Question

A \(1 \mathrm{~m}^{3}\) tank initially contains air at \(300 \mathrm{kPa}, 300 \mathrm{~K}\). Air slowly escapes from the tank until the pressure drops to 100 \(\mathrm{kPa}\). The air that remains in the tank undergoes a process described by \(p v^{1.2}=\) constant. For a control volume enclosing the tank, determine the heat transfer, in kJ. Assume ideal gas behavior with constant specific heats.

Short answer

The heat transfer is -123.7 kJ.

Step-by-step solution

Identify initial and final states

Initial State: \[P_1 = 300 \text{ kPa}, \, V_1 = 1 \text{ m}^3, \, T_1 = 300 \text{ K}\] Final State: \[P_2 = 100 \text{ kPa} \text{ and the process follows } p v^{1.2} = \text{constant}\]

Relate volumes using polytropic process

For a polytropic process, \(P_1 V_1^{1.2} = P_2 V_2^{1.2}\). Thus, we can solve for \(V_2\): \[V_2 = V_1 \left(\frac{P_1}{P_2}\right)^{\frac{1}{1.2}}\] Substitute the given values: \[V_2 = 1 \text{ m}^3 \left(\frac{300 \text{ kPa}}{100 \text{ kPa}}\right)^{\frac{1}{1.2}} = 2.143\text{ m}^3\]

Determine the final temperature

Using the ideal gas law, the specific volume \(v\) at state 1 is: \[v_1 = \frac{V_1}{\text{mass}} = \frac{RT_1}{P_1}\] And for state 2: \[v_2 = \frac{V_2}{\text{mass}} = \frac{RT_2}{P_2}\] Relating the temperatures using the polytropic process equation, and assuming \(R\) is constant, we get: \[T_2 = T_1 \left(\frac{V_2}{V_1}\right)^{1-1.2} = 300 \text{ K} \left(2.143\right)^{-0.2} = 251 \text{ K}\]

Calculate internal energy change

The internal energy change \(\Delta U\) can be calculated using: \[\Delta U = m c_v (T_2 - T_1)\] Calculating mass \(m\) using \(P_1 V_1 = m R T_1\): \[m = \frac{P_1 V_1}{R T_1} = \frac{300 \text{ kPa} \times 1 \text{ m}^3}{0.287 \text{ kJ/kg⋅K} \times 300 \text{ K}} = 3.484 \text{ kg}\] Specific heat at constant volume \(c_v\) for air is approximately \(0.718 \text{ kJ/kg⋅K}\). Hence: \[\Delta U = 3.484 \text{ kg} \times 0.718 \text{ kJ/kg⋅K} \times (251 \text{ K} - 300 \text{ K}) = -123.7 \text{ kJ}\]

Apply the first law of thermodynamics

The first law of thermodynamics for a control volume is given by: \[\Delta U = Q - W\] Since there is no work done (volume change at constant pressure), \(W = 0\). Therefore, the heat transfer \(Q\) is: \[Q = \Delta U = -123.7 \text{ kJ}\]

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in thermodynamics, describing the state of an ideal gas using the equation of state:\[ PV = nRT \]where \( P \) stands for pressure, \( V \) for volume, \( n \) for the number of moles, \( R \) for the universal gas constant, and \( T \) for temperature in Kelvin. This law helps us understand relationships between these properties in gaseous systems.In our exercise, we applied the ideal gas law at different states (initial and final) to determine quantities like mass and temperature change.
To apply the ideal gas law correctly, make sure you:

• Use consistent units for pressure (Pa, kPa), volume (m³), and temperature (K).
• Use the specific gas constant for air, which approximately is \( 0.287 \text{ kJ/kg·K} \).

The ideal gas law simplification assumes no intermolecular forces, thus only holds exactly true under specific conditions:

• High temperatures
• Low pressures

But it provides a good approximation for many practical situations.

Internal Energy Change

Internal energy change (\(\Delta U\)) is a measure of the energy gained or lost by a system due to changes in temperature. For an ideal gas, the formula to calculate this change is: \[ \Delta U = m c_v \Delta T \]where \(m\) represents the mass of the gas, \(c_v\) the specific heat at constant volume, and \(\Delta T\) the temperature change.Let's break down the steps to calculate internal energy change more fully:

• Firstly, determine the mass of the gas using: \( m = \frac{P_1 V_1}{R T_1} \).
• This mass helps in computing the change in internal energy when temperature changes from \( T_1 \) to \( T_2 \).
• For air, a common value for \( c_v \) is around 0.718 kJ/kgK.

In our exercise:

• \(m\) = 3.484 kg was calculated using the initial state properties.
• Then we used this to compute \( \Delta U = 3.484 \times 0.718 \times (251 - 300) = -123.7 \text{kJ} \).

A negative \( \Delta U \) indicates the system lost energy. Thus, internal energy changes directly correlate with temperature changes, considering specific heat and gas mass.

First Law of Thermodynamics

The First Law of Thermodynamics establishes the principle of energy conservation in thermodynamic systems. It states: \[ \Delta U = Q - W \]where \( \Delta U \) is the change in internal energy, \( Q \) is the heat transfer into the system, and \( W \) is the work done by the system.In our particular exercise, we found:

• \(W\) was zero because there was no work done (isobaric volume change).
• Thus, the change in internal energy equaled the heat transfer (\( Q = \Delta U = -123.7 \text{kJ} \)).

This law is crucial for solving many engineering and physics problems as it concretely links energy changes in a system to heat transfer and work done. To apply this law effectively:

• Always define the system boundary and assess if work or heat transfer occurs across it.
• Use appropriate signs: positive for heat added to the system and negative for heat lost.

By adhering to the first law, we ensure that energy is neither created nor destroyed but transformed from one form to another. This understanding allows us to solve for unknown quantities in diverse thermodynamic processes.