Question

A horizontal cylinder is separated into two compartments by an adiabatic, frictionless piston. One side contains \(0.2 \mathrm{m}^{3}\) of nitrogen and the other side contains \(0.1 \mathrm{kg}\) of helium, both initially at \(20^{\circ} \mathrm{C}\) and 95 kPa. The sides of the cylinder and the helium end are insulated. Now heat is added to the nitrogen side from a reservoir at \(500^{\circ} \mathrm{C}\) until the pressure of the helium rises to 120 kPa. Determine \((a)\) the final temperature of the helium, \((b)\) the final volume of the nitrogen, \((c)\) the heat transferred to the nitrogen, and \((d)\) the entropy generation during this process.

Short answer

In summary, for the given process involving a horizontal cylinder separated into a nitrogen and helium compartment: a) The final temperature of the helium is 356.46 K. b) The final volume of the nitrogen is 0.1 m³. c) The heat transferred to the nitrogen is 9931.82 J. d) The entropy generation during the process is 60.94 J/K.

Step-by-step solution

Find the initial number of moles for helium and nitrogen

We need to find the number of moles for helium and nitrogen. Using the given mass and volume for helium and nitrogen and their respective molar masses (\(M_{He} = 4.00 \mathrm{kg/mol}\) and \(M_{N2} = 28.02 \mathrm{kg/mol}\)), we can find their initial number of moles. \(n_{He} = \frac{m_{He}}{M_{He}} = \frac{0.1}{4.00} = 0.025\,\text{mol}\) For nitrogen, we can use the ideal gas law to find the initial number of moles: $PV=nRT \Rightarrow n_{N2} = \frac{PV}{RT} \\ n_{N2} = \frac{95\times10^3\text{Pa} \times 0.2\text{m}^3}{8.314\text{J/(mol}\text{K)}\times (20 + 273.15)\text{K}} \\ n_{N2} \approx 0.771\, \text{mol}$

Find the final temperature of the helium

Since the process is adiabatic, the work done by the gravity on the piston can be ignored. Thus, the adiabatic relationship can be expressed as: \(P_1V_1^\gamma = P_2V_2^\gamma\), where P is the pressure, V is the volume, and \(\gamma\) is the specific heat ratio. For helium, \(\gamma_{He} = 1.66\). The final pressure (`P_2`) of the helium is given as 120 kPa. As the ideal gas law holds true for each gas, the volume ratio (`V_1/V_2`) can be replaced by the temperature ratio (`T_1/T_2`). Rearrange the equation for the final temperature (`T_2`) of the helium: \(T_2 = \frac{P_2V_1^\gamma}{P_1V_1^\gamma} T_1\) \(T_2 = \frac{120 \times 10^3\text{Pa}}{95 \times 10^3\text{Pa}}\times \left(\frac{0.2}{0.3}\right)^{1.66} \times (20 + 273.15)\,\text{K}\) \(T_2 \approx 356.46\,\text{K}\) So, the final temperature of the helium is \(356.46\,\text{K}\).

Find the final volume of the nitrogen

As the total volume of the cylinder is fixed, the volume displaced by helium will be equal to the change in nitrogen volume. \(V_{N2,final} = V_{N2,initial} - V_{He,initial} \) \(V_{N2,final} = 0.2\,\text{m}^3 - 0.1\,\text{m}^3\) \(V_{N2,final} = 0.1\,\text{m}^3\) So, the final volume of the nitrogen is \(0.1\,\text{m}^3\).

Calculate the heat transferred to the nitrogen

We can use the relationship between heat and specific heat capacities, \(Q = mc_{p}\Delta T\) to calculate the heat transferred to nitrogen. First, we have to find the final temperature of nitrogen. We can find it by using the ideal gas law: \(T_{N2,final} = \frac{P_{N2,final}V_{N2,final}}{n_{N2}R}\) Since the final pressure on both sides of the piston is equal: \(P_{N2,final} = P_{He,final} = 120\,\text{kPa}\) Then, \(T_{N2,final} = \frac{120\times10^3\text{Pa} \times 0.1\text{m}^3}{0.771 \times 8.314\text{J/(mol}\text{K)}}\) \(T_{N2,final} \approx 592.92\,\text{K}\) Now we can calculate the heat transferred to the nitrogen using the specific heat capacity at constant pressure for nitrogen, \(c_{p,N2} = 29.12\,\text{J/(mol}\,\text{K)}\): \(Q_{N2} = n_{N2}c_{p,N2}(T_{N2,final} - T_{N2,initial})\) \(Q_{N2} = 0.771\times 29.12\,\text{J/(mol}\,\text{K)} \times (592.92 - (20+273.15))\,\text{K}\) \(Q_{N2} \approx 9931.82\,\text{J}\) So, the heat transferred to the nitrogen is approximately \(9931.82\,\text{J}\).

Calculate the entropy generation during the process

Entropy generation for an isolated system can be calculated as: \(\Delta S_{gen} = \Delta S_{N2} + \Delta S_{He}\) The entropy change for each gas can be calculated as: \(\Delta S = n \cdot c_{p} \cdot \ln{\frac{T_2}{T_1}} - n \cdot R \cdot \ln{\frac{P_2}{P_1}}\) We already have the values of \(T_1\), \(T_2\), \(P_1\), and \(P_2\) for both gases. The gas constant for either gas is \(R = 8.314\,\text{J/(mol}\,\text{K)}\). \(\Delta S_{N2} = 0.771 \times 29.12\,\text{J/(mol}\,\text{K)} \times \ln{\frac{592.92}{293.15}} - 0.771 \times 8.314\,\text{J/(mol}\,\text{K)} \times \ln{\frac{120\times10^3}{95\times10^3}}\) \(\Delta S_{N2} \approx 59.69\,\text{J/K}\) \(\Delta S_{He} = 0.025 \times 20.8\,\text{J/(mol}\,\text{K)} \times \ln{\frac{356.46}{293.15}} - 0.025 \times 8.314\,\text{J/(mol}\,\text{K)} \times \ln{\frac{120\times10^3}{95\times10^3}}\) \(\Delta S_{He} \approx 1.25\,\text{J/K}\) Now, we can find the entropy generation: \(\Delta S_{gen} = 59.69\,\text{J/K} + 1.25\,\text{J/K}\) \(\Delta S_{gen} \approx 60.94\,\text{J/K}\) So, the entropy generation during this process is approximately \(60.94\,\text{J/K}\).