Problem 5 - University Physics with Modern Physics

Question

During the time 0.305 mol of an ideal gas undergoes an isothermal compression at \(22.0^{\circ} \mathrm{C}, 518 \mathrm{~J}\) 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 \(p V\) -diagram for the process.

Step-by-Step Solution

Answer

(a) The initial pressure was approximately 0.754 atm. (b) The p-V diagram shows a hyperbola with increasing pressure and decreasing volume.

Step 1: Understand the Isothermal Process

In an isothermal process, the temperature remains constant. For an ideal gas, the equation for isothermal compression is given by \( W = nRT \ln \left(\frac{V_i}{V_f}\right) \), where \( W \) is the work done, \( n \) is the number of moles, \( R \) is the gas constant (8.314 J/(mol·K)), and \( V_i \) and \( V_f \) are the initial and final volumes, respectively. In this problem, since the temperature is constant, the initial and final states can be related to pressures and volumes using the ideal gas law \( PV = nRT \).

Step 2: Convert Temperature to Kelvin

The given temperature is \(22.0^{\circ} \mathrm{C}\). First, we need to convert this to Kelvin: \( T = 22.0 + 273.15 = 295.15 \, \mathrm{K} \).

Step 3: Use Work Done Formula for Isothermal Process

We are given that \( 518 \, \mathrm{J} \) of work is done on the gas. For isothermal work, we use the equation: \( W = nRT \ln \left(\frac{P_f}{P_i}\right) \). Here \( W = 518 \, \text{J} \), \( n = 0.305 \, \text{mol} \), and the final pressure \( P_f = 1.76 \, \text{atm} = 1.76 \times 101325 \, \text{Pa} \approx 178.332 \, \text{kPa} \).

Step 4: Solve for Initial Pressure \( P_i \)

Rearrange the equation to solve for \( P_i \):\[ W = nRT \ln \left(\frac{P_f}{P_i}\right) \]\[ \ln \left(\frac{P_f}{P_i}\right) = \frac{W}{nRT} \]\[ \frac{P_f}{P_i} = e^{\frac{W}{nRT}} \]\[ P_i = \frac{P_f}{e^{\frac{W}{nRT}}} \]Plug in the values:\[ P_i = \frac{178.332}{e^{\frac{518}{0.305 \times 8.314 \times 295.15}}} \]\[ P_i \approx \frac{178.332}{e^{0.847}} \]\[ P_i \approx \frac{178.332}{2.333} \approx 76.46 \, \text{kPa} \approx 0.754 \, \text{atm} \]

Step 5: Sketch the p-V Diagram

On a p-V diagram, isothermal processes appear as hyperbolas. The process should start at \( P_i = 0.754 \, \text{atm} \) and end at \( P_f = 1.76 \, \text{atm} \). The volume decreases as pressure increases, indicating compression. Plot these pressures on the y-axis against corresponding volumes on the x-axis to show a downward curving hyperbola from left to right.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics. It describes the relationship between pressure, volume, temperature, and the number of moles of a gas. The equation is expressed as:

• \( PV = nRT \)

- **P** stands for pressure, usually measured in pascals (Pa) or atmospheres (atm).- **V** represents volume, typically in cubic meters (m³) or liters (L).- **n** is the number of moles of the gas.- **R** is the universal gas constant, valued at 8.314 J/(mol·K).- **T** denotes temperature, which must always be in Kelvin for this equation.
The Ideal Gas Law helps us predict how changes in one variable (e.g., temperature) affect the others. It's especially useful in isothermal processes, where temperature remains constant, allowing us to directly relate pressure and volume changes.

Temperature Conversion to Kelvin

In thermodynamic equations, including those involving the Ideal Gas Law, temperature must always be expressed in Kelvin. This is because Kelvin is an absolute temperature scale starting from absolute zero. To convert a temperature from Celsius (\(^\circ C\)) to Kelvin (K), you simply add 273.15 to the Celsius temperature:

• \( T(K) = T(^{\circ}C) + 273.15 \)

For example, if the temperature is 22.0\(^\circ C\), it converts to 295.15 K as shown in the step-by-step solution.
Accurate temperature conversion is crucial as it ensures that thermodynamic calculations reflect real-world conditions properly. Using the Kelvin scale allows the equations to remain consistent and physically meaningful.

Work Done in Thermodynamics

In thermodynamics, work done by or on a gas is a way to describe energy transfer. For an isothermal process, where the temperature remains constant, the work done, \( W \), is calculated using:

• \( W = nRT \ln \left(\frac{P_f}{P_i}\right) \)

Here, **\( n \)** is the number of moles, **\( R \)** is the gas constant, and **\( T \)** is the temperature in Kelvin. **\( P_f \)** and **\( P_i \)** are the final and initial pressures, respectively.
In our problem, 518 J of work is done on the ideal gas by the surroundings. This indicates compression. The formula shows how work is linked to pressure changes during an isothermal process and the significance of keeping temperature and pressure in the right units for accuracy.

p-V Diagram

A p-V diagram, or pressure-volume diagram, is an essential tool in thermodynamics. It visually represents how a gas behaves during different processes. In an isothermal compression, like the one detailed in the exercise, the process appears as a hyperbola on this diagram. Here's why:

• As pressure increases, volume decreases, and the curve shifts downward.
• The area under the curve represents the work done on the gas.

To sketch the p-V diagram for this problem, plot the initial and final pressures on the y-axis, noting that 0.754 atm is the starting pressure and 1.76 atm is the ending pressure. Correspond these to their respective volumes on the x-axis. Given that it's a compression, the curve will decline as it moves rightward, depicting the inverse relationship between pressure and volume under constant temperature.