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Chapter 12: Problem 64

An insulated container is divided into two equal parts. One part contains an ideal gas at pressure \(P\) and temperature \(T\), while the other part is a perfect vacuum. If a hole is opened between the two parts, find the change in the temperature of the gas.

Short Answer

Expert verified
The change in temperature of the gas as it expands into the vacuum is given by: \(\Delta T = \left(\frac{2}{2^{\gamma}} - 1\right)T_1\)

Step by step solution

01

Identify the given information

We are given the initial pressure \(P\) and the initial temperature \(T\) of the gas in one part of the container. The other part is a vacuum with no pressure. As the gas expands into the vacuum, we need to find the change in temperature.
02

Apply the ideal gas law

The ideal gas law can be given as: \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles of the gas, \(R\) is the ideal gas constant, and \(T\) is the temperature. For our problem, we can rewrite the ideal gas law as: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) where the subscripts \(1\) and \(2\) denote the initial and final states, respectively, and we made use of the fact that the number of moles and the gas constant remain the same. Since the volume of the container doubles when the gas expands, we have: \(V_2 = 2V_1\)
03

Apply the adiabatic condition

Since the container is insulated and there is no heat exchange, the process is adiabatic. For an adiabatic expansion of an ideal gas, we have the following relation: \(P_1V_1^{\gamma} = P_2V_2^{\gamma}\) where \(\gamma\) is the adiabatic index, which is a constant for a particular gas, usually greater than 1.
04

Find the final pressure and temperature

First, we need to find the final pressure \(P_2\). Using the adiabatic condition and the fact that \(V_2 = 2V_1\), we can write: \(P_2 = \frac{P_1V_1^{\gamma}}{(2V_1)^{\gamma}}\) Now we can simplify this equation: \(P_2 = \frac{P_1}{2^{\gamma}}\) Next, we will use the ideal gas law in the form \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) to find the final temperature \(T_2\): \(T_2 = \frac{P_2V_2}{P_1V_1}T_1 = \frac{2V_1}{V_1}\frac{P_1}{2^{\gamma}}T_1 = \frac{(2P_1)}{(2^{\gamma}P_1)} T_1\)
05

Calculate the change in temperature

We now have the final temperature in terms of the initial temperature and pressure. We can find the change in temperature, \(\Delta T\), by subtracting the initial temperature from the final temperature: \(\Delta T = T_2 - T_1 = \frac{(2P_1)}{(2^{\gamma}P_1)} T_1 - T_1\) The change in temperature of the gas as it expands into the vacuum is: \(\Delta T = \left(\frac{2}{2^{\gamma}} - 1\right)T_1\)

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Most popular questions from this chapter

The internal energy \(U\) is a unique function of any state because of change in \(U\) (A) does not depend upon path. (B) depends upon path. (C) corresponds to an adiabatic process. (D) corresponds to an isothermal process.

Two moles of an ideal gas at \(300 \mathrm{~K}\) were cooled at constant volume so that the pressure is reduced to half the initial value. As a result of heating at constant pressure, the gas has expanded till it attains the original temperature. The total heat absorbed by gas, if \(R\) is the gas constant (A) \(150 R\) (B) \(300 R\) (C) \(75 R\) (D) \(100 R\)

Two rigid boxes containing different ideal gases are placed on a table. Box \(A\) contains one mole of nitrogen at temperature \(T_{0}\), while Box \(B\) contains one mole of helium at temperature (7/3) \(T_{0}\). The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature. (Ignore the heat capacity of boxes.) Then, the final temperature of the gases, \(T_{f}\), in terms of \(T_{0}\) is (A) \(T_{f}=\frac{5}{2} T_{0}\) (B) \(T_{f}=\frac{3}{7} T_{0}\) (C) \(T_{f}=\frac{7}{3} T_{0}\) (D) \(T_{f}=\frac{3}{2} T_{0}\)

Which of the following statements is correct for any thermodynamic system? (A) The internal energy changes in all processes. (B) Internal energy and entropy are state functions. (C) The change in entropy can never be zero. (D) The work done in an adiabatic process is always zero.

A cylindrical tube of uniform cross-sectional area \(A\) is fitted with two air tight frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the pressure of the gas is \(P_{0}\) and temperature is \(T_{0}\). Atmospheric pressure is also \(P_{0}\). Now the temperature of the gas is increased to \(2 T_{0}\), the tension in the wire will be (A) \(2 P_{0} A\) (B) \(P_{0} A\) (C) \(\frac{P_{0} A}{2}\) (D) \(4 P_{0} A\)
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