Question

Determine the change in the internal energy of helium, in \(\mathrm{kJ} / \mathrm{kg},\) as it undergoes a change of state from \(100 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) to \(600 \mathrm{kPa}\) and \(300^{\circ} \mathrm{C}\) using the equation of state \(P(v-a)=R T\) where \(a=0.01 \mathrm{m}^{3} / \mathrm{kg},\) and compare the result to the value obtained by using the ideal gas equation of state.

Short answer

Answer: The change in internal energy of helium using both the real gas and ideal gas equations of state is 876,020.10 J/kg.

Step-by-step solution

Convert temperatures and pressures to SI units

First, convert the given temperatures to Kelvin and pressures to Pascals: \(T_1 = 20^{\circ}C + 273.15 = 293.15 K\) \(T_2 = 300^{\circ}C + 273.15 = 573.15 K\) \(P_1 = 100 kPa \times 1000 = 100,000 Pa\) \(P_2 = 600 kPa \times 1000 = 600,000 Pa\)

Calculate initial and final specific volume using real gas equation

Using the real gas equation of state, we can find the initial and final specific volumes (v1 and v2). Rearrange the equation for both states, and solve for v1 and v2: \(v_1 = \frac{R T_1}{P_1} + a\) \(v_2 = \frac{R T_2}{P_2} + a\) The gas constant for helium R is 2077 J/kg K. \(v_1 = \frac{2077 \times 293.15}{100,000} + 0.01 = 0.07 \frac{m^3}{kg}\) \(v_2 = \frac{2077 \times 573.15}{600,000} + 0.01 = 0.02 \frac{m^3}{kg}\)

Determine internal energy change with real gas equation

For an ideal diatomic gas like helium, the change in internal energy (\(\Delta u\)) can be determined using: \(\Delta u = c_v \Delta T\) Since helium is a diatomic gas, its specific heat at constant volume (cv) is: \(c_v = \frac{3}{2} R\) Now, we can find the change in internal energy using the real gas equation of state: \(\Delta u = c_v \Delta T = \frac{3}{2} R (T_2 - T_1)\) \(\Delta u = \frac{3}{2} \times 2077 \times (573.15 - 293.15) = 876020.10 \frac{J}{kg}\)

Determine internal energy change using ideal gas equation

For the ideal gas equation of state, we can use the same formula for change in internal energy: \(\Delta u = c_v \Delta T = \frac{3}{2} R (T_2 - T_1)\) \(\Delta u = \frac{3}{2} \times 2077 \times (573.15 - 293.15) = 876020.10 \frac{J}{kg}\)

Comparing results

Now, we compare the change in internal energy calculated using the real gas equation and the ideal gas equation of state: For real gas equation: \(\Delta u = 876020.10 \frac{J}{kg}\) For ideal gas equation: \(\Delta u = 876020.10 \frac{J}{kg}\) The change in internal energy calculated using both equations is the same, which indicates that the real gas equation of state and the ideal gas equation of state give the same result for this particular set of conditions.

Key concepts

Real Gas Equation of State

When studying the behavior of gases, especially under varying pressures and temperatures, we need a mathematical model to describe their properties. Real gases deviate from ideal behavior due to molecular interactions and volume occupied by gas molecules. The real gas equation of state attempts to account for these deviations.

One commonly used real gas equation is the van der Waals equation, given by \( P(v - a) = RT \) where \( P \) is the pressure, \( v \) is specific volume, \( a \) accounts for the attraction between gas molecules, \( R \) is the gas constant, and \( T \) is temperature. The term \( a \) is introduced to correct for the lower pressure resultant of molecular attraction, and \( v-a \) essentially represents a 'corrected' volume.

This equation adapts the Ideal Gas Law, making it more accurate for real-world scenarios, especially where gases are under high pressures or low temperatures. By using this refined model, we can predict the behavior of gases like helium with higher precision than with the ideal model, as seen in our exercise.

Ideal Gas Equation of State

Moving on to the ideal gas equation of state, it describes the behavior of gases under the assumption that the molecules do not interact and that they occupy no space. The equation is mathematically represented by \( PV = nRT \) where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles of the gas, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.

For many applications, especially at standard temperature and pressure (STP), this equation serves as an excellent approximation, simplifying calculations and providing a basis for understanding more complex gas behavior. While it's less accurate for high-pressure or low-temperature conditions, in our exercise example involving helium, the ideal gas equation provided the same internal energy change as the real gas equation, reaffirming its utility for certain conditions.

Specific Heat at Constant Volume

Specific heat capacity at constant volume, denoted by \( c_v \) plays a crucial role in the calculation of the change in internal energy for a gas. It represents the amount of heat required to raise the temperature of a unit mass of a substance by one degree while maintaining a constant volume.

In the context of gases, \( c_v \) for monatomic gases like helium is particularly simple, given by \( c_v = \frac{3}{2} R \), where \( R \) is the gas constant. This constant arises from the kinetic theory of gases, which correlates the internal energy with temperature for monatomic gases. Thus, when we calculate the change in internal energy (\( \( \)Delta u = c_v \( \)Delta T \) ) using this specific heat, we can determine how much energy the gas has absorbed or released during a thermal process.

Thermodynamic Properties of Helium

Now, let's delve into the thermodynamic properties of helium, which are essential in understanding exercises like ours. Helium, being a noble gas, is characterized by its very low reactivity due to having a full valence electron shell. It is also a monatomic gas, meaning it consists of single atoms, rather than molecules.

Helium's unique properties affect its thermodynamic behavior. For instance, due to its low atomic mass, it has one of the highest thermal conductivity and specific heat capacities of any gas. Moreover, as a monatomic gas, its behavior as a real gas can often be closely approximated by the ideal gas law, particularly at standard conditions.

In the exercise, we used helium as our working substance, and irrespective of whether we applied the real gas equation or the ideal gas equation, the change in internal energy was remarkably consistent. This underlines helium's simplicity thermodynamically, making it an excellent subject for such theoretical exercises.