Question

Determine the change in the enthalpy of helium, in \(\mathrm{kJ} / \mathrm{kg},\) as it undergoes a change of state from \(150 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) to \(750 \mathrm{kPa}\) and \(380^{\circ} \mathrm{C}\) 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

Question: Calculate the change in enthalpy for helium gas undergoing a change of state from an initial pressure of 150kPa and temperature 20°C to a final pressure of 750kPa and temperature of 380°C using the equation of state P(v-a)=RT, and compare the results with the ideal gas equation of state. Given, R=2080 J/kg.K, a=0.01 m³/kg, and cp=5,192.6 J/kg.K. Answer: The change in enthalpy for helium gas (Δh) using the given equation of state is approximately *______ J/kg*. The change in enthalpy for helium gas (Δh') using the ideal gas equation of state is approximately *______ J/kg*. The difference between the two is approximately *______ J/kg*.

Step-by-step solution

Convert temperature to Kelvin

To work with the gas equations, we need to use the absolute temperature (in Kelvin). We can convert the given Celsius temperatures to Kelvin by adding 273.15. Initial temperature, \(T_1 = 20^\circ \mathrm{C} + 273.15 \mathrm{K} = 293.15 \mathrm{K}\) Final temperature, \(T_2 = 380^\circ \mathrm{C} + 273.15 \mathrm{K} = 653.15 \mathrm{K}\)

Calculate specific volumes

We are given the given equation of state \(P(v-a)=RT\). Let's find the specific volumes, \(v_1\) and \(v_2\), at both states (\(P_1, T_1\)) and (\(P_2, T_2\)) respectively. \(P_1 = 150\,\text{kPa} = 150,000\,\text{Pa}\), \(T_1 = 293.15\,\text{K}\), and \(a = 0.01\,\text{m}^3/\text{kg}\) \(P_2 = 750 \,\text{kPa} = 750,000\,\text{Pa}\), \(T_2 = 653.15\,\text{K}\) Using the equation of state \(P(v-a)=RT\), we can solve for \(v_1\) and \(v_2\): \(v_1 = \displaystyle \frac{P_1}{R T_1} + a\) \(v_2 = \displaystyle \frac{P_2}{R T_2} + a\) We know for helium, \(R = 2,080\,\text{J/kg.K}\). Plug in these values and calculate \(v_1\) and \(v_2\): \(v_1 = \displaystyle \frac{150,000}{2,080 \times 293.15} + 0.01\) \(v_2 = \displaystyle \frac{750,000}{2,080 \times 653.15} + 0.01\)

Calculate change in enthalpy

Now, we can find the change in enthalpy, \(\Delta h\), for the helium using the relation \(\Delta h = c_p \Delta T\), where \(c_p\) is the specific heat at a constant pressure and \(\Delta T\) is the change in temperature. For helium, the specific heat \(c_p = 5,192.6\,\text{J/kg.K}\). \(\Delta h = c_p \Delta T = c_p(T_2 - T_1) = 5,192.6 \times (653.15 - 293.15)\) Calculate \(\Delta h\) to get the change in enthalpy per kilogram of helium.

Calculate using the ideal gas equation of state

Next, we calculate the change in enthalpy using the ideal gas equation of state \(PV=RT\). Calculate the specific volumes \(v_1'\) and \(v_2'\) for initial and final states using the ideal gas equation of state. \(v_1' = \displaystyle \frac{R T_1}{P_1}\) \(v_2' = \displaystyle \frac{R T_2}{P_2}\) Now, find the change in enthalpy, \(\Delta h'\), using the ideal gas equation of state: \(\Delta h' = c_p(T_2 - T_1) = 5,192.6 \times (653.15 - 293.15)\)

Compare results

Finally, compare the change in enthalpy calculated using the given equation of state \(\Delta h\) and the ideal gas equation of state \(\Delta h'\) to find the difference between the two.

Key concepts

Real Gas Equation of State

In thermodynamics, the real gas equation of state is a more realistic model compared to the ideal gas law for describing the behavior of gases. Unlike the ideal gas law, it takes into account the interactions between gas particles and the finite size of these particles.

The van der Waals equation is a well-known real gas equation of state which includes two correction factors: one for the particle's volume (denoted as 'a') and another for the intermolecular forces ('b'). The given equation of state, \(P(v-a)=RT\), is a simplified version where 'a' is a constant that corrects for the volume occupied by the gas molecules.

To apply this equation for calculating various thermodynamic properties, such as in our exercise, you'll typically need to rearrange it to solve for the specific volume (v) at different pressures (P) and temperatures (T), as you've seen in the provided solution steps. This adjustment allows us to evaluate real gases' behavior under various conditions and to measure how their physical properties differ from ideal gases.

Ideal Gas Law

The ideal gas law is a fundamental equation in thermodynamics that describes the relationship between the pressure (P), volume (V), temperature (T), and the amount of gas (n) for an ideal gas. It is expressed as \(PV=nRT\), or in terms of specific volume (v) as \(PV=RT\) when dealing with a unit mass of the gas.

The ideal gas law assumes that the gas particles do not attract or repel each other (no intermolecular forces) and that they occupy no space (point particles). While these assumptions are not true for real gases, the ideal gas law can still provide reasonable approximations for many gases under a range of conditions, especially when the gas is at a high temperature and low pressure, where the particles have enough kinetic energy to overcome any intermolecular forces.

In the exercise, students are encouraged to compare calculations made using the ideal gas law to those using the real gas equation of state to acknowledge the latter's extra complexity and accuracy in certain scenarios.

Specific Heat

Specific heat, often symbolized as \( c_p \) for constant pressure or \( c_v \) for constant volume, is a substance's property that indicates the amount of heat required to change the temperature of a unit mass by one degree Celsius (or one Kelvin).

Helium, like other gases, has distinct specific heat values when kept at constant pressure or volume. This property is significant in calculating the energy change or enthalpy change \( (\Delta h) \) of a system during temperature changes. The equation \(\Delta h = c_p \Delta T\) leverages specific heat to determine the enthalpy change associated with a given temperature change \( \Delta T \).

For students grappling with these concepts, recognizing the role of specific heat in connecting temperature changes to energy changes in a system can clarify how substances absorb or release heat during thermodynamic processes.

Temperature Conversion

Accurate temperature conversion is crucial in thermodynamics to ensure consistency and correctness in calculations. The exercise involves converting Celsius to Kelvin, which is a simple yet vital step for working with gas equations.

The Kelvin scale is the SI unit for temperature and is one of the fundamental units in physics. It is preferred in scientific calculations because, unlike the Celsius or Fahrenheit scales, it begins at absolute zero – the theoretical lowest temperature possible.

To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. It's essential to remember this conversion factor when solving thermodynamic problems, as using the wrong temperature scale can significantly impact your results. As evidenced in the exercise, conversions are an integral part of working through thermodynamic problems and cannot be overlooked if one is to obtain accurate and meaningful results.