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Chapter 19: Problem 32

A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 \(\times\) 10\(^5 \)Pa and occupies a volume of 2.50 \(\times\) 10\(^{-3}\) m\(^3\). (a) Find the initial temperature of the gas in kelvins. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.

Short Answer

Expert verified
(a) Initial temperature: 300.5 K. (b) Final conditions: (i) Isothermal: 300.5 K, 0.50 脳 10鈦� Pa; (ii) Isobaric: 601.0 K, 1.00 脳 10鈦� Pa; (iii) Adiabatic: 238.7 K, 0.316 脳 10鈦� Pa.

Step by step solution

01

Understand Ideal Gas Law

The Ideal Gas Law is given by the equation \( PV = nRT \). Here, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant (8.314 J/(mol路K)), and \( T \) is the temperature in Kelvin.
02

Solve for Initial Temperature

To find the initial temperature \( T_i \), rearrange the ideal gas law: \( T_i = \frac{PV}{nR} \). Plug in the given values: \( P = 1.00 \times 10^5 \) Pa, \( V = 2.50 \times 10^{-3} \) m\(^3\), \( n = 0.100 \) mol, and \( R = 8.314 \) J/(mol路K), to get \( T_i = \frac{(1.00 \times 10^5)(2.50 \times 10^{-3})}{0.100 \, \cdot 8.314} \).
03

Calculate Initial Temperature

Perform the calculations: \( T_i = \frac{2.5 \times 10^2}{0.8314} \approx 300.5 \) K.
04

Solve Isothermal Expansion (i)

For isothermal expansion, the temperature remains constant. Therefore, \( T_f = T_i = 300.5 \) K. Using \( P_i V_i = P_f V_f \), rearrange to find \( P_f \): \( P_f = \frac{P_i V_i}{V_f} \). Substitute \( V_f = 2 \times 2.50 \times 10^{-3} = 5.00 \times 10^{-3} \) m\(^3\) to get \( P_f = \frac{1.00 \times 10^5 \cdot 2.50 \times 10^{-3}}{5.00 \times 10^{-3}} = 0.50 \times 10^5 \) Pa.
05

Solve Isobaric Expansion (ii)

For an isobaric process, the pressure remains constant. So \( P_f = P_i = 1.00 \times 10^5 \) Pa. Using the ideal gas law \( \frac{T_f}{T_i} = \frac{V_f}{V_i} \), calculate \( T_f = T_i \cdot \frac{V_f}{V_i} = 300.5 \cdot \frac{5.00 \times 10^{-3}}{2.50 \times 10^{-3}} = 601.0 \) K.
06

Solve Adiabatic Expansion (iii)

For a monatomic gas, the adiabatic condition is given by \( PV^\gamma = \text{constant} \), where \( \gamma = \frac{5}{3} \). Using \( T_f = T_i \cdot \left(\frac{V_i}{V_f}\right)^{\gamma-1} \), calculate \( T_f = 300.5 \cdot \left(\frac{2.50 \times 10^{-3}}{5.00 \times 10^{-3}}\right)^{2/3} \approx 238.7 \) K. For \( P_f \), use \( P_f = P_i \cdot \left(\frac{V_i}{V_f}\right)^\gamma \approx 0.316 \times 10^5 \) Pa.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isothermal Process
In an isothermal process, the temperature remains constant throughout the expansion or compression of the gas. This means, that for an isothermal expansion, the initial and final temperatures are equal. According to the ideal gas law, when temperature is constant,
  • The product of pressure (\(P\)) and volume (\(V\)) remains constant: \(PV = ext{constant}\).
  • For the expansion of the gas to twice its initial volume, \(V_f = 2V_i\).
  • Using this property, we can find the final pressure:\(P_f = \frac{P_i V_i}{V_f}\).
The most important aspect of the isothermal process is the constancy of temperature; hence, internal energy changes only through work done by or on the gas. This process requires perfect thermal equilibrium with the surroundings to keep the temperature unchanged.
Isobaric Process
In an isobaric process, the pressure remains constant. This phenomenon affects other gas properties as they change according to the ideal gas law. Since the pressure does not change:
  • The relationship between temperature and volume becomes directly proportional.
  • Using the equation \(\frac{T_f}{T_i} = \frac{V_f}{V_i}\), it shows that the final temperature is determined by the change in volume.
For the gas to expand to twice its initial volume, the ideal gas law simplifies the calculation for the final state properties given constant pressure. During such processes, thermal energy is absorbed or released to maintain constant pressure, leading to a change in the gas's volume and temperature.
Adiabatic Expansion
Adiabatic processes are distinctive because they occur without the transfer of heat between the system and its surroundings. For a monatomic ideal gas, the key equations include:
  • The adiabatic equation: \(PV^\gamma = ext{constant}\), where \(\gamma = \frac{5}{3}\) (ratio of specific heats for a monatomic gas).
  • Temperature changes even though no heat enters or leaves the system.
  • Final temperature can be determined with \(T_f = T_i \left(\frac{V_i}{V_f}\right)^{\gamma-1}\).
Since adiabatic processes occur without heat exchange, changes in internal energy are driven solely by work done on or by the gas. This process is significant in understanding how temperature and pressure can drastically change together in closed systems.

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Most popular questions from this chapter

A cylinder contains 0.250 mol of carbon dioxide (\(CO_2\)) gas at a temperature of 27.0\(^\circ\)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\(^\circ\)C. Assume that the CO\(_2\) may be treated as an ideal gas. (a) Draw a \(pV\)-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?

In another test, the valve of a 500-L cylinder full of the gas mixture at 2000 psi (gauge pressure) is opened wide so that the gas rushes out of the cylinder very rapidly. Why might some \(N_2O\) condense during this process? (a) This is an isochoric process in which the pressure decreases, so the temperature also decreases. (b) Because of the rapid expansion, heat is removed from the system, so the internal energy and temperature of the gas decrease. (c) This is an isobaric process, so as the volume increases, the temperature decreases proportionally. (d) With the rapid expansion, the expanding gas does work with no heat input, so the internal energy and temperature of the gas decrease.

Two moles of an ideal monatomic gas go through the cycle \(abc\). For the complete cycle, 800 J of heat flows out of the gas. Process \(ab\) is at constant pressure, and process \(bc\) is at constant volume. States \(a\) and \(b\) have temperatures \(T_a\) = 200 K and \(T_b\) = 300 K. (a) Sketch the \(pV\)-diagram for the cycle. (b) What is the work \(W\) for the process \(ca\)?

You have a cylinder that contains 500 L of the gas mixture pressurized to 2000 psi (gauge pressure). A regulator sets the gas flow to deliver 8.2 \(L\)/min at atmospheric pressure. Assume that this flow is slow enough that the expansion is isothermal and the gases remain mixed. How much time will it take to empty the cylinder? (a) 1 h; (b) 33 h; (c) 57 h; (d) 140 h.

During an isothermal compression of an ideal gas, 410 J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?
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