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Air as an ideal gas is compressed from a state where the pressure is \(0.1 \mathrm{MPa}\) and the temperature is \(27^{\circ} \mathrm{C}\) to a state where the pressure is \(0.5 \mathrm{MPa}\) and the temperature is \(207^{\circ} \mathrm{C}\). Can this process occur adiabatically? If yes, determine the work per unit mass of air, in \(\mathrm{kJ} / \mathrm{kg}\), for an adiabatic process between these states. If no, determine the direction of the heat transfer.

Short Answer

Expert verified
The process cannot be adiabatic; heat must be added to the system.

Step by step solution

01

- Determine initial and final temperatures in Kelvin

Convert the given temperatures from degrees Celsius to Kelvin using the formula \( T(K) = T(°C) + 273.15 \). Initial temperature: \( T_1 = 27 + 273.15 = 300.15 \mathrm{K} \). Final temperature: \( T_2 = 207 + 273.15 = 480.15 \mathrm{K} \).
02

- Calculate specific heat ratio (γ) for air

For air, the specific heat ratio (γ) is typically given as \( \frac{C_p}{C_v} = 1.4 \).
03

- Check if the process is adiabatic

For an adiabatic process, the relation \( \frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}} \) holds true. substituting the given values: \( \frac{480.15}{300.15} = \left( \frac{0.5}{0.1} \right)^{\frac{1.4-1}{1.4}} \).Simplifying this, \( 1.6 = 5^{0.2857} \), which results in \( 1.6 \approx 1.8 \). Since the values do not match exactly, the process cannot be adiabatic.
04

- Determining the direction of heat transfer

Since the process cannot be adiabatic, heat is transferred. The temperature of the air increases meaning heat is added to the system.

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Key Concepts

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

ideal gas law
The Ideal Gas Law is a fundamental equation in thermodynamics that describes the behavior of an ideal gas. It is expressed as:
\[ PV = nRT \]
Where:
- \( P \) is the pressure of the gas
- \( V \) is the volume of the gas
- \( n \) is the number of moles of gas
- \( R \) is the universal gas constant
- \( T \) is the temperature of the gas in Kelvin
It helps us understand how changes in temperature, pressure, and volume affect a gas. In the problem, we use the ideal gas law to track changes in states of air, assuming air behaves like an ideal gas. This equation is crucial when we assess how gases transition from one state to another under different conditions. Because the process involves changes in temperature and pressure, we utilize the ideal gas law to understand these interactions deeply.
specific heat ratio
The specific heat ratio, often denoted by \( \gamma \) (gamma), is the ratio of the specific heat capacity at constant pressure \( C_p \) to the specific heat capacity at constant volume \( C_v \). Mathematically, it is represented as:
\[ \gamma = \frac{C_p}{C_v} \]
For air and many other diatomic gases, the value of \( \gamma \) is typically around 1.4. This ratio is significant in determining how a gas behaves during different thermodynamic processes, especially adiabatic processes.
In our exercise, we employ this ratio to check if given pressures and temperatures of air illustrate an adiabatic process:
\[ \frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^\frac{\gamma - 1}{\gamma} \]
By substituting our values, we find that the expected ratio does not align perfectly. This discrepancy suggests the process is not entirely adiabatic, indicating heat transfer is involved.
heat transfer
Heat transfer is the process by which thermal energy flows from one body or substance to another due to temperature differences. In our exercise, we analyze if heat transfer occurs by contrasting the ideal ratios for an adiabatic process.
Since the provided conditions do not match the requirements for an adiabatic process:
\[ \frac{480.15}{300.15} eq \left(\frac{0.5}{0.1}\right)^{0.2857} \approx 1.8 \]
It's clear that the process is not adiabatic. The mismatch reveals that the system must have some heat transfer. Since the air's temperature increases, we can deduce that heat is added to the system. Tracking heat flow helps us understand energy conversion and the efficiency of processes in thermodynamics.

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Most popular questions from this chapter

Complete the following involving reversible and irreversible cycles: (a) Reversible and irreversible power cycles each discharge energy \(Q_{\mathrm{C}}\) to a cold reservoir at temperature \(T_{\mathrm{C}}\) and receive energy \(Q_{\mathrm{H}}\) from hot reservoirs at temperatures \(T_{\mathrm{H}}\) and \(T_{\mathrm{H}}^{\prime}\), respectively. There are no other heat transfers. Show that \(T_{\mathrm{H}}^{\prime}>T_{\mathrm{H}}\). (b) Reversible and irreversible refrigeration cycles each discharge energy \(Q_{\mathrm{H}}\) to a hot reservoir at temperature \(T_{\mathrm{H}}\) and receive energy \(Q_{C}\) from cold reservoirs at temperatures \(T_{C}\). and \(T_{C}^{\prime}\), respectively. There are no other heat transfers. Show that \(T_{\mathrm{C}}^{\prime}>T_{\mathrm{C}}\). (c) Reversible and irreversible heat pump cycles each receive energy \(Q_{\mathrm{C}}\) from a cold reservoir at temperature \(T_{\mathrm{C}}\) and discharge energy \(Q_{\mathrm{H}}\) to hot reservoirs at temperatures \(T_{\mathrm{H}}\) and \(T_{\mathrm{H}}^{\prime}\), respectively. There are no other heat transfers. Show that \(T_{\mathrm{H}}^{\prime}

Air enters a compressor operating at steady state with a volumetric flow rate of \(8 \mathrm{~m}^{3} / \mathrm{min}\) at \(23^{\circ} \mathrm{C}, 0.12 \mathrm{MPa}\). The air is compressed isothermally without internal irreversibilities, exiting at \(1.5 \mathrm{MPa}\). Kinetic and potential energy effects can be ignored. Evaluate the work required and the heat transfer, each in \(\mathrm{kW}\).

An insulated cylinder is initially divided into halves by a frictionless, thermally conducting piston. On one side of the piston is \(1 \mathrm{~m}^{3}\) of a gas at \(300 \mathrm{~K}, 2\) bar. On the other side is \(1 \mathrm{~m}^{3}\) of the same gas at \(300 \mathrm{~K}, 1\) bar. The piston is released and equilibrium is attained, with the piston experiencing no change of state. Employing the ideal gas model for the gas, determine (a) the final temperature, in \(\mathrm{K}\). (b) the final pressure, in bar. (c) the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{kg}\).

Nitrogen \(\left(\mathrm{N}_{2}\right)\) at \(3.8 \mathrm{~atm}\) and \(170^{\circ} \mathrm{C}\) enters an insulated turbine operating at steady state and expands to \(1 \mathrm{~atm}\). If the isentropic turbine efficiency is \(83.2 \%\), determine the temperature at the turbine exit, in \({ }^{\circ} \mathrm{C}\), using the ideal gas model for the nitrogen and ignoring kinetic and potential energy changes.

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