Question

An insulated \(40-\mathrm{ft}^{3}\) rigid tank contains air at 50 psia and \(120^{\circ} \mathrm{F}\). A valve connected to the tank is now opened, and air is allowed to escape until the pressure inside drops to 25 psia. The air temperature during this process is maintained constant by an electric resistance heater placed in the tank. Determine the electrical work done during this process.

Short answer

Question: Calculate the electrical work done during the process of air release from an insulated rigid tank of 40 ft³ volume. The tank initially contains air at 50 psia and 120°F temperature. The air is released until the pressure in the tank drops to 25 psia, and the temperature is maintained constant throughout the process. Answer: The electrical work done during this process is approximately -69,292 J.

Step-by-step solution

Find the initial mass of air in the tank

To find the initial mass of air in the tank, we can use the ideal gas law: \(PV = mRT\) where, P = pressure (50 psia = 344738.47 Pa), V = volume (40 ft³ = 1132.673 Liter), m = mass (kg), R = specific gas constant for air (287 J/kg·K), T = temperature (120°F = 322.04 K). Rearranging the equation for mass: \(m_{initial} = \frac{PV}{RT}\) Plugging in the values: \(m_{initial} = \frac{(344738.47)(1132.673)} {(287)(322.04)} = 13.033\, kg\)

Find the final mass of air in the tank

Since the tank is insulated and the air temperature remains constant throughout the process, we can assume the process to be isothermal. We can use the ideal gas law for the final state of the tank: \(P_{final}V = m_{final}RT\) where, \(P_{final}\) = final pressure (25 psia = 172369.236 Pa). Rearranging the equation for final mass: \(m_{final} = \frac{P_{final}V}{RT}\) Plugging in the values: \(m_{final} = \frac{(172369.236)(1132.673)} {(287)(322.04)} = 6.516\, kg\)

Find the change in internal energy (ΔU)

Since the process is isothermal and the temperature remains constant, there is no change in internal energy: \(\Delta U = 0\)

Calculate the work done during the process

To find the work done during the process, we can use the first law of thermodynamics: \(\Delta U = Q - W\) Here, \(\Delta U = 0\) (as calculated in Step 3), and since the tank is insulated, there is no heat transfer (Q = 0). Therefore, the electrical work done (W) is: \(W = -\Delta U\)

Use additional information about air release

Since the released air is not required during the process, we can find the total electrical work done as: \(W = W_{cylinder}\) The work done for an isothermal process is given by: \(W = -nR(T_{final}\ln{(\frac{T_{initial}}{T_{final}})} - P_{final}(V_{final} - V_{initial}))\) Now, since, \(T_{initial} = T_{final} = 322.04 K\), the first term becomes zero. And, \(m_{initial}R = m_{final}R\) \(P_{initial}(V_{initial}) = m_{initial}RT_{initial}\) \(P_{final}(V_{final}) = m_{final}RT_{final}\) Thus, we can find the work by solving for it: \(W = -P_{final}(V_{final} - V_{initial})\) Plug in values: \(W = -172369.236 ( \frac{m_{final}R}{P_{final}} - \frac{m_{initial}R}{P_{initial}} )\) \(W = -172369.236 (\frac{(6.516)(287)}{172369.236} - \frac{(13.033)(287)}{344738.47})\) \(W = -69292\, J\) Hence, the electrical work done during this process is approximately -69,292 J (note the negative sign indicating work is done on the system).