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Magazine Chapter 3: Problem 83
A piston-cylinder assembly contains propane, initially at \(27^{\circ} \mathrm{C}, 1\) bar, and a volume of \(0.2 \mathrm{~m}^{3}\). The propane undergoes a process to a final pressure of 4 bar, during which the pressure- volume relationship is \(p V^{\mathrm{L} .1}=\) constant. For the propane, evaluate the work and heat transfer, each in kJ. Kinetic and potential energy effects can be ignored.
Short Answer
Expert verified
Work done by propane is -2.2 kJ, and heat transfer is -0.24 kJ.
Step by step solution
01
- Define Initial State
Identify the initial conditions for the propane. Initial temperature is given as \(27^{\circ}C\), initial pressure \( p_1 \) is 1 bar, and initial volume \( V_1 \) is 0.2 \( m^3 \).
02
- Convert Units if Necessary
Convert the initial temperature to Kelvin if necessary for calculations. Temperature \( T_1 = 27 + 273.15 = 300.15 K \).
03
- Apply Pressure-Volume Relationship
Use the given relationship \( p V^{1.1} = constant \). This implies \( p_1 V_1^{1.1} = p_2 V_2^{1.1} \). Plug in the known values to find \( V_2 \). \[ 1 \times 0.2^{1.1} = 4 \times V_2^{1.1} \]. Solve for \( V_2 \).
04
- Calculate Final Volume
Solve the equation for \( V_2 \). \[ V_2 = \left( \frac{0.2^{1.1}}{4} \right)^{1/1.1} = 0.105 \; (approx.) \; m^3 \]
05
- Work Done by Propane
Work for a polytropic process (where \( p V^n = constant \)) is given by \( W = \frac{p_2 V_2 - p_1 V_1}{1-n} \). Here, \( n = 1.1 \). Substitute the values, \[ W = \frac{4 \times 0.105 - 1 \times 0.2}{1 - 1.1} \]. Solve for \( W \).
06
- Substitute and Calculate Work
Continue from the formula: \[ W = \frac{4 \times 0.105 - 1 \times 0.2}{-0.1} = \frac{0.42 - 0.2}{-0.1} = \frac{0.22}{-0.1} = -2.2 \; kJ \] The work done by the propane is \( -2.2 \) kJ (work done on the system, hence negative).
07
- Identify Final Temperature
Using ideal gas law, calculate the final temperature \( T_2 \). \[ T_2 = T_1 \times \frac{p_2 V_2}{p_1 V_1} = 300 \times \frac{4 \times 0.105}{1 \times 0.2} = 630 \; K \]
08
- Calculate Heat Transfer
Using first law of thermodynamics: \( Q = \Delta U + W \). For an ideal gas, \( \Delta U = m c_v \Delta T \). Propane specific heat \( c_v \) is needed. Estimate \( m \) (mass) using \( p_1 V_1 = m R T_1 \).
09
- Solve for Mass
\( m = \frac{p_1 V_1}{R T_1} \). \( R \) for propane = 0.1886 kJ/kgK. \[ m = \frac{1 \times 0.2}{0.1886 \times 300} = 0.00354 \; kg \] Using \( \Delta T = 630 - 300 = 330 \; K \).
10
- Calculate Internal Energy Change
\( \Delta U = m c_v \Delta T \). With \( c_v = 1.68 kJ/kgK \) for propane, \[ \Delta U = 0.00354 \times 1.68 \times 330 = 1.96 \; kJ \]
11
- Find Heat Transfer
\( Q = \Delta U + W \). \[ Q = 1.96 - 2.2 = -0.24 \; kJ \] The heat transfer is \( -0.24 \) kJ.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Piston-Cylinder Assembly
A piston-cylinder assembly is a common apparatus in thermodynamics used to demonstrate how gases behave under varying conditions. We'll use this to explore how propane behaves inside a confined space. Picture a cylinder with a movable piston that can slide up and down, changing the volume inside the cylinder. The propane gas inside the cylinder will either expand or compress when the piston moves, altering the pressure and volume. This setup is fundamental for studying different thermodynamic processes because it allows clear observation of pressure-volume changes.
Polytropic Process
The polytropic process is a type of thermodynamic process that follows the relationship \(pV^n = \text{constant}\). In our case, the process has an exponent \(n = 1.1\). This means that the product of pressure \(p\) and volume raised to the power of 1.1 remains constant throughout. Polytropic processes are important as they can represent various real-world scenarios more accurately than simple isothermal or adiabatic processes. By understanding how pressure and volume interact in such a process, we can calculate critical properties like work done and heat transfer.
Work and Heat Transfer Calculations
Calculating work and heat transfer in thermodynamic processes involves several steps. When dealing with a polytropic process, we use the formula \(W = \frac{p_2 V_2 - p_1 V_1}{1-n}\) to find work done. Here, \(p_1\) and \(V_1\) are initial pressure and volume, while \(p_2\) and \(V_2\) are their final values respectively. We also need to apply the First Law of Thermodynamics, which states \(Q = \text{螖U} + W\). \(Q\) represents heat transfer, \(螖U\) the change in internal energy, and \(W\) the work done. To find \(螖U\), for an ideal gas, it's \(螖U = m c_v 螖T\), where \(m\) is the mass, \(c_v\) is the specific heat capacity at constant volume, and \(螖T\) is the change in temperature.
Ideal Gas Law
The Ideal Gas Law is a pivotal concept in thermodynamics, described by the equation \(pV = nRT\). Here, \(p\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. This law helps us relate pressure, volume, and temperature for ideal gases like our propane. Through knowing any three of these variables, the fourth can be calculated. In our scenario, we used the Ideal Gas Law to find the final temperature of propane after the process.
