Question

How much work is done in compressing one mole of an ideal gas from a starting volume of \(1 \mathrm{~L}\) to a final volume of \(250 \mathrm{~mL}\) at a constant temperature of 293 K? What is the change in entropy? Assume that the process occurs near-equilibrium (reversibly).

Short answer

Work done: 3385.3 J, Change in entropy: -11.52 J/K.

Step-by-step solution

Understanding the Work Done in Compression

For compression of an ideal gas under constant temperature (isothermal process), the work done is given by the formula: \[ W = -nRT \, \ln \left(\frac{V_f}{V_i}\right) \] Here, \(n\) is the number of moles, \(R\) is the gas constant \(8.314 \, \text{J/mol K}\), \(T\) is the temperature, \(V_f\) is the final volume, and \(V_i\) is the initial volume.

Convert Volumes to the Same Units

The given volumes are \(V_i = 1 \, \text{L} = 1000 \, \text{mL}\) and \(V_f = 250 \, \text{mL}\). We need to use these in the same units when applying the formula for work done.

Calculate the Work Done

Substitute the known values into the formula:\[ W = -(1\,\text{mol})(8.314\,\text{J/mol K})(293\,\text{K}) \ln \left(\frac{250}{1000}\right) \]Calculate the natural logarithm:\[ \ln\left(\frac{250}{1000}\right) = \ln(0.25) \approx -1.386 \]Thus,\[ W = -(1)(8.314)(293)(-1.386) \approx 3385.3 \, \text{J} \]

Understanding Change in Entropy

The change in entropy for an ideal gas in a reversible isothermal process is given by:\[ \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \]Here, the variables have the same meaning as in the work done formula.

Calculate the Change in Entropy

Substitute the known values into the formula for entropy:\[ \Delta S = (1\,\text{mol})(8.314\,\text{J/mol K}) \ln \left(\frac{250}{1000}\right) \]Using the previously calculated natural logarithm:\[ \Delta S = (1)(8.314)(-1.386) = -11.52 \, \text{J/K} \]

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental principle in thermodynamics and relates the pressure, volume, and temperature of an ideal gas. It is represented by the equation:\[ PV = nRT \]where:

• \( P \) is the pressure of the gas
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the ideal gas constant (\(8.314 \, \text{J/mol K}\))
• \( T \) is the temperature in Kelvin

This law assumes that the gas is ideal, meaning it behaves perfectly under certain conditions. It helps in understanding how gases react under different temperature and pressure scenarios in theoretical conditions. In reality, gases might deviate from this behavior, but for many practical purposes, the ideal gas law serves as an excellent approximation.

Isothermal Process

An isothermal process is one where the temperature remains constant throughout. During such a process with an ideal gas, the relationship between pressure and volume of the gas is inverse, as described by Boyle's Law. In the case of compression, as the volume decreases, the pressure increases to maintain the temperature constant.An important characteristic of an isothermal process is that all the heat added to the system is used to perform work, and there is no change in the internal energy if you are dealing with an ideal gas. Since the temperature is constant, the ideal gas law simplifies the analysis. In our exercise, the gas was compressed isothermally from an initial volume of \(1 \, \text{L}\) to \(250 \, \text{mL}\), while maintaining a temperature of 293 K.

Entropy Change

In thermodynamics, entropy is a measure of disorder or randomness. The change in entropy during a process indicates the degree of irreversibility and the dispersal of energy. For an ideal gas undergoing a reversible isothermal process, the change in entropy \(\Delta S\) can be calculated using:\[ \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \]This equation shows how entropy is related to the natural logarithm of the volume ratio. In our case, when compressing the gas from \(1000 \, \text{mL}\) to \(250 \, \text{mL}\), the entropy change was found to be negative, \(-11.52 \, \text{J/K}\), indicating a decrease in disorder as the gas becomes more compact.

Work Done in Compression

The work done on or by a gas during compression or expansion in a thermodynamic process is an important aspect of energy transfer. For a reversible isothermal compression of an ideal gas, the work done \(W\) is given by:\[ W = -nRT \ln\left(\frac{V_f}{V_i}\right) \]Here, the negative sign indicates that work is done on the gas, which is typical for compression.During the compression from \(1 \, \text{L} \) to \(250 \, \text{mL} \) at a constant temperature of 293 K, the work done in this process was calculated to be approximately \(3385.3 \, \text{J} \). This energy reflects the input needed to compress the gas to a smaller volume and maintain the constant temperature throughout the process.