Question

Using the ideal gas model with constant specific heats, determine the mixture temperature, in \(\mathrm{K}\), for each of two cases: (a) Initially, \(0.6 \mathrm{kmol}\) of \(\mathrm{O}_{2}\) at \(500 \mathrm{~K}\) is separated by a partition from \(0.4 \mathrm{kmol}\) of \(\mathrm{H}_{2}\) at \(300 \mathrm{~K}\) in a rigid insulated vessel. The partition is removed and the gases mix to obtain a final equilibrium state. (b) Oxygen \(\left(\mathrm{O}_{2}\right)\) at \(500 \mathrm{~K}\) and a molar flow rate of \(0.6 \mathrm{kmol} / \mathrm{s}\) enters an insulated control volume operating at steady state and mixes with \(\mathrm{H}_{2}\) entering as a separate stream at \(300 \mathrm{~K}\) and a molar flow rate of \(0.4 \mathrm{kmol} / \mathrm{s}\). A single mixed stream exits. Kinetic and potential energy effects can be ignored.

Short answer

For case (a), the final mixture temperature can be determined using the specific heats and initial temperatures and moles. For case (b), the final temperature is found similarly, but using flow rates.

Step-by-step solution

Title - Identify System Properties

We need to determine the mixture temperature for two cases. First, note the given properties:(a) Initially, 0.6 kmol of \(\text{O}_2\) at 500 K and 0.4 kmol of \(\text{H}_2\) at 300 K in a rigid insulated vessel.(b) Oxygen (\(\text{O}_2\)) at 500 K with molar flow rate 0.6 kmol/s and hydrogen (\(\text{H}_2\)) at 300 K with molar flow rate 0.4 kmol/s in a steady-state insulated control volume.

- Use Ideal Gas Model Assumptions

Since we are using the ideal gas model with constant specific heats, the internal energy of each gas can be approximated using \(\text{C}_v\), the molar specific heat at constant volume.

- Calculate Mixture Temperature for Case (a)

For a closed, insulated vessel:The internal energy before mixing is:\( U_{\text{initial}} = n_{\text{O}_2} \cdot C_{v, \text{O}_2} \cdot T_{\text{O}_2} + n_{\text{H}_2} \cdot C_{v, \text{H}_2} \cdot T_{\text{H}_2} \)Since there is no heat transfer and it is a rigid vessel, the total internal energy remains constant:\(U_{\text{final}} = U_{\text{initial}}\)Therefore, for the final mixture temperature \(T_{\text{final}}\):\( n_{\text{total}} \cdot C_v \cdot T_{\text{final}} = n_{\text{O}_2} \cdot C_{v, \text{O}_2} \cdot 500 \ + n_{\text{H}_2} \cdot C_{v, \text{H}_2} \cdot 300 \)Where \(n_{\text{total}} = n_{\text{O}_2} + n_{\text{H}_2} = 0.6 \text{kmol} + 0.4 \text{kmol} = 1.0 \text{kmol}\)Solve for \(T_{\text{final}} \)

- Apply Constant Specific Heat Values

Use the given or standard \(\text{C}_v\) values for \(\text{O}_2\) and \(\text{H}_2\): \(\text{C}_{v, \text{O}_2}= 21.047 \text{ J/(mol·K)} \) and \(\text{C}_{v, \text{H}_2} = 20.183 \text{ J/(mol·K)}\).Plug these values into the equation and solve for final mixture temperature \(T_{\text{final, (a)}} \)

- Calculate Mixture Temperature for Case (b)

For a steady-state insulated control volume:The enthalpy before mixing is:\(H_{\text{initial}} = \dot{n}_{\text{O}_2} \cdot C_{p, \text{O}_2} \cdot T_{\text{O}_2} + \dot{n}_{\text{H}_2} \cdot C_{p, \text{H}_2} \cdot T_{\text{H}_2}\)Enthalpy must remain constant since the system is insulated:\(H_{\text{final}} = H_{\text{initial}}\)Therefore, for the final mixture temperature \(T_{\text{final}}\):\( \dot{n}_{\text{total}} \cdot C_p \cdot T_{\text{final}} = \dot{n}_{\text{O}_2} \cdot C_{p, \text{O}_2} \cdot 500 + \dot{n}_{\text{H}_2} \cdot C_{p, \text{H}_2} \cdot 300 \)Where \(\dot{n}_{\text{total}} = \dot{n}_{\text{O}_2} + \dot{n}_{\text{H}_2} = 0.6 \text{kmol/s} + 0.4 \text{kmol/s} = 1.0 \text{kmol/s}\)Solve for \(T_{\text{final, (b)}} \). Use the given or standard \(\text{C}_{p}\) values for \(\text{O}_2 \) (29.355 J/(mol·K)) and \(\text{H}_2 \) (28.836 J/(mol·K)).

Key concepts

mixture temperature

When different gases mix, their individual temperatures usually change until they reach a new equilibrium temperature. This temperature is called the mixture temperature and is found using energy balance principles.
For example, in a closed and insulated system, we use the internal energy of each gas before mixing and assume no energy is lost or gained by the system. This tells us that the total internal energy before mixing is equal to the total internal energy after mixing.
We can calculate the final mixture temperature by ensuring the sum of the products of the specific heat and temperature of each gas, multiplied by their amounts, remains constant.

constant specific heats

Specific heats are properties of gases that tell us how much energy is needed to change their temperature. For ideal gases, these can be considered constant for a wide range of temperatures.
The molar specific heat at constant volume (\(C_v\)) is used when there is no change in volume, and the molar specific heat at constant pressure (\(C_p\)) is used when the pressure remains constant.
In our example, we use given or standard values for the specific heats of oxygen and hydrogen to make our calculations.

steady-state control volume

A steady-state control volume is a system where conditions do not change over time. This means all fluxes (inflows and outflows) are balanced.
For mixture calculations, it means the total energy entering the system is equal to the total energy exiting the system. In an insulated control volume, this energy balance directly translates to the final mixture temperature.
In the given problem, we assume that the molar flow rates and specific heats determine the enthalpy of the system at steady state.

insulated vessel

An insulated vessel means no heat is exchanged with the environment. This simplifies our energy balance because we assume the total internal energy remains constant within the vessel.
The insulation ensures we only need to consider the internal energy contributions of the gases inside. This constraint helps us predict how the temperature changes when the gases mix.

internal energy

Internal energy is the total energy contained within a system. It includes kinetic and potential energy at the molecular level.
For ideal gases, internal energy depends on temperature and is calculated using the molar specific heat at constant volume (\(C_v\)).
In both cases of our exercise, the internal energy before mixing must equal the internal energy after mixing due to the insulation constraint. This principle allows us to solve for the final mixture temperature.

enthalpy

Enthalpy is a measure of the total heat content in a system. It includes internal energy plus the product of pressure and volume.
In steady-state processes involving ideal gases, enthalpy can be simplified using the molar specific heat at constant pressure (\(C_p\)) and temperature.
This concept helps us balance the energy in scenarios where there might be changes in pressure or volume, such as in an open system with a continuous flow. In our example, enthalpy balance is used to determine the final mixture temperature in the control volume.