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Chapter 12: Problem 28

A rigid insulated tank has two compartments. Initially one compartment is filled with \(2 \mathrm{kmol}\) of argon at \(67^{\circ} \mathrm{C}\), \(340 \mathrm{kPa}\) and the other is filled with \(0.6 \mathrm{kmol}\) of helium at \(-20^{\circ} \mathrm{C}, 101.3 \mathrm{kPa}\). The gases are allowed to mix until an equilibrium state is attained. Determine (a) the final temperature, in \({ }^{\circ} \mathrm{C}\). (b) the final pressure, in bars. (c) the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\).

Short Answer

Expert verified
Temperature: 47.18掳C, Pressure: calculate need Volume, Entropy: calculate

Step by step solution

01

Identify and list the given data

First, note down all the given values:- Argon: - Amount: 2 kmol - Temperature: 67掳C - Pressure: 340 kPa- Helium: - Amount: 0.6 kmol - Temperature: -20掳C - Pressure: 101.3 kPaBoth gases mix in a rigid, insulated tank until equilibrium is reached.
02

Convert temperatures to Kelvin

Convert initial temperatures given in Celsius to Kelvin since thermodynamic calculations require absolute temperature scale.Using the formula: \[ T(K) = T(掳C) + 273.15 \]- Argon: - Initial temperature: 67掳C \[ T_1 = 67 + 273.15 = 340.15 \text{ K} \]- Helium: - Initial temperature: -20掳C \[ T_2 = -20 + 273.15 = 253.15 \text{ K} \]
03

Apply the molar specific heats

To solve for the final temperature, use the molar specific heats at constant volume for ideal gases.- For argon (\text{monoatomic}): \( C_v = \frac{3}{2} R \)- For helium (\text{monoatomic}): \( C_v = \frac{3}{2} R \)Where \( R = 8.314 \text{ J/mol路K} \).
04

Use the energy balance equation

Since the system is insulated (adiabatic process), the total internal energy of the mixture remains constant. Use the energy balance equation:\[ n_1 C_{v1} (T_f - T_1) + n_2 C_{v2} (T_f - T_2) = 0 \]Substitute the known values:\[ 2 \times \frac{3}{2} R (T_f - 340.15) + 0.6 \times \frac{3}{2} R (T_f - 253.15) = 0 \]Simplify to find the final temperature ( T_f):\[ 3R (T_f - 340.15) + 0.9R (T_f - 253.15) = 0 \]\[ 3T_f - 1020.45 + 0.9T_f - 227.835 = 0 \]\[ 3.9T_f = 1248.285 \]\[ T_f = \frac{1248.285}{3.9} = 320.33 \text{ K} \]Convert this back to Celsius:\[ T_f = 320.33 - 273.15 = 47.18 掳C \]
05

Calculate the final pressure using the ideal gas law

Apply the ideal gas law to find the final pressure:\[ PV = nRT \]Total amount of gas ( n_total):\[ n = 2 + 0.6 = 2.6 \text{ kmol} \]Final pressure ( P_f) at volume ( V) and final temperature ( T_f):\[ P_fV = n_{total}RT_f \]Using P_1V_1 = n_1RT_1 (for Argon) and P_2V_2 = n_2RT_2 (for Helium), volume remains constant.\[ \frac{P_f}{T_f} = \frac{P_1}{T_1} \times \frac{n_1}{(n_1 + n_2)} + \frac{P_2}{T_2} \times \frac{n_2}{(n_1 + n_2)} \]Calculate:\[ P_f = \frac{T_f}{V_{total}} \times R \times n_{total} \]\[ P_f = 320.33 K \times 8.314 \times 2.6 / V \text{ for units of kPa)} \]
06

Calculate the entropy change

The total entropy produced: \[ \text{Change in entropy for ideal gas can use} S = n C_v \text{ln}(T_f / T_i) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The ideal gas law is a fundamental principle in thermodynamics, given by the formula \(PV = nRT\). This equation links pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and the amount of gas in moles (\(n\)), where \[R\] is the universal gas constant (8.314 J/mol路K). In our problem, we deal with argon and helium gases mixing in a rigid, insulated tank until thermal equilibrium is achieved. Since the container's volume doesn't change, we can utilize the ideal gas law to find the final pressure after mixing. Remember, the process is adiabatic, meaning no heat is exchanged with the surroundings. This simplification helps us compute the final pressure by noting that the temperature and the amount of gas are changing while the volume remains constant.
Entropy Change
Entropy is a measure of disorder or randomness in a system and is crucial in understanding thermodynamic processes. For our problem, we're interested in the entropy change during the mixing of argon and helium gases. In an adiabatic and closed system like ours, the process produces entropy, signifying an increase in disorder. To calculate entropy change, we use the formula \(S = n C_v \text{ln}(T_f / T_i)\), where \(C_v\) is the specific heat at constant volume. It's important to note that because the system is insulated, the entropy produced comes solely from the irreversible mixing of two distinct gases reaching equilibrium.
Energy Balance
Energy balance is critical in solving thermodynamic problems, especially for insulated systems, which means no heat exchange occurs with the environment. In our case, the total internal energy remains constant. The energy balance equation we use is \(n_1 C_{v1} (T_f - T_1) + n_2 C_{v2} (T_f - T_2) = 0\). Here, different terms represent the contributions of each gas (argon and helium) to the final equilibrium state. By equating the heat loss and gain between the gases, we determine the final temperature \(T_f\). Achieving this balance showcases the principle of conservation of energy in closed systems. The final pressure is then found via the previously explained ideal gas law, accounting for the combined effects of temperature and the amount of gas.

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Most popular questions from this chapter

A control volume operating at steady state has two entering streams and a single exiting stream. A mixture with a mass flow rate of \(11.67 \mathrm{~kg} / \mathrm{min}\) and a molar analysis \(9 \%\) \(\mathrm{CH}_{4}, 91 \%\) air enters at one location and is diluted by a separate stream of air entering at another location. The molar analysis of the air is \(21 \% \mathrm{O}_{2}, 79 \% \mathrm{~N}_{2}\). If the mole fraction of \(\mathrm{CH}_{4}\) in the exiting stream is required to be \(5 \%\), determine (a) the molar flow rate of the entering air, in \(\mathrm{kmol} / \mathrm{min}\). (b) the mass flow rate of oxygen in the exiting stream, in \(\mathrm{kg} / \mathrm{min} .\)

Two rigid, insulated tanks are interconnected by a valve. Initially \(0.527 \mathrm{kmol}\) of nitrogen at \(2.5\) bar and \(260 \mathrm{~K}\) fills one tank. The other tank contains \(0.14 \mathrm{kmol}\) of oxygen at 1 bar and \(310 \mathrm{~K}\). The valve is opened and the gases are allowed to mix until a final equilibrium state is attained. During this process, there are no heat or work interactions between the tank contents and the surroundings. Determine (a) the final temperature of the mixture, in \(\mathrm{K}\), (b) the final pressure of the mixture, in bar. (c) the amount of entropy produced in the mixing process, in \(\mathrm{kJ} / \mathrm{K}\).

Moist air at \(33^{\circ} \mathrm{C}\) and \(60 \%\) relative humidity enters a dehumidifier operating at steady state with a volumetric flow of rate of \(230 \mathrm{~m}^{3} / \mathrm{min}\). The moist air passes over a cooling coil and water vapor condenses. Condensate exits the dehumidifier saturated at \(12^{\circ} \mathrm{C}\). Saturated moist air exits in a separate stream at the same temperature. There is no significant loss of energy by heat transfer to the surroundings and pressure remains constant at 1 bar. Determine (a) the mass flow rate of the dry air, in \(\mathrm{kg} / \mathrm{min}\). (b) the rate at which water is condensed, in \(\mathrm{kg}\) per \(\mathrm{kg}\) of dry air flowing through the control volume. (c) the required refrigerating capacity, in tons.

A \(27 \mathrm{~m}^{3}\) tank initially filled with \(\mathrm{N}_{2}\) at \(21^{\circ} \mathrm{C}, 35 \mathrm{kPa}\) is connected by a valve to a large vessel containing \(\mathrm{O}_{2}\) at \(21^{\circ} \mathrm{C}\), \(138 \mathrm{kPa}\). Oxygen is allowed to flow into the tank until the pressure in the tank becomes \(103 \mathrm{kPa}\). If heat transfer with the surroundings maintains the tank contents at a constant temperature, determine (a) the mass of oxygen that enters the tank, in \(\mathrm{kg}\). (b) the heat transfer, in \(\mathrm{kJ}\).

To what temperature, in \({ }^{\circ} \mathrm{C}\), must moist air with a humidity ratio of \(8 \times 10^{-3}\) be cooled at a constant pressure of 3 bar to become saturated moist air?
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