Question

Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow either gas to leak into the other side. Initially, one side of the piston contains \(1 \mathrm{m}^{3}\) of \(\mathrm{N}_{2}\) gas at \(500 \mathrm{kPa}\) and \(80^{\circ} \mathrm{C}\) while the other side contains \(1 \mathrm{m}^{3}\) of \(\mathrm{He}\) gas at \(500 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine \((a)\) the final equilibrium temperature in the cylinder and ( \(b\) ) the wasted work potential during this process. What would your answer be if the piston were not free to move? Take \(T_{0}=25^{\circ} \mathrm{C}\)

Short answer

Additionally, what would the result be if the piston were not free to move? Answer: (a) The final equilibrium temperature can be found using the expression: $$\frac{m_{N2}C_{vN2}T_{1N2}}{m_{N2}C_{vN2} + m_{He}C_{vHe}} = T_{2}$$, by calculating the initial internal energy of both gases and using the energy conservation principle. (b) The wasted work potential can be determined using the expression: $$W_{wasted} = (U_{1N2} + U_{1He}) - U_{2}(T_{0})$$, by finding the difference between the initial and final available work. (c) If the piston were not free to move, the final equilibrium temperature would be the same as the initial temperature of the gas, and no work potential would be wasted.

Step-by-step solution

(Step 1: Calculate initial internal energy of both gases)

First, we will calculate the initial internal energy \((U_{1N2})\) of the \(\mathrm{N}_{2}\) gas, and the initial internal energy \((U_{1He})\) of the \(\mathrm{He}\) gas. The internal energy for a gas can be calculated using the following expression: $$U_{1} = mC_{v}T_{1}$$ We are given the specific heat capacities at constant volume \((C_{v})\) for both gases at room temperature, as well as their initial temperatures \((T_{1N2}\) and \(T_{1He})\) and volumes \((V_{1N2}\) and \(V_{1He})\). The mass \((m)\) of the gases can be determined using the ideal gas law: $$PV = nRT$$ Which can be rearranged to find mass: $$m = \frac{P_{1}V_{1}}{RT}$$ For both gases we have the same initial pressure and volume.

(Step 2: Calculate final internal energy of both gases)

The energy conservation states the internal energy of both gases at the end of the process are equal. Therefore, we can write the following equation for the final internal energy: $$U_{2N2} = U_{2He}$$ Then substituting the initial internal energy expressions from Step 1 and rearranging to find the final equilibrium temperature \((T_{2})\): $$\frac{m_{N2}C_{vN2}T_{1N2}}{m_{N2}C_{vN2} + m_{He}C_{vHe}} = T_{2}$$ Now we plug in the values for \(m_{N2}\), \(C_{vN2}\), \(T_{1N2}\), \(m_{He}\), and \(C_{vHe}\) using the ideal gas law and specific heat capacities provided, and solve for \(T_{2}\).

(Step 3: Calculate the wasted work potential)

To determine the wasted work potential during the process, we need to calculate the difference between the initial and final available work. The available work is the maximum work that could be obtained from the system to the surroundings at temperature \(T_{0}\). The wasted work potential can be expressed as: $$W_{wasted} = (U_{1N2} + U_{1He}) - U_{2}(T_{0})$$ Plugging in the values for \(U_{1N2}\), \(U_{1He}\), and \(U_{2}(T_{0})\), we can compute the wasted work potential.

(Step 4: Consider the case of a stationary piston)

If the piston were not free to move, then no work would be done on the gas, and the gas temperature would not change, remaining at its initial value. Consequently, the final equilibrium temperature would be the same as the initial temperature of the gas, and no work potential would be wasted. Therefore, the main output of this exercise is: (a) the final equilibrium temperature found in Step 2, and (b) the wasted work potential found in Step 3.