Problem 50 - University Physics with Modern Physics

Question

Two moles of helium are initially at a temperature of \(27.0^{\circ} \mathrm{C}\) and occupy a volume of 0.0300 \(\mathrm{m}^{3} .\) The helium first expands at constant pressure until its volume has doubled. Then it expands adiabatically until the temperature returns to its initial value. Assume that the helium can be treated as an ideal gas. (a) Draw a diagram of the process in the \(p V\) -plane. (b) What is the total heat supplied to the helium in the process? (c) What is the total change in internal energy of the helium? (d) What is the total work done by the helium? (e) What is the final volume of the helium?

Step-by-Step Solution

Answer

The total heat supplied is 12471 J, change in internal energy is 0 J, total work done is 4988.4 J, and final volume is 0.1697 m³.

Step 1: Understanding the Initial Conditions

We start with two moles of helium initially at a temperature of \(27.0^{\circ} \mathrm{C}\) (which is \(300 \mathrm{K}\)) and a volume of \(0.0300 \mathrm{m}^{3}\). The pressure can be found using the ideal gas law \(PV = nRT\). With \(n = 2\) moles and \(R = 8.314 \, \mathrm{J/(mol \cdot K)}\), we find the initial pressure.

Step 2: Finding Initial Pressure

Using \(P_1V_1 = nRT_1\) for the initial state:\[ P_1 \times 0.0300 = 2 \times 8.314 \times 300 \]Solving for \(P_1\), we find:\[ P_1 = \frac{2 \times 8.314 \times 300}{0.0300} = 166280 \, \mathrm{Pa} \] or \(166.28 \, \mathrm{kPa}\).

Step 3: Process 1 - Constant Pressure Expansion

The helium expands at constant pressure until its volume doubles. The work done \(W_1\) during this is given by \(W_1 = P_1 \Delta V\), where \(\Delta V = V_2 - V_1 = 0.0300 \, \mathrm{m}^3 \). Therefore:\[W_1 = 166280 \times 0.0300 = 4988.4 \, \mathrm{J}\].

Step 4: Adiabatic Expansion

In an adiabatic process, \(Q = 0\), meaning no heat is exchanged, and changes occur in temperature and volume to keep \(TV^{\gamma-1}\) a constant (\(\gamma = \frac{5}{3}\) for helium). With \(T\) returning to 300 K, use \(T_3 V_3^{\gamma-1} = T_2 V_2^{\gamma-1}\) to find \(V_3\).

Step 5: Solving for Final Volume

Using \((T_3/T_2) = (V_2/V_3)^{\gamma-1}\), rearrange to find \(V_3\).\[ V_3 = V_2 \left( \frac{T_2}{T_3} \right)^{\frac{1}{\gamma-1}} \].Solving gives:\[ V_3 = 0.0600 \left( \frac{T_2}{300} \right)^{\frac{1}{\frac{5}{3} - 1}} \].Using \(T_2 = 600 \mathrm{K}\) from isothermal calculations:\[ V_3 = 0.0600 \times 2^{1.5} = 0.1697 \, \mathrm{m}^3\].

Step 6: Calculating the Total Heat Supplied

The only heat exchange occurs during the constant pressure expansion. Using \(Q = nC_p \Delta T\), with \(C_p = \frac{5}{2}R\) for helium:\[ Q = 2 \times \frac{5}{2} \times 8.314 \times (600 - 300)\], which gives \(12471 \, \mathrm{J}\).

Step 7: Calculating the Total Change in Internal Energy

The change in internal energy \( \Delta U = nC_v \Delta T \) with \( C_v = \frac{3}{2}R\):\[ \Delta U = 2 \times \frac{3}{2} \times 8.314 \times (300 - 300) = 0 \mathrm{J} \].

Step 8: Calculating the Total Work Done by the Helium

The total work done \(W\) is the sum of work done in each process. Since only the first process involves pressure-volume work:\(W = W_1 = 4988.4 \, \mathrm{J}\).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in thermodynamics that describes the relationship between the pressure, volume, temperature, and amount of an ideal gas. It is expressed as:

• \( PV = nRT \)

Where:

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

In the given exercise, helium gas is treated as an ideal gas, allowing the use of the Ideal Gas Law to find the initial pressure and solve various parts of the problem. It helps predict how the gas will behave under different conditions and forms the basis for understanding processes like expansion and compression.

Adiabatic Process

An adiabatic process is one in which no heat is exchanged between the system and its surroundings. In other words, the heat transfer \( Q = 0 \). During an adiabatic process, changes in temperature and volume occur without any heat loss or gain.

• A key feature of adiabatic processes is the relationship \( TV^{\gamma-1}= \text{constant} \)

Here, \( \gamma \) is the heat capacity ratio \( \frac{C_p}{C_v} \), which is \( \frac{5}{3} \) for helium because it is a monoatomic ideal gas.
In the exercise, after expanding at constant pressure, the helium undergoes adiabatic expansion. This means volume and temperature changes are governed by the adiabatic relationship until the temperature returns to its initial value. By leveraging the relationship between temperature and volume during adiabatic expansion, the final volume, after all changes, is determined.

Internal Energy

Internal energy refers to the energy contained within a system that is associated with the random motion of molecules. For ideal gases, it depends only on temperature. The change in internal energy is given by:

• \( \Delta U = nC_v \Delta T \)

Where:

• \( \Delta U \) is the change in internal energy
• \( n \) is the number of moles
• \( C_v \) is the molar specific heat at constant volume, given as \( \frac{3}{2}R \) for monoatomic gases like helium
• \( \Delta T \) is the change in temperature

In this exercise, the temperature of the helium returns to its original state after the adiabatic expansion, indicating \( \Delta T = 0 \). Therefore, there is no change in internal energy, and \( \Delta U = 0 \) J. This concept helps understand energy conservation during ideal gas transformations.

Work Done by Gas

The work done by a gas during a thermodynamic process can be determined by considering the pressure and change in volume. For a process at constant pressure, the work done \( W \) is given by:

• \( W = P \Delta V \)

Where:

• \( P \) is the constant pressure during the process
• \( \Delta V \) is the change in volume

In the provided exercise, the helium expands at constant pressure first, leading to the calculation of the work done as \( W = 4988.4 \) J. This work is significant because it represents the energy transferred from the gas by doing work on its surroundings. Understanding work done by gas helps in analyzing energy transformations and mechanical work conversions within thermodynamic cycles.