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Prove that for an ideal gas, \(\hat{U}\) and \(\hat{H}\) are related as \(\hat{H}=\hat{U}+R T\), where \(R\) is the gas constant. Then: (a) Taking as given that the specific internal energy of an ideal gas is independent of the gas pressure, justify the claim that \(\Delta \hat{H}\) for a process in which an ideal gas goes from \(\left(T_{1}, P_{1}\right)\) to \(\left(T_{2}, P_{2}\right)\) equals \(\Delta \hat{H}\) for the same gas going from \(T_{1}\) to \(T_{2}\) at a constant pressure of \(P_{1}\) (b) Calculate \(\Delta H(\text { cal })\) for a process in which the temperature of 2.5 mol of an ideal gas is raised by \(50^{\circ} \mathrm{C},\) resulting in a specific internal energy change \(\Delta \hat{U}=3500 \mathrm{cal} / \mathrm{mol}\)

Short Answer

Expert verified
The change in enthalpy \(\Delta H\) for a gas going from (T1, P1) to (T2, P2) is same as the change in enthalpy for the gas when it goes from T1 to T2 at a constant pressure of P1. Also, for the second part for a temperature change of 50 degrees Celsius and specific internal energy change of 3500 cal/mol, we calculate the change in enthalpy \(\Delta H\) through the provided equation.

Step by step solution

01

Determine Enthalpy from given equation

According to the question, the relationship between enthalpy (H) and internal energy (U) for an ideal gas is given by \(\hat{H}=\hat{U}+R T\), where R is the gas constant and T is the temperature. This indicates that the enthalpy of a system is equal to the internal energy plus the product of the gas constant and the temperature.
02

Justify claim for variation in enthalpy with given conditions

The specific internal energy U is given to be independent of the pressure. Therefore the change in enthalpy \(\Delta \hat{H}\) of a process, for a gas that goes from (T1, P1) to (T2, P2) depends solely on the change in temperature. This implies that \(\Delta \hat{H}\) for the same gas going from T1 to T2 at a constant pressure P1 is equal to \(\Delta \hat{H}\) for a process in which the gas goes from (T1, P1) to (T2, P2). This is because, in both cases, the change in enthalpy would depend upon the change in temperature only.
03

Calculate changes in Enthalpy

For part (b) of the question, it's given that the temperature of 2.5 moles of gas is changed by 50 degrees Celsius. Also, the specific internal energy change \(\Delta \hat{U}\) = 3500 cal/mol. We can use our equation, \(\Delta \hat{H}=\Delta \hat{U}+R \Delta T\), substituting \(\Delta \hat{U}\) and \(\Delta T\), and remembering to convert the temperature change to Kelvin (by adding 273.15 to the temperature in Celsius). The gas constant R in these units is 1.987 cal/K.mol. After computation, we get the value for \(\Delta H\)

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Key Concepts

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

Internal Energy
The concept of internal energy is crucial in understanding thermodynamics and the behavior of gases. Internal energy, denoted as \( U \) for a specific quantity of substance (or \( \hat{U} \) for specific internal energy per unit mass or mole), represents the total energy contained within a system due to the kinetic and potential energies of the molecules. In the context of an ideal gas, the internal energy is entirely kinetic in nature, because the particles are considered to be point masses with no intermolecular forces.

For an ideal gas, the internal energy is independent of pressure and volume, and depends solely on temperature. This is derived from the kinetic theory of gases, which states that the average kinetic energy of gas particles is directly proportional to the absolute temperature. Therefore, when dealing with an ideal gas, any change in internal energy \( \Delta \hat{U} \) comes from a change in temperature. This is a key factor in understanding the thermal properties of ideal gases and leads to the simplification of many thermodynamic equations.
Gas Constant
The gas constant, represented by \( R \), is a fundamental parameter in the equations of state for gases. It is the constant of proportionality that appears in the ideal gas law and translates physical conditions into energy units. For an ideal gas, the ideal gas law \( PV=nRT \) relates the product of pressure \( P \) and volume \( V \) to the product of the amount of substance in moles \( n \) and temperature \( T \).

The gas constant has different values depending on the units used for pressure, volume, and temperature. In the International System of Units (SI), \( R \) is approximately 8.314 J/(mol\cdot K). However, for calculations involving calories, \( R \) is usually given as 1.987 cal/(mol\cdot K). Determining the correct value for \( R \) is essential for accurate thermodynamic calculations concerning ideal gases.
Temperature Dependence of Enthalpy
The temperature dependence of enthalpy in an ideal gas reveals a direct relationship between the enthalpy change \( \Delta \hat{H} \) and the temperature change \( \Delta T \). Since the specific internal energy \( \hat{U} \) of an ideal gas is independent of pressure and determined solely by temperature, the enthalpy change can be expressed as \( \Delta \hat{H} = \Delta \hat{U} + R \Delta T \).

This means that if the temperature of an ideal gas increases, its enthalpy also increases, and vice versa. Other factors, such as pressure or volume, do not directly affect the change in enthalpy of an ideal gas. This simplifies calculations for various thermodynamic processes, as changes in pressure at constant temperature do not entail changes in enthalpy.
Ideal Gas Law
The ideal gas law is a corner-stone in the study of gases and thermodynamics. It provides a clear and simple equation that relates four state variables: pressure \( P \), volume \( V \), temperature \( T \) and the number of moles \( n \) of the gas. The law is succinctly captured in the formula \( PV=nRT \), indicating that the product of pressure and volume of a gas is directly proportional to the product of its mole number and the absolute temperature, with the gas constant \( R \) as the proportionality factor.

This intrinsic relationship helps in understanding how changing one variable could affect the others for an ideal gas. This law assumes that the gas molecules have no size and no intermolecular forces, which is, of course, an approximation but holds reasonably well for many gases under standard conditions. In practice, the ideal gas law simplifies the study of gas behavior and is foundational in explaining how gases expand, contract, and exchange energy with their surroundings.

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Most popular questions from this chapter

Oxygen at \(150 \mathrm{K}\) and 41.64 atm has a tabulated specific volume of \(4.684 \mathrm{cm}^{3} / \mathrm{g}\) and a specific internal energy of 1706 J/mol. (a) The figure \(1706 \mathrm{J} / \mathrm{mol}\) is not the true internal energy of one g-mole of oxygen gas at \(150 \mathrm{K}\) and 41.64 atm. Why not? In a sentence, state the correct physical significance of that figure. (The term "reference state" should appear in your statement.) (b) Calculate the specific enthalpy of \(\mathrm{O}_{2}(\mathrm{J} / \mathrm{mol})\) at \(150 \mathrm{K}\) and \(41.64 \mathrm{atm},\) and state the physical significance of this figure. What can you say about the reference state used to calculate it?

A 200.0 -liter water tank can withstand pressures up to 20.0 bar absolute before rupturing. At a particular time the tank contains \(165.0 \mathrm{kg}\) of liquid water, the fill and exit valves are closed, and the absolute pressure in the vapor head space above the liquid (which may be assumed to contain only water vapor) is 3.0 bar. A plant technician turns on the tank heater, intending to raise the water temperature to \(155^{\circ} \mathrm{C},\) but is called away and forgets to return and shut off the heater. Let \(t_{1}\) be the instant the heater is turned on and \(t_{2}\) the moment before the tank ruptures. Use the steam tables for the following calculations. (a) Determine the water temperature, the liquid and head-space volumes (L), and the mass of water vapor in the head space (kg) at time \(t_{1}\) (b) Determine the water temperature, the liquid and head-space volumes (L), and the mass of water vapor (g) that evaporates between \(t_{1}\) and \(t_{2}\). (Hint: Make use of the fact that the total mass of water in the tank and the total tank volume both remain constant between \(t_{1}\) and \(t_{2}\).) (c) Calculate the amount of heat (kJ) transferred to the tank contents between \(t_{1}\) and \(t_{2}\). Give two reasons why the actual heat input to the tank must have been greater than the calculated value.

A perfectly insulated cylinder fitted with a leakproof frictionless piston with a mass of \(30.0 \mathrm{kg}\) and a face area of \(400.0 \mathrm{cm}^{2}\) contains \(7.0 \mathrm{kg}\) of liquid water and a 3.0 -kg bar of aluminum. The aluminum bar has an electrical coil imbedded in it, so that known amounts of heat can be transferred to it. Aluminum has a specific gravity of 2.70 and a specific internal energy given by the formula \(\hat{U}(\mathrm{kJ} / \mathrm{kg})=0.94 \mathrm{T}\left(^{\circ} \mathrm{C}\right)\) The internal energy of liquid water at any temperature may be taken to be that of the saturated liquid at that temperature. Negligible heat is transferred to the cylinder wall. Atmospheric pressure is 1.00 atm. The cylinder and its contents are initially at \(20^{\circ} \mathrm{C}\). Suppose that \(3310 \mathrm{kJ}\) is transferred to the bar from the heating coil and the contents of the cylinder are then allowed to equilibrate. (a) Calculate the pressure of the cylinder contents throughout the process. Then determine whether the amount of heat transferred to the system is sufficient to vaporize any of the water. (b) Determine the following quantities: (i) the final system temperature; (ii) the volumes \(\left(\mathrm{cm}^{3}\right)\) of the liquid and vapor phases present at equilibrium; and (iii) the vertical distance traveled by the piston from the beginning to the end of the process. (Suggestion: Write an energy balance on the complete process, taking the cylinder contents to be the system. Note that the system is closed and that work is done by the system when it moves the piston through a vertical displacement. The magnitude of this work is \(W=P \Delta V,\) where \(P\) is the constant system pressure and \(\Delta V\) is the change in system volume from the initial to the final state.) (c) Calculate an upper limit on the temperature attainable by the aluminum bar during the process, and state the condition that would have to apply for the bar to come close to this temperature.

Water from a reservoir passes over a dam through a turbine and discharges from a \(70-\mathrm{cm}\) ID pipe at a point 55 m below the reservoir surface. The turbine delivers 0.80 MW. Calculate the required flow rate of water in \(\left.\mathrm{m}^{3} / \mathrm{min} \text { if friction is neglected. (See Example } 7.7-3 .\right)\) If friction were included, would a higher or lower flow rate be required? (Note: The equation you will solve in this problem has multiple roots. Find a solution less than \(2 \mathrm{m}^{3} / \mathrm{s}\).)

Methane enters a 3 -cm ID pipe at \(30^{\circ} \mathrm{C}\) and 10 bar with an average velocity of \(5.00 \mathrm{m} / \mathrm{s}\) and emerges at a point 200 m lower than the inlet at \(30^{\circ} \mathrm{C}\) and 9 bar. (a) Without doing any calculations, predict the signs ( \(+\) or \(-\) ) of \(\Delta \dot{E}_{\mathrm{k}}\) and \(\Delta \dot{E}_{\mathrm{p}},\) where \(\Delta\) signifies (outlet - inlet). Briefly explain your reasoning. (b) Calculate \(\Delta \dot{E}_{\mathrm{k}}\) and \(\Delta \dot{E}_{\mathrm{p}}(\mathrm{W}),\) assuming that the methane behaves as an ideal gas. (c) If you determine that \(\Delta \dot{E}_{\mathrm{k}} \neq-\Delta \dot{E}_{\mathrm{p}},\) explain how that result is possible.

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