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Chapter 20: Problem 46

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

Expert verified
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

01

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, \(\Delta S = nC_v \ln{\frac{T_2}{T_1}}\)
02

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 \(p_2\).
03

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 \(T_2\): \(T_2 = \frac{p_2V_2}{p_1V_1} T_1\)
04

Calculate the change in entropy

Now that we have found \(T_2\), we can calculate the change in entropy using the entropy change formula: \(\Delta S = nC_v \ln{\frac{T_2}{T_1}}\)
05

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.

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