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Chapter 8: Problem 1

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).]

Short Answer

Expert verified
The internal energy of formaldehyde vapor at \(0^{\circ}C\) and \(200^{\circ}C\) can be calculated using the given expression. The calculated values are not the true values of the specific internal energy because it does not include the changes in state. The heat required to raise the temperature from \(0^{\circ}C\) to \(200^{\circ}C\) can be found using the energy balance and verification can be done using the formula of heat capacity at constant volume.

Step by step solution

01

Calculate Specific Internal Energies

Substitute \(T\) with \(0^{\circ} C\) and \(200^{\circ} C\) in the given formula to get internal energies at these temperatures. Also, look for the reference temperature where \(\hat{U} = 0\).
02

Discuss Physical Significance

The physical significance of the calculated quantity is the amount of energy required to heat formaldehyde vapor from the reference temperature to the given temperature at constant volume. It is not the true value because it does not account for the changes in state (from liquid to vapor).
03

Calculate Heat Required

The heat (q) required to raise the tempreature from \(0^{\circ} C\) to \(200^{\circ} C\) can be calculated by using the relation \(q = \Delta U = U(200^{\circ} C) - U(0^{\circ} C)\), where \(U(T)\) is the internal energy calculated using given formula.
04

Derive and Use Heat Capacity Formula

From the definition of heat capacity at constant volume (\(Cv\)), \(Cv = (\Delta U / \Delta T)\). Integration of this formula from \(T1\) to \(T2\) gives the heat required. Cross verify this value with the one calculated in step 3.

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Key Concepts

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

Closed System Energy Balance
The concept of a closed system energy balance is central to understanding how energy transforms within a system that does not exchange mass with its surroundings. Consider a sealed container of gas; even though heat may enter or leave, the mass of the gas remains constant. When we apply this to calculating the change in the internal energy of a substance like formaldehyde vapor, the first step is to quantify the initial and final internal energy levels under constant volume conditions.

For instance, when raising the temperature of formaldehyde vapor from 0掳C to 200掳C, the change in specific internal energy (\(\frac{\text{J}}{\text{mol}}\)) can be found using the provided temperature-dependent function for \(\text{\(\text{U}\)}}\). You simply calculate the internal energy at both temperatures, and their difference reflects the net energy transfer in the form of heat (assuming no work is done, as is the case at constant volume). This straightforward approach is an illustration of the conservation of energy principle, fundamental to all closed system energy balances.
Heat Capacity at Constant Volume
Heat capacity at constant volume (\(C_v(T)\)) is an intrinsic property of a material that indicates how much heat energy is required to raise the temperature of a unit amount of substance by one degree Celsius at constant volume.

Derivation of \(C_v\)

From thermodynamics, the heat capacity is defined by the rate of change of internal energy with temperature. Mathematically, this is presented as \(C_v = \frac{\text{d}U}{\text{d}T}\), where \(U\) is internal energy and \(T\) is temperature. For a temperature-dependent heat capacity, this would involve integrating the expression over the temperature range of interest.

The close relationship between heat capacity and internal energy is why calculations from steps involving heat capacity should corroborate the findings from an energy balance, emphasizing the interconnected nature of these concepts.
Temperature Effects on Internal Energy
Internal energy is deeply affected by temperature changes, particularly in gases. The relationship between the two can often be described by a polynomial where each term represents energy contributions at different temperature conditions. In the case of formaldehyde vapor, the given formula shows a linear and a squared term as it relates to temperature, indicating the energy changes more rapidly at higher temperatures.

Understanding the Polynomial

The linear term (proportional to \(T\)) suggests a direct change with temperature, while the squared term (proportional to \(T^2\)) reveals a progressively greater impact as temperature increases. Together, these terms manifest the specific internal energy variations, which guide us in calculating the thermal energy required for a temperature-induced process, such as heating formaldehyde gas from one temperature to another. This understanding is crucial for accurately gauging the energy dynamic in response to temperature fluctuations.

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Most popular questions from this chapter

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?

The brakes on an automobile act by forcing brake pads, which have a metal support and a lining, to press against a disk (rotor) attached to the wheel. Friction between the pads and the disk causes the car to slow or stop. Each wheel has an iron brake disk with a mass of \(15 \mathrm{lb}_{\mathrm{m}}\) and two brake pads, each having a mass of \(11 \mathrm{b}_{\mathrm{m}}\). (a) Suppose an automobile is moving at 55 miles per hour when the driver suddenly applies the brakes and brings the car to a rapid halt. Take the heat capacity of the disk and brake pads to be \(0.12 \mathrm{Btu} /\left(\mathrm{lb}_{\mathrm{m}} \cdot^{\circ} \mathrm{F}\right)\) and assume that the car stops so rapidly that heat transfer from the disk and pads has been insignificant. Estimate the final temperature of the disk and pads if the car is (i) a Toyota Camry, which has a mass of about \(3200 \mathrm{Ib}_{\mathrm{m}},\) or (ii) a Cadillac Escalade, which has a mass of about \(5.900 \mathrm{lb}_{\mathrm{m}}.\) (b) Why are the linings on brake pads no longer made of asbestos? Your answer should provide information on specific issues or concerns caused by the use of asbestos.

On a cold winter day the temperature is \(2^{\circ} \mathrm{C}\) and the relative humidity is \(15 \% .\) You inhale air at an average rate of \(5500 \mathrm{mL} / \mathrm{min}\) and exhale a gas saturated with water at body temperature, roughly \(37^{\circ} \mathrm{C} .\) If the mass flow rates of the inhaled and exhaled air (excluding water) are the same, the heat capacities \(\left(C_{p}\right)\) of the water-free gases are each \(1.05 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right),\) and water is ingested into the body as a liquid at \(22^{\circ} \mathrm{C},\) at what rate in \(\mathrm{J} /\) day do you lose energy by breathing? Treat breathing as a continuous process (inhaled air and liquid water enter, exhaled breath exits) and neglect work done by the lungs.

Propane gas enters a continuous adiabatic heat exchanger \(^{17}\) at \(40^{\circ} \mathrm{C}\) and \(250 \mathrm{kPa}\) and exits at \(240^{\circ} \mathrm{C}\). Superheated steam at \(300^{\circ} \mathrm{C}\) and 5.0 bar enters the exchanger flowing countercurrently to the propane and exits as a saturated liquid at the same pressure. (a) Taking as a basis 100 mol of propane fed to the exchanger, draw and label a process flowchart. Include in your labeling the volume of propane fed \(\left(\mathrm{m}^{3}\right),\) the mass of steam fed \((\mathrm{kg}),\) and the volume of steam fed \(\left(\mathrm{m}^{3}\right)\) (b) Calculate values of the labeled specific enthalpies in the following inlet-outlet enthalpy table for this process. $$\begin{array}{|l|cc|cc|} \hline \text { Species } & n_{\text {in }} & \hat{H}_{\text {in }} & n_{\text {out }} & \hat{H}_{\text {out }} \\ \hline \mathrm{C}_{3} \mathrm{H}_{8} & 100 \mathrm{mol} & \hat{H}_{\mathrm{a}}(\mathrm{kJ} / \mathrm{mol}) & 100 \mathrm{mol} & \hat{H}_{\mathrm{c}}(\mathrm{kJ} / \mathrm{mol}) \\ \mathrm{H}_{2} \mathrm{O} & m_{\mathrm{w}}(\mathrm{kg}) & \hat{H}_{\mathrm{b}}(\mathrm{kJ} / \mathrm{kg}) & m_{\mathrm{w}}(\mathrm{kg}) & \hat{H}_{\mathrm{d}}(\mathrm{kJ} / \mathrm{kg}) \\ \hline \end{array}$$ (c) Use an energy balance to calculate the required mass feed rate of the steam. Then calculate the volumetric feed ratio of the two streams ( \(\mathrm{m}^{3}\) steam fed \(/ \mathrm{m}^{3}\) propane fed). Assume ideal-gas behavior for the propane but not the steam and recall that the exchanger is adiabatic. (d) Calculate the heat transferred from the water to the propane ( \(k J / m^{3}\) propane fed). (Hint: Do an energy balance on either the water or the propane rather than on the entire heat exchanger.) (e) Over a period of time, scale builds up on the heat-transfer surface, resulting in a lower rate of heat transfer between the propane and the steam. What changes in the outlet streams would you expect to see as a result of the decreased heat transfer?

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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