Question

During the time 0.305 mol of an ideal gas undergoes an isothermal compression at 22.0\(^\circ\)C, 392 J of work is done on it by the surroundings. (a) If the final pressure is 1.76 atm, what was the initial pressure? (b) Sketch a \(pV\)-diagram for the process.

Short answer

(a) The initial pressure was approximately 1.87 atm. (b) Sketch the curve as hyperbolic from high volume, low pressure to low volume, high pressure on the pV-diagram.

Step-by-step solution

Understanding Isothermal Compression

In an isothermal process, the temperature remains constant. For an ideal gas undergoing isothermal compression, the internal energy doesn't change, meaning the work done on the gas is equal to the heat transferred out. We'll use the ideal gas equation to find the initial pressure.

Using the Ideal Gas Equation

For an ideal gas, the equation is given by \( PV = nRT \). Since the temperature is constant, \( P_1V_1 = P_2V_2 \) applies. The work done on the gas \( W = nRT \ln\left(\frac{V_2}{V_1}\right) \) helps find the ratio \( \frac{V_2}{V_1} \). Use \( W = 392 \text{ J} \) and the gas constant \( R = 8.314 \text{ J/mol K} \) to calculate \( T \) in Kelvin, \( n = 0.305 \text{ mol} \), and \( T = 22.0 + 273.15 \text{ K} = 295.15 \text{ K} \).

Calculating Temperature in Kelvin

Convert the given temperature in Celsius to Kelvin: \( T = 22.0 + 273.15 = 295.15 \text{ K} \). This allows the use of consistent units throughout the calculations.

Finding the Pressure-Volume Ratio

Calculate the volume ratio from work done: \[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \implies 392 = 0.305 \times 8.314 \times 295.15 \times \ln\left(\frac{V_2}{V_1}\right) \]. Solve for \( \ln\left(\frac{V_2}{V_1}\right) \), which provides \( \ln\left(\frac{V_2}{V_1}\right) \approx 0.058 \).

Solving for the Volume Ratio

From the previous step, \( \ln\left(\frac{V_2}{V_1}\right) = 0.058 \) implies \( \frac{V_2}{V_1} = e^{0.058} \approx 1.06 \). Since \( P_1V_1 = P_2V_2 \), then \( P_1 = P_2 \frac{V_2}{V_1} \).

Calculating Initial Pressure

Using the ratio \( \frac{V_2}{V_1} = 1.06 \) and \( P_2 = 1.76 \text{ atm} \), find \( P_1 = 1.76 \times 1.06 \approx 1.87 \text{ atm} \).

Drawing the pV-Diagram

In a \( pV \)-diagram, plot the pressure on the y-axis and volume on the x-axis. The curve will show a hyperbolic shape (since the product \( PV \) is constant), starting from a higher volume and lower pressure (\( P_1 \)) to a lower volume and higher pressure (\( P_2 \)).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in thermodynamics that describes the behavior of ideal gases. It is represented by the formula: \( PV = nRT \), 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, approximately 8.314 J/mol·K
• \( T \) is the temperature in Kelvin

This law assumes that the gas particles do not interact and that the volume of the gas particles is negligible compared to the volume of the container. It is especially useful for predicting the behavior of gases under various conditions of pressure, volume, and temperature. When applying the Ideal Gas Law, it is crucial to ensure that all units are consistent, particularly the temperature which must be in Kelvin. This law applies very well to many common gases at low pressure and high temperature.

Isothermal Process

An isothermal process is a thermodynamic process that occurs at a constant temperature. This means that the heat transferred into or out of the system adjusts to keep the temperature constant. For ideal gases, there are some important implications:

• The internal energy of the gas remains unchanged because internal energy depends on temperature.
• The work done by or on the gas equals the heat exchanged during the process; in symbols, \( Q = W \).
• The equation \( P_1V_1 = P_2V_2 \) holds true, demonstrating that the product of pressure and volume remains constant during the process.

During an isothermal compression, like the one presented in the exercise, the gas is compressed by the surroundings doing work on it, while any heat produced is simultaneously transferred out of the gas to maintain constant temperature. Such processes are graphically represented by a hyperbolic curve on a Pressure-Volume diagram, indicating that as the volume decreases, the pressure increases.

Work Done on Gas

Work done on a gas during an isothermal process is an important concept in thermodynamics. During such a process, the total work done can be calculated by:\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]where:

• \( W \) is the work done
• \( n \) is the number of moles of gas
• \( R \) is the gas constant
• \( T \) is the temperature held constant in Kelvin
• \( V_1 \) and \( V_2 \) are the initial and final volumes respectively

It shows that work done on the gas depends on the ratio of final volume to initial volume. When calculating the work done, the sign convention can be important: work done on the gas is considered positive, while work done by the gas is considered negative. Understanding these principles helps in analyzing various thermodynamic processes and predicting how a system will respond under isothermal conditions.

Pressure-Volume Diagram

A Pressure-Volume (PV) diagram is a graphical representation of the relationship between the pressure and volume of a gas. In these diagrams:

• The y-axis typically represents pressure.
• The x-axis represents volume.
• Each point on the curve represents a different state of the gas during the process.

For an isothermal process involving an ideal gas, the PV diagram displays a hyperbolic curve. This curve reflects the fact that during isothermal compression (or expansion), the product \( PV \) stays constant: as the volume decreases, pressure increases proportionately.To sketch a PV diagram for an isothermal process like the one in the exercise:

• Start at the initial pressure \( P_1 \) and volume \( V_1 \).
• The curve will slope downwards towards the final pressure \( P_2 \) and volume \( V_2 \).
• The curve follows a hyperbolic shape due to the inverse relationship between pressure and volume when temperature remains constant.

By analyzing these diagrams, we gain insight into how gases behave under different thermodynamic conditions, which is crucial for designing and evaluating engineering systems.