes how pressure and volume change with respect to one another during a process. For this specific exercise, the P-V relationship is given by the equation \( p \cdot V = \text{constant} \). This represents an isothermal process for an ideal gas where temperature remains constant.
In an isothermal process, the product of pressure and volume does not change. This implies that if the volume increases, the pressure must decrease to maintain the constant product, and vice versa.
This relationship can be visualized as a hyperbolic curve on a P-V diagram, where each point on the curve represents a specific state defined by pressure and volume.

Work Done in Thermodynamics

Work done in thermodynamics, particularly in a P-V diagram, can be thought of as the area under the curve representing the process. For an isothermal process involving an ideal gas, the work done by the gas during expansion or compression can be calculated using:
\[ \text{Work} = nRT \ln \left( \frac{V_2}{V_1} \right) \]
Here, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature. However, considering the provided conditions, we can use the constant value from the P-V relationship. In this case, \( nRT = pV = 50 \text{kPa} \cdot \text{m}^3 \), simplifying the formula to:
\[ \text{Work} = 50 \text{kPa} \ln \left( \frac{V_2}{V_1} \right) \]
Converting this to kJ (since 1 kPa·m³ = 1 kJ) simplifies the calculation and makes it easier to understand and apply.

Ideal Gas Law

The ideal gas law is a fundamental equation in thermodynamics given by \( pV = nRT \), 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.
• T is the temperature of the gas.

In this exercise, the constant product of pressure and volume (\( p \cdot V = \text{constant} \)) suggests that the temperature remains constant throughout the process. This aligns with the isothermal process assumption. Understanding the ideal gas law helps in comprehending the changes in state variables during various thermodynamic processes and serves as a foundation for more complex concepts.

Natural Logarithm Calculation

Natural logarithms (denoted as \( \ln \)) play a crucial role in calculating the work done in isothermal processes. In our exercise, we calculated the work done during the gas expansion using the natural logarithm function. Given:
\[ \text{Work} = 50 \text{kJ} \ln \left( \frac{V_2}{V_1} \right) \]
We needed to evaluate \( \ln \left( 5 \right) \) since \( \frac{V_2}{V_1} = \frac{0.5 \text{m}^3}{0.1 \text{m}^3} = 5 \). Using common logarithm tables or a calculator, we find that: \( \ln(5) \approx 1.609 \).
Therefore, the work done by the gas becomes:
\[ \text{Work} = 50 \text{kJ} \cdot 1.609 = 80.45 \text{kJ} \]
This step is vital, as it mathematically quantifies the energy involved in changing the gas's state during the expansion process.