An insulated tank that contains 1 kg of O2 at 15^∘ C and 300 kPa is connected to a 2-m^3 uninsu

An insulated tank that contains 1 kg of O2 at 15^∘ C and...

An insulated tank that contains $1 \mathrm{~kg}$ of $\mathrm{O}_{2}$ at $15^{\circ} \mathrm{C}$ and $300 \mathrm{kPa}$ is connected to a $2-\mathrm{m}^{3}$ uninsulated tank that contains $\mathrm{N}_{2}$ at $50^{\circ} \mathrm{C}$ and $500 \mathrm{kPa}$. The valve connecting the two tanks is opened, and the two gases form a homogeneous mixture at $25^{\circ} \mathrm{C}$. Determine $(a)$ the final pressure in the tank, $(b)$ the heat transfer, and $(c)$ the entropy generated during this process. Assume $T_{0}=25^{\circ} \mathrm{C}$. Answers: (a) $445 \mathrm{kPa},$ (b) $187 \mathrm{~kJ},$ (c) $0.962 \mathrm{~kJ} / \mathrm{K}$

Key Concepts

Thermodynamic Equilibrium

Thermodynamic equilibrium is the state at which a system’s macroscopic properties, such as temperature, pressure, and composition, are uniform throughout. In the mixing of gases from different initial states, reaching thermodynamic equilibrium means that the system stabilizes into a homogeneous state, where these properties are equalized across the entire system.

Entropy Generation

Entropy generation quantifies the irreversibility of a process, reflecting the degree to which energy becomes unavailable for work due to dissipative effects. In mixing problems, differences in temperature and pressure between the gases lead to an increase in entropy, which must be accounted for to fully understand the thermodynamic efficiency of the process.

Conservation of Energy

This principle, rooted in the first law of thermodynamics, is essential for analyzing the exchange of energy in a system. It states that the total energy of an isolated system remains constant, even as energy transfers in various forms (such as heat or work) occur between its subsystems. In the context of mixing two gases, conservation of energy guides the calculation of the final energy state of the combined system.

Ideal Gas Law

The ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas, is a cornerstone of thermodynamic analysis for gases. It provides a framework for determining the initial conditions of each gas and calculating the final state parameters after mixing, assuming the behavior of an ideal gas under the given conditions.

Heat Transfer

Heat transfer involves the movement of thermal energy from regions of higher temperature to regions of lower temperature. In a process where gases initially at different temperatures are allowed to mix, heat transfer is responsible for adjusting the temperature of the gases to reach a common equilibrium, even when the system is not perfectly insulated.

Transcript

00:04 The volume of oxygen tank, mrt by p, 1 into 0 .2598 into 288 upon 300.

00:27 So it is to be 0 .25 meter cube.

00:32 Total volume will be volume of oxygen tank plus nitrogen tank.

00:38 So it is 2 plus 0 .25 that is 2 .25 meter cube.

00:43 The mass of nitrogen you may calculate pv upon rd, 510 into 2 upon .2962 into 323.

01:02 So it is to be 10 .43 kg.

01:06 The number of moles, number of moles of the gases, number of moles of oxygen, mass upon molar mass, 1 upon 32, that is 0 .0 .3.

01:36 1 to 5 kilo mole that of nitrogen 10 .43 upon 28 that is 03725 kilo mole the final pressure will be n r t by v number of moles 0 .31 .03125 0 .031 plus 0 .315 plus 0 .315 plus 0 .315 plus 0 .3725 into 8 .314 into 298 25 upon 2 .25 so it is to be equal to 445 kilo pascal.

02:44 This is the answer of part a and now part b heat transfer q will be change in internal energy of nitrogen plus changing internal energy of oxygen minus mcb tm minus t1 mcv t1 mcv t1 this is for oxygen and this is for nitrogen substituting the value you will get the answer minus one point 1 into 0 .658 into 298 minus 288 plus 10 .43 into 0 .743 into 298 minus 323.

03:57 So 187 .2 kilojew.

04:02 This is the transfer of heat.

04:04 Now in c part we have to find mole fraction of oxygen, number of moles upon number of moles of mixture.

04:14 0 .03125 upon 0 .03125 plus 0 .3725.

04:33 So it is to be 0 .077...

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