A piston-cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 °C. The air undergoes a process to a state where the pressure is 1.0 bar, during which the pressure-volu... A piston-cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 °C. The air undergoes a process to a state where the pressure is 1.0 bar, during which the pressure-volume relationship is pV = constant. Assume ideal gas behavior for the air. Determine the work and heat transfer, in kJ.
Understand the Problem
The problem describes a thermodynamic process involving air within a piston-cylinder assembly. Given the initial conditions (mass, pressure, and temperature) and the final pressure, along with the relationship pV = constant and the ideal gas assumption, we need to determine the work done and heat transfer during the process.
Answer
Work done: $W = 301.3 \text{ kJ}$ Heat transfer: $Q = 301.12 \text{ kJ}$
Answer for screen readers
Work done: $W = 301.3 \text{ kJ}$ Heat transfer: $Q = 301.12 \text{ kJ}$
Steps to Solve
- Convert units and list given values
The initial pressure $p_1 = 2.0 \text{ bar} = 200 \text{ kPa}$, the final pressure $p_2 = 1.0 \text{ bar} = 100 \text{ kPa}$, the initial temperature $T_1 = 30^\circ \text{C} = 303.15 \text{ K}$, and the mass $m = 5.0 \text{ kg}$. Also, $pV = \text{constant}$.
- Calculate the specific gas constant for air
The specific gas constant for air is $R = 0.287 \text{ kJ/kg.K}$.
- Determine the initial specific volume
Using the ideal gas law, $p_1v_1 = RT_1$, we can find the initial specific volume $v_1$ $$ v_1 = \frac{RT_1}{p_1} = \frac{0.287 \frac{\text{kJ}}{\text{kg.K}} \times 303.15 \text{ K}}{200 \text{ kPa}} = 0.435 \frac{\text{m}^3}{\text{kg}} $$
- Find the final specific volume
Since $pV = \text{constant}$, we also have $p_1V_1 = p_2V_2$ or $p_1v_1 = p_2v_2$. Then, the final specific volume $v_2$ can be determined by: $$ v_2 = \frac{p_1v_1}{p_2} = \frac{200 \text{ kPa} \times 0.435 \frac{\text{m}^3}{\text{kg}}}{100 \text{ kPa}} = 0.870 \frac{\text{m}^3}{\text{kg}} $$
- Calculate the work done
The work done for a process where $pV = \text{constant}$ is given by: $$ W = \int_{V_1}^{V_2} p ,dV = p_1V_1 \ln{\frac{V_2}{V_1}} = m p_1v_1 \ln{\frac{v_2}{v_1}} $$ Substituting the given values: $$ W = 5.0 \text{ kg} \times 200 \text{ kPa} \times 0.435 \frac{\text{m}^3}{\text{kg}} \times \ln{\frac{0.870 \frac{\text{m}^3}{\text{kg}}}{0.435 \frac{\text{m}^3}{\text{kg}}}} = 301.3 \text{ kJ} $$
- Determine the final temperature
From the ideal gas law $p_2v_2 = RT_2$ $$ T_2 = \frac{p_2v_2}{R} = \frac{100 \text{ kPa} \times 0.870 \frac{\text{m}^3}{\text{kg}}}{0.287 \frac{\text{kJ}}{\text{kg.K}}} = 303.1 \text{ K} $$ $T_2 = 303.1 \text{ K}$ which is about $30^\circ \text{C}$
- Calculate the change in internal energy
The change in internal energy is given by $\Delta U = mc_v(T_2 - T_1)$. For air, $c_v = 0.718 \text{ kJ/kg.K}$. $$ \Delta U = 5.0 \text{ kg} \times 0.718 \frac{\text{kJ}}{\text{kg.K}} \times (303.1 - 303.15) \text{ K} = -0.1795 \text{ kJ} \approx 0 \text{ kJ} $$ Since the change in temperature is very small, the change in internal energy is approximately zero.
- Calculate the heat transfer
From the first law of thermodynamics, $\Delta U = Q - W$ $$ Q = \Delta U + W = -0.1795 \text{ kJ} + 301.3 \text{ kJ} = 301.12 \text{ kJ} $$
Work done: $W = 301.3 \text{ kJ}$ Heat transfer: $Q = 301.12 \text{ kJ}$
More Information
The work done is positive, indicating that work is done by the system. The heat transfer is also positive, meaning heat is added to the system. For an isothermal process (constant temperature) with an ideal gas, the change in internal energy is zero, and the heat transfer equals the work done.
Tips
A common mistake is not converting the units to a consistent system (e.g., kPa, m$^3$, kg, K) before performing calculations. Also, using the wrong gas constant or specific heat value for air is a common mistake. Students might also incorrectly apply the first law of thermodynamics or the work equation for the given process.
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