Question

An insulated piston-cylinder device contains \(100 \mathrm{L}\) of air at \(400 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). A paddle wheel within the cylinder is rotated until \(15 \mathrm{kJ}\) of work is done on the air while the pressure is held constant. Determine the final temperature of the air. Neglect the energy stored in the paddle wheel.

Short answer

Answer: The final temperature of the air is approximately 18.43°C.

Step-by-step solution

Identify the formula for the First Law of Thermodynamics

We will use the First Law of Thermodynamics for a closed system, which is given by: \[ \Delta U = Q - W \] where \(\Delta U\) is the change in internal energy, \(Q\) is the heat transfer, and \(W\) is the work done. Since the system is well-insulated, there is no heat transfer, so we have: \[ \Delta U = -W \]

Express internal energy change in terms of temperature

We can express the change in internal energy as: \[ \Delta U = m C_\textrm{v} \Delta T \] where \(m\) is the mass of the air, \(C_\textrm{v}\) is the specific heat at constant volume, and \(\Delta T\) is the change in temperature. Now substitute this expression into the First Law of Thermodynamics: \[ m C_\textrm{v} \Delta T = - W \]

Find the mass of air using Ideal Gas Law

We can find the mass of the air using the Ideal Gas Law: \[ PV = nRT \] where \(P\) 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. Now, we have: \[ PV = \frac{m}{M} R T \implies m = \frac{PVM}{R T} \] where, \(P = 400\times 10^3 \,\mathrm{Pa}\), \(V = 100 \times 10^{-3} \, \mathrm{m^3}\), \(M\) is the molar mass of air which is approximately \(29 \, \mathrm{g\,mol^{-1}}\), \(R = 8.314 \, \mathrm{J\,mol^{-1}\,K^{-1}}\), and \(T = 25^\circ \mathrm{C} = 298.15\,\mathrm{K}\). Substitute these values into the mass equation above, and then solve for the mass: \[ m = \frac{(400\times 10^3 \,\mathrm{Pa})(100 \times 10^{-3} \, \mathrm{m^3})(29 \, \mathrm{g\,mol^{-1}})}{(8.314 \, \mathrm{J\,mol^{-1}\,K^{-1}})(298.15\,\mathrm{K})} \] Solve for \(m\): \[ m \approx 4.810\,\mathrm{kg} \]

Solve for the change in temperature

Now we have: \[ m C_\textrm{v} \Delta T = - W \] Substitute the known values into this equation: \[ (4.810 \,\mathrm{kg})(0.717 \, \mathrm{kJ \, kg^{-1} \, K^{-1}}) \Delta T = - 15 \, \mathrm{kJ} \] Now, solve for the change in temperature, \(\Delta T\): \[ \Delta T \approx -6.574^\circ \, \mathrm{C} \]

Calculate the final temperature of the air

Finally, we can find the final temperature \(T_f\) by adding the initial temperature \(T_i\) and the change in temperature \(\Delta T\): \[ T_f = T_i + \Delta T \] Substituting the values: \[ T_f = 25^\circ \, \mathrm{C} - 6.574^\circ \, \mathrm{C} \approx 18.426^\circ \, \mathrm{C} \] So, the final temperature of the air is approximately \(18.43^\circ \mathrm{C}\).