Question

A vertical piston-cylinder device initially contains \(0.2 \mathrm{m}^{3}\) of air at \(20^{\circ} \mathrm{C}\). The mass of the piston is such that it maintains a constant pressure of \(300 \mathrm{kPa}\) inside. Now a valve connected to the cylinder is opened, and air is allowed to escape until the volume inside the cylinder is decreased by one-half. Heat transfer takes place during the process so that the temperature of the air in the cylinder remains constant. Determine \((a)\) the amount of air that has left the cylinder and (b) the amount of heat transfer.

Short answer

Answer: Half of the initial mass of air has left the cylinder, and the amount of heat transfer during the process is 30,000 J (heat is added to the system).

Step-by-step solution

Write down the given information

. Initially, the cylinder contains 0.2 m³ of air at 20°C, and the pressure is constant at 300 kPa. The volume inside will be decreased by half.

Calculate the initial and final volume

. Since the volume of the air inside the cylinder is decreased by half, we can find the initial and final volume as: Initial volume, \(V_1 = 0.2 \, \text{m}^3\) Final volume, \(V_2 = 0.5 \cdot V_1 = 0.1 \, \text{m}^3\)

Use ideal gas law to find the initial and final mass of air

. Since temperature and pressure are constant throughout the process, we can use the ideal gas law to find the initial and final mass of air: Initial state: \(m_1 = \frac{P_1 V_1}{RT_1}\) Final state: \(m_2 = \frac{P_2 V_2}{RT_2}\) Note that the temperature is constant (\(T_1 = T_2 = T\)), and so is the pressure (\(P_1 = P_2 = P\)). Therefore, we can find the mass ratio as: \(\frac{m_2}{m_1} = \frac{V_2}{V_1}\)

Calculate the mass of air that has left the cylinder

. Since we found the mass ratio in step 3, we can now find the mass of air that has left the cylinder: \(\Delta m = m_1 - m_2 = m_1 (1 - \frac{V_2}{V_1})\) Calculate the value using the given values: \(\Delta m = m_1 (1 - \frac{0.1}{0.2}) = 0.5 m_1\) The amount of air that has left the cylinder is half of the initial mass of air.

Calculate the amount of heat transfer using the first law of thermodynamics

. Since the process is isothermal (constant temperature), the change in internal energy is zero: \(\Delta U = 0\). We can use the first law of thermodynamics to find the heat transfer, \(Q\): \(Q - W = \Delta U\) Since \(\Delta U = 0\): \(Q = W\) We know that the work done during the process is \(W = -P(V_2 - V_1)\): \(Q = -P(V_2 - V_1)\) Calculate the heat transfer using the given values: \(Q = -(300 \, \text{kPa})(0.1 \, \text{m}^3 - 0.2 \, \text{m}^3) = -300(10^3 \, \text{Pa})(-0.1 \, \text{m}^3) = 30,000 \, \text{J}\) So, the amount of heat transfer during the process is 30,000 J (positive sign indicates heat is added to the system).