Problem 2 - University Physics with Modern Physics

Question

Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. The initial temperature of the gas is \(27.0^{\circ} \mathrm{C}\) and the pressure is constant. As part of a machine design project, calculate the final temperature of the gas after it has done \(2.40 \times 10^{3}\) J of work.

Step-by-Step Solution

Answer

The final temperature of the gas is approximately 348.10 K.

Step 1: Understanding the Given Information

We know the initial temperature of the gas is \(27.0^{\circ} \mathrm{C}\). Convert this to Kelvin by adding 273, which gives us an initial temperature \(T_1 = 300\, \mathrm{K}\). We have 6 moles of gas, and the gas has done \(2.40 \times 10^{3}\, \mathrm{J}\) of work under constant pressure.

Step 2: Applying the Ideal Gas Work Formula

The formula for work done by an ideal gas under constant pressure is given by \(W = nR(T_2 - T_1)\). Rearrange the formula to find the final temperature \(T_2\). We get \(T_2 = \frac{W}{nR} + T_1\).

Step 3: Solving for Final Temperature

Substitute the given work \(W = 2.40 \times 10^3\, \mathrm{J}\), number of moles \(n = 6\), the gas constant \(R = 8.314\, \mathrm{J/(mol \, K)}\), and initial temperature \(T_1 = 300\, \mathrm{K}\) into the equation: \[T_2 = \frac{2.40 \times 10^3}{6 \times 8.314} + 300\] Calculate this to find \(T_2\).

Step 4: Calculating Results

Perform the calculation: \[T_2 = \frac{2.40 \times 10^3}{49.884} + 300\]\[T_2 = 48.10 + 300\]\[T_2 \approx 348.10 \, \mathrm{K}\]. The final temperature of the gas after doing the work is approximately \(348.10\, \mathrm{K}\).

Key Concepts

Constant Pressure Process

In a constant pressure process, the pressure of the gas remains unchanged throughout the heating or cooling phase. This steady pressure condition is important when dealing with gases in closed systems with movable pistons.
For example, in this exercise, as the gas expands or contracts, pushing the piston, its pressure remains constant.
Since gas particles take up more space when heated, the piston moves while maintaining the pressure constant, allowing us to use specific formulas to calculate the effect of temperature changes. A key feature of constant pressure processes is that they lead naturally to work being done by or on the system.

• When a gas expands at constant pressure, it does work on its surroundings.
• Conversely, when a gas is compressed, the surroundings do work on the gas.

Understanding the constant pressure process helps us apply the right equations, such as the Ideal Gas Law and other relevant thermodynamic relations, to solve for variables like work and temperature change.

Work Done by Gas

Work done by a gas under a constant pressure is a critical concept when dealing with thermodynamic processes. The equation for work, in this context, is given by:\[ W = nR(T_2 - T_1) \] Here:

• \( W \) represents the work done by the gas.
• \( n \) is the number of moles of gas.
• \( R \) is the gas constant, approximately 8.314 J/(mol K).
• \( T_2 \) and \( T_1 \) are the final and initial temperatures in Kelvin, respectively.

When the gas does positive work on the piston, as in our exercise, it means the gas expands against the piston. The piston movement is an outcome of the energy exerted by the gas transitioning between two temperature states.
Calculating the work done gives us insight into the amount of energy transferred due to this expansion, linking to the final temperature obtained by the gas, when pressure is constant.

Temperature Conversion

Temperature conversion is crucial when dealing with gases and thermodynamic processes as most formulas use Kelvin as the temperature unit. In this exercise, the initial temperature provided in Celsius needs conversion to Kelvin to maintain consistency with the Ideal Gas Law. To convert degrees Celsius to Kelvin:

• Add 273.15 to the Celsius temperature.
• So, a temperature of 27°C becomes 300.15 K. However, for simplification in calculations, it is often rounded to 300 K.

Kelvin is preferred in scientific calculations because it starts at absolute zero, where theoretically, particles have minimal thermal motion. This ensures there are no negative values, which helps with mathematical consistency.
Using Kelvin in thermodynamic calculations allows us to effectively apply laws like the appropriate equation to solve for temperature changes during the constant pressure process.