A rigid container has 1 kg CO2 at 300 K and kg argon at 400 K, both at 150 kPa. Now they are all

A rigid container has 1 kg CO2 at 300 K and kg argon at 400...

A rigid container has $1 \mathrm{kg} \mathrm{CO}_{2}$ at $300 \mathrm{K}$ and $\mathrm{kg}$ argon at $400 \mathrm{K},$ both at $150 \mathrm{kPa}$. Now they are allowed to mix without any heat transfer. What is final $T, P ?$

Key Concepts

Mixing of Gases

Mixing different gases involves combining them to reach a new thermodynamic equilibrium. In the context of ideal gas mixtures, the final state is determined by both the individual gas properties and the overall constraints such as constant volume and total mass, where the gases share a common temperature and pressure after mixing.

Adiabatic Process

An adiabatic process is one in which there is no heat transfer into or out of the system. When two gases mix in a rigid, adiabatic container, the process is internally insulated, and the only energy exchanges are due to work done by or on the system, which in a rigid container is zero. This ensures that the internal energy redistribution determines the final temperature.

Conservation of Energy

In an isolated or adiabatic process, the total energy of the system remains constant. This principle is applied to determine the final equilibrium state by equating the initial and final total internal energies of the gases, despite differences in initial temperatures.

Internal Energy of Ideal Gases

For ideal gases, the internal energy depends solely on temperature and the specific heat capacities, independent of pressure or volume changes. This concept is crucial when analyzing energy balances in processes such as mixing, especially under adiabatic conditions where no heat is exchanged with the surroundings.

Ideal Gas Law

This law relates the pressure, volume, temperature, and number of moles of an ideal gas. It provides the foundation for analyzing the state of a gas and is critical for determining how individual gas properties interact in a container. By assuming ideal gas behavior, one can use PV = nRT to relate the conditions before and after mixing.

Transcript

00:01 Hey, everybody.

00:01 So for this one, it's asking, now they are allowed to mix with heat transfer.

00:06 What is the final tp? so for this particular one, we need to find out the energy requires.

00:18 We have u2 minus u1 equals zero.

00:21 And that's for the energy equation, okay? and for you want to, we have mass of that minus.

00:36 Times the u2 minus u1, the carbon.

00:45 So we have mass of the argon, u2 minus u1, argon equals zero.

00:54 Then we're just going to keep just solving this down.

00:57 So we have the cb of here, t2 minus t1, co2, plus m -a -r times t2 minus t1.

01:17 Oops, and i forgot.

01:19 So we'll just put the cv over here.

01:22 Make it a little easier.

01:24 And now we're just gonna keep just simplifying right the second.

01:30 So from this, we can have t times the cm, co2, cv, co2, plus m -a -r -c -v -a -r minus m -c -o -2 times c -v -c -o -2 -t -c -o -2 minus m -a -2 -2 -m -a -r times c -v -a -r equals 0.

02:03 And what we can do now, and actually it looks like i did this, okay.

02:13 So what we can do now is we have t2 is equal to 1 times 0 .653 plus 1 from the mass times 0 .312.

02:28 2.

02:31 Okay.

02:31 And this was minus 1, the mass again, times 0 .653 times the temperature, which is 300...

Please give Ace some feedback