Problem 19 - University Physics with Modern Physics

Question

A cylinder contains 0.250 mol of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas at a temperature of \(27.0^{\circ} \mathrm{C}\) . The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 Atm on the gas. The gas is heated until its temperature increases to \(127.0^{\circ} \mathrm{C}\) . Assume that the \(\mathrm{CO}_{2}\) may be treated as an ideal gas. (a) Draw a \(p V\) -diagram for this process. (b) How much work is done by the gas in this process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 atm?

Step-by-Step Solution

Answer

(a) pV-diagram: horizontal line. (b) 207.85 J. (c) On the piston. (d) 519.625 J. (e) 727.475 J. (f) 103.925 J.

Step 1: Analyze the Problem

We are given a gas cylinder containing 0.250 mol of CO2 at an initial temperature of 27.0°C with a constant pressure of 1.00 atm, which is heated to 127.0°C. We will apply the ideal gas law and thermodynamic equations to solve this problem. Key variables:- Initial temperature \( T_1 = 300 \, K \)- Final temperature \( T_2 = 400 \, K \)- Universal gas constant \( R = 8.314 \, J/(mol \cdot K) \)

Step 2: Draw a pV-Diagram

Since the pressure is constant, the process is isobaric and represented by a horizontal line in the \( pV \)-diagram. As the temperature increases, the volume will increase (Charles's Law). The horizontal line extends rightward from the initial to the final states, indicating an increase in volume.

Step 3: Calculate Work Done by the Gas

For an isobaric process, work done \( W = P \Delta V \). Use the ideal gas equation in terms of volume change:\[ W = nR(T_2 - T_1) \] Given:\( n = 0.250 \, mol \)\( T_1 = 300 \, K \)\( T_2 = 400 \, K \)\[ W = 0.250 \times 8.314 \times (400 - 300) = 207.85 \, J \]

Step 4: Identify Work Done On

The work done by the gas is on the piston as the gas expands, pushing the piston upward.

Step 5: Calculate Change in Internal Energy

The change in internal energy \( \Delta U \) can be calculated using the formula:\[ \Delta U = nC_V\Delta T \]The molar heat capacity at constant volume \( C_V = \frac{5}{2} R \):\[ \Delta U = 0.250 \times \frac{5}{2} \times 8.314 \times 100 = 519.625 \, J \]

Step 6: Calculate Heat Supplied

Using the first law of thermodynamics:\[ \Delta U = Q - W \]Solve for \( Q \):\[ Q = \Delta U + W = 519.625 + 207.85 = 727.475 \, J \]

Step 7: Calculate Work at 0.50 atm Pressure

If the pressure is 0.50 atm, work done on the piston is reduced proportionally to the pressure. Repeat the work calculation with the reduced pressure (1 atm = 101.3 kPa, therefore 0.5 atm = 50.65 kPa):\[ W = P \Delta V \rightarrow W = 0.5 \times 207.85 = 103.925 \, J \]

Key Concepts

Isobaric Process

An isobaric process is where the pressure remains constant while other parameters like volume and temperature change. In thermodynamics, this type of process is particularly important as it often represents real-world situations, especially in systems with pistons, like engines.

When a gas undergoes an isobaric process, the work done by the gas can be calculated using the equation: W = P \Delta V

Here, \(W\) is the work done, \(P\) is the constant pressure, and \(\Delta V\) is the change in volume. In an isobaric process, if the volume increases, the gas does positive work as it expands; if it decreases, the gas does negative work. In the context of the ideal gas law \(PV=nRT\), keeping \(P\) constant means changes in the volume \(V\) are directly related to temperature changes \(T\).

Thermodynamics

Thermodynamics is the study of energy, heat, and the laws governing their interactions. This branch of physics deals with how these forms of energy transform between one another and how they affect matter. One crucial aspect of thermodynamics is understanding different types of processes such as isobaric, isochoric, isothermal, and adiabatic processes, all of which describe energy exchanges under specific conditions.

In a thermodynamic system, particularly in isobaric systems, the state of the gas can be represented by parameters such as pressure, volume, and temperature, which can be shown on a \( pV \)-diagram. Here, any change in one state function affects the others, like changing temperature (resulting, in this context, to an increase in volume while pressure remains static). The area under a \( pV \)-diagram curve represents the work done by or on the gas during the process.

First Law of Thermodynamics

The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed. Instead, it can only be transferred or transformed from one form to another. Mathematically, this is expressed in most thermodynamic processes as: \Delta U = Q - W Where \(\Delta U\) is the change in internal energy of the system, \(Q\) is the heat added to the system, and \(W\) is the work done by the system.

This equation is central for understanding how energy flows within a system, whether it be a gas in a piston or any closed thermodynamic system. It allows us to compute changes in state functions during processes like heating or cooling. In our real-world example of a cylinder with CO₂ gas, the work done by the gas, any heat supplied, and the resulting change in its internal energy can be perfectly analyzed using this fundamental law.