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Chapter 8: Problem 44

(a) Determine the specific enthalpy ( \(\mathrm{kJ} / \mathrm{mol}\) ) of \(n\) -pentane vapor at \(200^{\circ} \mathrm{C}\) and 2.0 atm relative to n-pentane liquid at \(20^{\circ} \mathrm{C}\) and \(1.0 \mathrm{atm}\), assuming ideal-gas behavior for the vapor. Show clearly the process path you construct for this calculation and give the enthalpy changes for each step. State where you used the ideal-gas assumption.

Short Answer

Expert verified
The specific enthalpy change of n-pentane has been found by adding the enthalpy change during change of phase and the change of state. Therefore, the result will be the sum of the values found in steps 2 and 3. Note that there is no enthalpy change associated with pressure change for an ideal gas.

Step by step solution

01

Identifying the Initial and Final States

Start by identifying the initial and final states of the n-pentane. Here, the initial state is liquid n-pentane at \(20^{\circ}C\) and \(1.0 atm\), and the final state is n-pentane vapor at \(200^{\circ}C\) and \(2.0 atm\).
02

Determination of the Enthalpy Change due to Vapourization

Next, determine the enthalpy change in transforming the n-pentane from its initial state to its gaseous state at the same temperature (i.e., \(20^{\circ}C\)). This is the enthalpy of vaporization, and it can be found in a standard thermodynamic table.
03

Enthalpy Change due to Heating

Now, find the enthalpy change when the n-pentane gas is heated from \(20^{\circ}C\) to \(200^{\circ}C\). This involves finding the specific heat capacity of n-pentane in its vapor state at constant pressure (Cp) from standard tables and using the formula 螖H = Cp螖T where 螖T is the change in temperature in Kelvin. Assume that Cp does not change with temperature.
04

Pressure Change

Since we are assuming n-pentane behaves as an ideal gas, changing the pressure from \(1.0atm\) to \(2.0atm\) at constant temperature does not affect the enthalpy. This is a consequence of the ideal gas assumption.
05

Total Enthalpy Change

Finally, find the total enthalpy change by adding the enthalpy changes calculated in steps 2 and 3. This is the specific enthalpy change of the n-pentane.

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Key Concepts

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

Specific Enthalpy
Specific enthalpy is a measure of the energy content of a substance per unit mass and is usually expressed in kilojoules per mole (kJ/mol). It is an essential concept in chemical thermodynamics when analyzing energy changes during a process. In our context, we're looking at the energy change involved in turning liquid n-pentane at 20掳C and 1 atm into vapor at 200掳C and 2 atm.
To calculate specific enthalpy, we need to understand the process steps involved: starting with the n-pentane liquid phase and ending in the vapor phase. Each phase transition and temperature change adds to the specific enthalpy.
  • Start with the liquid at the initial temperature and pressure.
  • Transform the liquid into vapor, which involves the enthalpy of vaporization.
  • Heat the vapor from initial to final temperature to determine additional enthalpy change.
Understanding these transformations helps us calculate the specific enthalpy of the system.
Ideal Gas Behavior
Ideal gas behavior is a simplifying assumption used in many calculations of gas properties. It assumes gases have no intermolecular forces and occupy no volume, behaving according to the ideal gas law: \[ PV = nRT \]where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the gas constant, and \( T \) is temperature. These assumptions allow us to predict and calculate properties more straightforwardly, particularly when dealing with enthalpy changes.
In the problem at hand, we assume that n-pentane vapor behaves ideally, especially when considering the pressure change from 1 atm to 2 atm.
  • Although real gases deviate from ideal behavior, at moderate pressures and high temperatures, these deviations are minimal.
  • Under the ideal gas assumption, changes in pressure at constant temperature don't impact enthalpy.
Using ideal gas behavior simplifies the calculations and provides a reasonable approximation for many gases under standard conditions.
Enthalpy of Vaporization
The enthalpy of vaporization is the energy required to transform a liquid into a gas at constant temperature and pressure. It's a crucial step in understanding the thermodynamics of phase changes.
In our problem, you start with liquid n-pentane at 20掳C and 1 atm, and you need to vaporize it while maintaining the temperature.
  • This step involves looking up n-pentane's enthalpy of vaporization in a thermodynamic data table.
  • This value represents the amount of energy needed to break intermolecular forces holding the n-pentane molecules in the liquid state.
Once you have the enthalpy of vaporization, you can calculate the energy change associated with this phase transformation, making it an integral part of the overall enthalpy change.
Enthalpy Change
Enthalpy change represents the total heat content change in a system during a process at constant pressure. It's a cumulative measure that includes all the individual steps of transformations and temperature changes.
For our n-pentane example, enthalpy change involves several layers:
  • First, vaporize the liquid n-pentane, which involves the enthalpy of vaporization.
  • Next, heat the vapor from the initial temperature to the final temperature, considering the specific heat capacity.
  • Pressure change does not affect the enthalpy due to the ideal gas assumption.
By adding these enthalpy changes together, you obtain the total specific enthalpy change in the transition from the initial to the final state. This understanding is essential for practical applications like energy management and thermodynamic cycle analysis.

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Most popular questions from this chapter

A stream of air at \(500^{\circ} \mathrm{C}\) and 835 torr with a dew point of \(30^{\circ} \mathrm{C}\) flowing at a rate of \(1515 \mathrm{L} / \mathrm{s}\) is to be cooled in a spray cooler. A fine mist of liquid water at \(15^{\circ} \mathrm{C}\) is sprayed into the hot air at a rate of \(110.0 \mathrm{g} / \mathrm{s}\) and evaporates completely. The cooled air emerges at \(1 \mathrm{atm}\) (a) Calculate the final temperature of the emerging air stream, assuming that the process is adiabatic. (Suggestion: Derive expressions for the enthalpies of dry air and water at the outlet air temperature, substitute them into the energy balance, and use a spreadsheet to solve the resulting fourth-order polynomial equation.) (b) At what rate (kW) is heat transferred from the hot air feed stream in the spray cooler? What becomes of this heat? (c) In a few sentences, explain how this process works in terms that a high school senior could understand. Incorporate the results of Parts (a) and (b) in your explanation.

Fish and wildlife managers have determined that a sudden temperature increase greater than \(5^{\circ} \mathrm{C}\) would be harmful to the marine ecosystem of a river. Warmer waters contain less dissolved oxygen and cause organisms in a river to increase their metabolism; if the temperature increase is sudden, the organisms do not have time to adapt to the new environment and likely will die. (Changes in river temperatures of five degrees and more due to seasonal temperature variations are common, but those temperature changes are gradual.) A proposed chemical plant plans to use river water for process cooling. The river flows at a rate of \(15.0 \mathrm{m}^{3} / \mathrm{s}\) at a temperature of \(15^{\circ} \mathrm{C}\), and a fraction of it will be diverted to the plant. Preliminary calculations reveal that the cooling water will remove \(5.00 \times 10^{5} \mathrm{kJ} / \mathrm{s}\) of heat from the plant. A portion of the extracted water will evaporate from the plant into the atmosphere, and the remainder will be returned to the river at a temperature of \(35^{\circ} \mathrm{C}\). (a) Draw and completely label a flowchart of the process and prove that there is enough information available to calculate all of the unknown stream flow rates on the chart. (b) Estimate the fraction of the river flow that must be diverted to the plant and the percentage of the cooling water that evaporates. Assume that water has a constant heat capacity of \(4.19 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) and a heat of vaporization roughly that of water at the normal boiling point, and also assume that the specific enthalpy of the water vapor relative to liquid water at \(15^{\circ} \mathrm{C}\) equals the heat of vaporization. (c) Write (but don't evaluate) an expression for the enthalpy change neglected by the assumption about the specific enthalpy of the steam.

A sheet of cellulose acetate film containing 5.00 wt\% liquid acetone enters an adiabatic dryer where \(90 \%\) of the acetone evaporates into a stream of dry air flowing over the film. The film enters the dryer at \(T_{\mathrm{f} 1}=35^{\circ} \mathrm{C}\) and leaves at \(T_{\mathrm{f} 2}\left(^{\circ} \mathrm{C}\right) .\) The air enters the dryer at \(T_{\mathrm{al}}\left(^{\circ} \mathrm{C}\right)\) and 1.01 atm and exits the dryer at \(T_{\mathrm{a} 2}=49^{\circ} \mathrm{C}\) and 1 atm with a relative saturation of \(40 \% . C_{p}\) may be taken to be \(1.33 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) for dry film and \(0.129 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\) for liquid acetone. Make a reasonable assumption regarding the heat capacity of dry air. The heat of vaporization of acetone may be considered independent of temperature. Take a basis of \(100 \mathrm{kg}\) film fed to the dryer for the requested calculations. (a) Estimate the feed ratio [liters dry air (STP)/kg dry film]. (b) Derive an expression for \(T_{\mathrm{al}}\) in terms of the film temperature change, \(\left(T_{\mathrm{f} 2}-35\right),\) and use it to answer Parts (c) and (d). (c) Calculate the film temperature change if the inlet air temperature is \(120^{\circ} \mathrm{C}\). (d) Calculate the required value of \(T_{\mathrm{al}}\) if the film temperature falls to \(34^{\circ} \mathrm{C},\) and the value if it rises to \(36^{\circ} \mathrm{C}.\) (e) If you solved Parts (c) and (d) correctly, you found that even though the air temperature is consistently higher than the film temperature in the dryer, so that heat is always transferred from the air to the film, the film temperature can drop from the inlet to the outlet. How is this possible?

Estimate the heat of vaporization of diethyl ether at its normal boiling point using Trouton's rule and Chen's rule and compare the results with a tabulated value of this quantity. Calculate the percentage error that results from using each estimation. Then estimate \(\Delta \hat{H}_{\mathrm{v}}\) at \(100^{\circ} \mathrm{C}\) using Watson's correlation.

The specific internal energy of formaldehyde (HCHO) vapor at 1 atm and moderate temperatures is given by the formula $$\hat{U}(\mathrm{J} / \mathrm{mol})=25.96 T+0.02134 T^{2}$$ where \(T\) is in \(^{\circ} \mathrm{C}\) (a) Calculate the specific internal energies of formaldehyde vapor at \(0^{\circ} \mathrm{C}\) and \(200^{\circ} \mathrm{C}\). What reference temperature was used to generate the given expression for \(\hat{U} ?\) (b) The value of \(\hat{U}\) calculated for \(200^{\circ} \mathrm{C}\) is not the true value of the specific internal energy of formaldehyde vapor at this condition. Why not? (Hint: Refer back to Section 7.5a.) Briefly state the physical significance of the calculated quantity. (c) Use the closed system energy balance to calculate the heat (J) required to raise the temperature of 3.0 mol HCHO at constant volume from 0^0 C to 200^'C. List all of your assumptions. (d) From the definition of heat capacity at constant volume, derive a formula for \(C_{v}(T)\left[\mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]\) Then use this formula and Equation \(8.3-6\) to calculate the heat \((\) J) required to raise the temperature of 3.0 mol of HCHO(v) at constant volume from 0^ C to 200^'C. [You should get the same result you got in Part (c).]
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