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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

An aqueous slurry at \(30^{\circ} \mathrm{C}\) containing \(20.0 \mathrm{wt} \%\) solids is fed to an evaporator in which enough water is vaporized at 1 atm to produce a product slurry containing 35.0 wt\% solids. Heat is supplied to the evaporator by feeding saturated steam at 2.6 bar absolute into a coil immersed in the liquid. The steam condenses in the coil, and the slurry boils at the normal boiling point of pure water. The heat capacity of the solids may be taken to be half that of liquid water. (a) Calculate the required steam feed rate ( \(\mathrm{kg} / \mathrm{h}\) ) for a slurry feed rate of \(1.00 \times 10^{3} \mathrm{kg} / \mathrm{h}\). (b) Vapor recompression is often used in the operation of an evaporator. Suppose that the vapor (steam) generated in the evaporator described above is compressed to 2.6 bar and simultaneously heated to the saturation temperature at 2.6 bar, so that no condensation occurs. The compressed steam and additional saturated steam at 2.6 bar are then fed to the evaporator coil, in which isobaric condensation occurs. How much additional steam is required? (c) What more would you need to know to determine whether or not vapor recompression is economically advantageous in this process?

The heat required to raise the temperature of \(m\) (kg) of a liquid from \(T_{1}\) to \(T_{2}\) at constant pressure is $$ Q=\Delta H=m \int_{T_{1}}^{T_{2}} C_{p}(T) d T $$ In high school and in first-year college physics courses, the formula is usually given as $$ Q=m C_{p} \Delta T=m C_{p}\left(T_{2}-T_{1}\right) $$ (a) What assumption about \(C_{p}\) is required to go from Equation 1 to Equation \(2 ?\) (b) The heat capacity \(\left(C_{p}\right)\) of liquid \(n\) -hexane is measured in a bomb calorimeter. A small reaction flask (the bomb) is placed in a well- insulated vessel containing \(2.00 \mathrm{L}\) of liquid \(n-\mathrm{C}_{6} \mathrm{H}_{14}\) at \(T=300 \mathrm{K} .\) A combustion reaction known to release \(16.73 \mathrm{kJ}\) of heat takes place in the bomb, and the subsequent temperature rise of the system contents is measured and found to be \(3.10 \mathrm{K}\). In a separate experiment, it is found that \(6.14 \mathrm{kJ}\) of heat is required to raise the temperature of everything in the system except the hexane by \(3.10 \mathrm{K}\). Use these data to estimate \(C_{p}[\mathrm{kJ} /(\mathrm{mol} \cdot \mathrm{K})]\) for liquid \(n\) -hexane at \(T \approx 300 \mathrm{K},\) assuming that the condition required for the validity of Equation 2 is satisfied. Compare your result with a tabulated value.

A gas stream containing \(n\) -hexane in nitrogen with a relative saturation of \(90 \%\) is fed to a condenser at \(75^{\circ} \mathrm{C}\) and 3.0 atm absolute. The product gas emerges at \(0^{\circ} \mathrm{C}\) and 3.0 atm at a rate of \(746.7 \mathrm{m}^{3} / \mathrm{h}\). (a) Calculate the percentage condensation of hexane (moles condensed/mole fed) and the rate \((\mathrm{kW})\) at which heat must be transferred from the condenser. (b) Suppose the feed stream flow rate and composition and the heat transfer from the condenser are the same as in Part (a), but the condenser and outlet stream pressure is only 2.5 atm instead of 3.0 atm. How would the outlet stream temperatures and flow rates and the percentage condensations of hexane calculated in Parts (a) and (b) change (increase, decrease, no change, no way to tell)? Don't do any calculations, but explain your reasoning.

A stream of air at \(77^{\circ} \mathrm{F}\) and 1.2 atm absolute flowing at a rate of \(225 \mathrm{ft}^{3} / \mathrm{h}\) is blown through ducts that pass through the interior of a large industrial motor. The air emerges at \(500^{\circ} \mathrm{F}\). Calculate the rate at which the air is removing heat generated by the motor. What assumption have you made about the pressure dependence of the specific enthalpy of air?

The heat capacity at constant pressure of hydrogen cyanide is given by the expression $$ C_{p}\left[J /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]=35.3+0.0291 T\left(^{\circ} \mathrm{C}\right) $$ (a) Write an expression for the heat capacity at constant volume for HCN, assuming ideal-gas behavior. (b) Calculate \(\Delta \hat{H}(\mathrm{J} / \mathrm{mol})\) for the constant- pressure process $$ \mathrm{HCN}\left(\mathrm{v}, 25^{\circ} \mathrm{C}, 0.80 \mathrm{atm}\right) \rightarrow \mathrm{HCN}\left(\mathrm{v}, 200^{\circ} \mathrm{C}, 0.80 \mathrm{atm}\right) $$(c) Calculate \(\Delta \hat{U}(\mathrm{J} / \mathrm{mol})\) for the constant- volume process $$\mathrm{HCN}\left(\mathrm{v}, 25^{\circ} \mathrm{C}, 50 \mathrm{m}^{3} / \mathrm{kmol}\right) \rightarrow \mathrm{HCN}\left(\mathrm{v}, 200^{\circ} \mathrm{C}, 50 \mathrm{m}^{3} / \mathrm{kmol}\right)$$ (d) If the process of Part (b) were carried out in such a way that the initial and final pressures were each 0.80 atm but the pressure varied during the heating, the value of \(\Delta \hat{H}\) would still be what you calculated assuming a constant pressure. Why is this so?
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