Question

An ideal gas is enclosed in a cylinder with a movable piston at the top. The walls of the cylinder are insulated, so no heat can enter or exit. The gas initially occupies volume \(V_{1}\) and has pressure \(p_{1}\) and temperature \(T_{1}\). The piston is then moved very rapidly to a volume of \(V_{2}=3 V_{1}\). The process happens so rapidly that the enclosed gas does not do any work. Find \(p_{2}, T_{2},\) and the change in entropy of the gas.

Short answer

Question: During an adiabatic process, a gas is compressed such that its final volume is one-third of its initial volume. Determine the final pressure, final temperature, and change in entropy of the gas. Answer: To find the final pressure, final temperature, and change in entropy of the gas, use the step by step solution outlined above. The first formula calculates the final pressure, the second calculates the final temperature, and the last calculates the change in entropy. By inputting the initial values and working through the solution steps, you can determine the final pressure, final temperature, and change in entropy of the enclosed gas under the adiabatic process conditions.

Step-by-step solution

Write down the relevant formulas

The ideal gas equation is given by: \(PV=nRT\) Since the process is adiabatic, \(\Delta Q = 0\) Also, for an adiabatic process, \(p_{1}V_{1}^\gamma = p_{2}V_{2}^\gamma\) For entropy change, \(\Delta S = nC_v \ln{\frac{T_2}{T_1}}\)

Use the given information and ideal gas equation

Since \(V_2 = 3V_1\), we can plug this into the adiabatic process equation: \(p_1V_1^\gamma = p_2(3V_1)^\gamma\) Since the question asks for the final pressure \(p_2\), we want to solve for \(p_2\). \(p_2 = \frac{V_1^\gamma}{(3V_1)^\gamma} p_1\) With this equation, we can find the value of \(p_2\).

Find the final temperature

To find the final temperature \(T_2\), note that \(nR\) is constant in the ideal gas equation. So, \(\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}\) We've already found \(p_2\). We can plug in the values of \(V_1\) and \(V_2\), then solve for \(T_2\): \(T_2 = \frac{p_2V_2}{p_1V_1} T_1\)

Calculate the change in entropy

Now that we have found \(T_2\), we can calculate the change in entropy using the entropy change formula: \(\Delta S = nC_v \ln{\frac{T_2}{T_1}}\)

Conclusion

In this problem, we used the ideal gas equation, the adiabatic process equation, and entropy change formula to find \(p_2\), \(T_2\), and the change in entropy of the enclosed ideal gas. By carefully following the steps outlined above, the solution can be obtained.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in physics and chemistry, especially when dealing with gases. It describes the relationship between the pressure (\(P\)), volume (\(V\)), and temperature (\(T\)) of an ideal gas, which can be expressed as \(PV = nRT\). Here, \(n\) is the number of moles of the gas, and \(R\) is the universal gas constant.

The law assumes the gas being discussed behaves ideally, meaning that the gas particles have no volume and no intermolecular forces act between them. For most real gases under normal conditions, this assumption holds approximately true, allowing the Ideal Gas Law to serve as a good approximation.

In our exercise, the initial conditions of the gas—volume \(V_1\), pressure \(p_1\), and temperature \(T_1\)—are known, and the law helps relate these initial conditions to the final ones after changes in volume (and hence pressure and temperature) during adiabatic expansion.

Entropy Change

Entropy measures the dispersal of energy in a system and is often linked closely to the concept of disorder. The change in entropy (\(\Delta S\)) is a critical concept in thermodynamics, especially when assessing processes like expansion or compression of gases.

An important aspect of entropy change is that, in an isolated system, the total entropy either increases or remains unchanged. For our case, the formula \(\Delta S = nC_v \ln\left(\frac{T_2}{T_1}\right)\) is used to calculate the change in entropy of the gas. Here, \(C_v\) is the heat capacity at constant volume, and \(\ln\) denotes the natural logarithm.

Since the process is adiabatic, there is no heat exchange (\(\Delta Q = 0\)), and any entropy change is solely due to internal transformations within the gas system, observed through the temperature change from \(T_1\) to \(T_2\).

Adiabatic Expansion

An adiabatic expansion refers to a process where a gas expands without exchanging heat with its surroundings. In an ideal adiabatic process, the walls of the system are perfectly insulated, as noted in the problem statement.

During such an expansion, the gas' own internal energy is used to do work on the system, which results in a drop in temperature unless energy is added by external pressure. This concept is governed by the equation \(p_1V_1^\gamma = p_2V_2^\gamma\), where \(\gamma\) is the adiabatic index or ratio of heat capacities (\(C_p/C_v\)).

Because of this property, the process affects both the pressure and temperature of the gas. As the volume of the gas increases, the pressure decreases, and the temperature may drop depending on the work done during the expansion. This connectivity is precisely why the adiabatic process can be utilized to derive \(p_2\) without direct heat input or measurement.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships and conversions between heat and other forms of energy. It provides a comprehensive framework for analyzing energy transfers in systems, particularly involving processes like the expansion or compression of gases, heat engines, and refrigerators.

There are four laws of thermodynamics, but in this context, the first and second laws are especially important.

• The first law, or the law of energy conservation, notes that the energy change in a closed system is equal to the heat added minus the work done by the system. In adiabatic processes, \(Q = 0\), so the work done is directly tied to changes in internal energy.
• The second law introduces the concept of entropy, emphasizing that the entropy of an isolated system can only increase over time.

In our exercise, these laws help us understand why the system behaves as it does when the gas undergoes rapid expansion without external heat interaction, thereby explaining the drop in temperature and change in entropy.