Question

A piston-cylinder device contains 4 kg of argon at \(250 \mathrm{kPa}\) and \(35^{\circ} \mathrm{C}\). During a quasi-equilibrium, isothermal expansion process, \(15 \mathrm{kJ}\) of boundary work is done by the system, and \(3 \mathrm{kJ}\) of paddle-wheel work is done on the system. Determine the heat transfer for this process.

Short answer

Answer: The heat transfer during the quasi-equilibrium, isothermal expansion process is 12 kJ.

Step-by-step solution

Calculate the total work done during the process

First, we will find the total work done during the process by adding the boundary work done by the system and the paddle-wheel work done on the system. The work done is given by: $$W_\text{total} = W_\text{boundary} + W_\text{paddle-wheel}$$ We are given \(W_\text{boundary} = 15\,\text{kJ}\) and \(W_\text{paddle-wheel} = -3\,\text{kJ}\) (negative because it is done on the system). So, we have: $$W_\text{total} = 15\,\text{kJ} - 3\,\text{kJ} = 12\,\text{kJ}$$

Apply the first law of thermodynamics for the isothermal process

Since the process is isothermal, there is no change in internal energy (ΔU = 0). Therefore, we can write the first law of thermodynamics as: $$Q = \Delta U + W_\text{total}$$ Substitute ΔU = 0 and the total work done calculated in step 1: $$Q = 0 + 12\,\text{kJ}$$

Calculate the heat transfer

Now we can calculate the heat transfer (Q) for the process: $$Q = 12\,\text{kJ}$$ So, the heat transfer during this quasi-equilibrium, isothermal expansion process is 12 kJ.