Question

Methane \(\left(\mathrm{CH}_{4}\right)\) undergoes an isentropic expansion from an initial state where the temperature is \(1000 \mathrm{~K}\) and the pressure is 5 bar to a final state where the temperature is \(T\) and the pressure is \(p\). Using the ideal gas model together with \(\bar{c}_{\mathrm{p}}(T)\) from Table A-21, determine (a) \(p\) when \(T=500 \mathrm{~K}\) (b) \(T\) when \(p=1\) bar. (c) Check the results of parts (a) and (b) using \(I T\).

Short answer

(a) p = 4.365 bar, (b) T = 45 K

Step-by-step solution

Identify Initial and Final States

The initial state of the methane is given as: \(T_1 = 1000 \mathrm{~K}\) and \(p_1 = 5\) bar. The final state variables are \(T\) and \(p\). Two scenarios are provided: one where \(T_2 = 500 \mathrm{~K}\) and another where \(p_2 = 1\) bar.

Understand Isentropic Relation

For an isentropic process of an ideal gas, we can use the relation: \[\left(\frac{T_2}{T_1}\right)^{\frac{\bar{R}}{\bar{c}_p(T)}} = \frac{p_2}{p_1}\] where \(\bar{R}=8.314 \frac{J}{mol \cdot K}\) for methane and \(\bar{c}_p(T)\) is the specific heat capacity at constant pressure, which is a function of temperature.

Calculate \(p_2\) for \(T_2 = 500 \mathrm{~K}\)

Using the specific heat capacity from Table A-21, we find \(\bar{c}_p(1000 \mathrm{~K}) = 35.1 \frac{J}{mol \cdot K}\) and \(\bar{c}_p(500 \mathrm{~K}) = 25.1 \frac{J}{mol \cdot K}\). For an approximation, using average \(\bar{c}_p\), we can take \(\bar{c}_p = 30.1 \frac{J}{mol \cdot K}\). Substitute into the isentropic relation: \[ \left( \frac{500}{1000} \right)^{\frac{8.314}{30.1}} = \frac{p_2}{5}\]. Therefore, solving for \(p_2\): \[ p_2 = 5 \times \left( \frac{1}{2} \right)^{\frac{8.314}{30.1}} = 5 \times 0.873 = 4.365 \text{ bar} \]

Calculate \(T_2\) for \(p_2 = 1 \text{ bar}\)

Using the same isentropic relation: \[\left( \frac{T_2}{1000} \right)^{\frac{8.314}{30.1}} = \frac{1}{5}\]. Solving for \(T_2\): \[ \left( \frac{T_2}{1000} \right)^{0.276} = 0.2 \]. Thus: \[\frac{T_2}{1000} = 0.2^{3.62}\]. So, \[ T_2 = 1000 \times 0.0448 = 45 \text{ K} \].

Validate Results using Information Technology (I T)

Use software or online calculators for thermodynamic properties of methane and the isentropic process to validate intermediate steps. Compare the calculated pressure and temperature with the result from the steps.

Key concepts

ideal gas model

To understand the isentropic expansion in this exercise, we first need to grasp the *Ideal Gas Model*. This model assumes that gases consist of a vast number of very small particles moving in random directions and that these particles do not interact except for perfectly elastic collisions.
In the context of the Ideal Gas Law, the equation is: \[ PV = nRT \] where

• \(P\) is the pressure,
• \(V\) is the volume,
• \(n\) is the amount of substance,
• \(R\) is the ideal gas constant,
• \(T\) is the absolute temperature.

Here, methane can be treated as an ideal gas, making use of this law crucial for solving related problems.

specific heat capacity

*Specific Heat Capacity* is another essential concept for this exercise. It quantifies the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. For gases, we often talk about the specific heat capacity at constant pressure, denoted as \( \bar{c}_p(T) \).
For methane, \( \bar{c}_p(T) \) can vary with temperature. For example, in the given solution:

• \( \bar{c}_p(1000 \text{ K}) = 35.1 \frac{J}{mol \cdot K} \)
• \( \bar{c}_p(500 \text{ K}) = 25.1 \frac{J}{mol \cdot K} \)

By averaging these values, we use \( \bar{c}_p = 30.1 \frac{J}{mol \cdot K} \). This approximation simplifies calculations for problems involving temperature changes and isentropic relations.

thermodynamic properties

*Thermodynamic Properties* describe the fundamental aspects of substances that dictate the behavior in thermodynamic systems. These include temperature, pressure, volume, internal energy, and entropy.
For this problem, we focus on the properties of methane during the isentropic expansion process. Key properties are:

• Initial and final temperatures \( T_1 \) and \( T_2 \)
• Initial and final pressures \( p_1 \) and \( p_2 \)

Thermodynamic relationships help us connect these properties, as seen in the isentropic process equations.

isentropic process

Finally, understanding the *Isentropic Process* is vital. This is a thermodynamic process where entropy remains constant. For an ideal gas undergoing an isentropic process, relations between temperature and pressure are described by the equation: \[ \frac{T_2}{T_1} = \frac{p_2}{p_1}^{\frac{\bar{R}}{\bar{c}_p(T)}} \textrm{with } \bar{R} = 8.314 \frac{J}{mol \bullet K} \] In this exercise involving methane, this equation allowed us to solve for unknown pressures and temperatures in different cases:

• When \( T_2 = 500 \text{ K} \)
• When \( p_2 = 1 \text{ bar} \)

Such calculations underline the importance of understanding isentropic processes in thermodynamics.