One day our star will get old, and it will get to rest (and how we envy it that right now). We've mentioned how, once it runs out of Hydrogen, fusion becomes too energy-consuming, and it will stop producing energy that can cancel out it's gravity.
This means that hydrostatic equilibrium is lost and the star will collapse, causing the temperature around the core to increase so that Hydrogen burns outside the core. This outward pressure leads to a higher radius and what we call a Subgiant (2). The luminosity increases a bit from the energy in the shell where H->He, but not much as the radius also increases, causing a lower surface temperature.
Soon, energy will be transported through convection, not radiation as previously, and as this is much more effective, the luminosity will increase, and the star will be a Red Giant (2), with a radius between 10 to 100 times the original.
Our star is \(3.5M_\bigodot\) (3.5 solar masses), and so the temperature will eventually be high enough 
to begin the process of burning Helium to Carbon and Oxygen, but in lower mass stars something interesting takes place. The core will contract to a point where quantum physics sets an upper limit on the number of electrons within a certain volume with a certain momentum, there can be. This is called electron degeneracy*, and the core is then electron degenerate. At this point, it is not dependent on temperature, so increasing temperature does not mean the radius gets bigger.
Helium burning begins simultaneously everywhere in the core, releasing a huge amount of energy in a short time, which causes an explosive onset of Helium, known as a Helium (core) Flash, lasting only a few seconds. This huge amount of energy breaks the electron
degeneracy and pressure is again dependent on temperature, allowing the core to expand.
The expanding core pushes hydrogen out where it cools and hydrogen burning ceases. The star will again constact in order to have hydrostatic equilibrium, and will increase in temperature, moving left (3). After a while the mean molecular weight in the core has increased so much that the pressure from gravity causes the core to contract, the temperature in the core increases and the star expands. As the surface temperature decreases, the star moves right (4).
At this point the Helium is exhausted, and nuclear energy production ceases. Similar to what happened when Hydrogen was exhausted, the core contracts untill Helium can burn in the shell, and energy is produced. The radius increases and energy is transported by convective means, increasing the luminosity quite a bit. This is the Asymptocic Giant(5) (luminosity class II), which has a radius of around 1000 times the original.
Our star has a mass of \(M<8M_\bigodot\), and so the next phase is the core again contracting until it becomes electron degenerate, pushing against gravity. As the star contracts, the temperature in the other parts of the star increases: Hydrogen burning again becomes more efficient than Helium burning. Helium falls down where it doesn't burn, but the density here becomes very high and partially degenerate, which leads to a Helium Shell Flash.
This lifts the Hydrogen burning shell further away from the center. The star contracts until Hydrogen again burns, and the process repeats. These outer layers are seen surrounding the star, and soon all that is left is the degenerate Oxygen and Carbon core. This previous core now make up the surface of what is called a White Dwarf(6-7), which will cool off eventually.
How large will our star be at this point? We can use the equation of hydrostatic equilibrium, \(\frac{\partial P}{\partial r} = -\rho g\), to get an estimate of the radius \(R\), assuming uniform density:
\(\frac{P}{R}\approx\frac{GM}{\pi R^34/3}=\frac{3GM^2}{4\pi R^5}\)
where \(P\) is the degenerate pressure, \(G\) is the gravitational constant, \(M\) is the mass of the white dwarf star, and \(R\) is the radius. The expression for the degenerated pressure is
\(R = (\frac{3}{2\pi})^{4/3}\frac{h^2}{20m_eG}(\frac{Z}{Am_H})^{5/3}M^{-1/3}\)
where \(h\) is Plancks constant, \(m_e\) is the mass of an electron, \(Z\) is the numper of protons per nucleus and \(A\) is the number of nucleons per nucleus. A nucleus typically contains the same number of neutrons as protons such that the total number of nucleons is twice the number of protons and \(Z/A=0.5\). We know that the mass of a white dwarf is \(M_{WD}=\frac{M}{8M_\bigodot}M_{Chandresekhar}\), where \(M_\bigodot\) is one solar mass (the mass of the Sun), and \(M_{Chandresekhar}=1.4M_\bigodot\) is the upper limit on the weight of a white dwarf star. If the star has a mass greater than this at the white dwarf stage, gravity will be to great for degeneration and the star will continue to collapse. In our case, the mass of the white dwarf star is \(M_{WD}=1.20\cdot 10^{30}\)kg, meaning that the radius becomes \(R_{WD}=1692.371\)km.
To compare, the Earth has a radius of 6371 km. That's a really small star, especially when we compare it to the original radius of the star, which was 1 746 710km, but it's a realistic number! The original mass of our star was \(6.86\cdot 10^{30}\)kg, meaning our star has lost quite a bit of mass to the Helium flashes and fusion, but imagine all that mass in an object the size of the earth!
Exactly how dense is this star? We can use the standard expression for density, \(\rho=\frac{M}{V}=\frac{M_{WD}}{4/3\pi R_{WD}^3}\), 
and see that the density of our white dwarf star is \( 59.17\cdot 10^9kg/m^3\). The average density of a white dwarf is \(10^9kg/m^3\), which isn't too far off from ours, so we suspect that the number could be correct, even if ours is a little larger. To put this number into perspective, only one liter of our white dwarf star would weigh \(59\: 171\: 079 kg\)! THAT CAN'T BE RIGHT? <-- That right there is the note I made to myself so I'd remember to check this insane number. But everything I do (including checking how many needles go in a liter) seems to point to this being correct, or at least not too far off.
The gravitational acceleration of such a star is \(a = \frac{GM}{R^2}= 27\: 996\: 232m/s^2\), a rather large number. The average gravity on the surface of a white dwarf is 350 000 times that on the earth, making somewhere around 3 493 000 m/s2. But maybe our star is just really massive? It is was rather large, after all.
Now that there is no fusion in the core anymore, our white dwarf star will slowly lose its energy, and eventually end its life as a brown dwarf star, witn a core consisting of Carbon and Oxygen. But this didn't have to be the case. If the star was much bigger, \(M>8M_\bigodot\), the temperature and density in the core will be much greater in order to keep hydrostatic equilibrium. The star, like ours, will begin shrinking when the Hydrogen and Helium have been exhausted, which again increases temperature and pressure. However, this star is massive enough to begin fusion of even heavier elements! This means that electron degeneration will not set in as soon as in less massive stars. Basically, fusion of an element begins and continues until it has been exhausted, the star collapses, the core heats up, and fusion of the next element begins (while fusion of the earlier elements begin in different 'shells'). The closer to the core, the heavier the element.
This process cannot continue forever, because when Iron is formed the star will need to add energy to the element in order to burn it, and that is just not feasible. The process would require increasing the mass per nucleon as the number of nucleons increase, a process called fission. And stars just don't do fission. They are too famous, and don't want to spend their energy like that.
At this point the star will lose its hydrostatic equilibrium, same as for smaller stars, and begin collapsing. Electron degeneracy is not strong enough to resist the massive mass, and the star won't stop collapsing! Oh no! This is new!? Suddenly we have to move form electron degeneracy, to neutron degeneracy, and finally the core can stop collapsing. Unfortulately for those nearby, this leads to the outer layers of the star suddenly being pushed out again, in a spectacular explosion called a supernova. Depending on the mass of the star, what is left can be a neutron star, or (for stars where the neutron degeneracy isn't strong enough) even a black hole.
* Real quick, let us explain properly what exactly electron degeneracy is. It's mentioned quickly, just for the basic understanding, but for the extra curious, here it is!
You may be familiar with what we call bosons and fermions, the two types of fundamental particles known in the entire universe. These have a sort of angular momentum to them, which we call spin. Particles with spins that come in half-integer multiples are known as fermions, while particles that come in integer multiples are bosons.
This is important for many reasons, one of them being that the Pauli exclusion priniciple only applies to fermions. This is the principle that leads to the electron degenerate state. It states that in any quantum system, no two fermions can occupy the same quantum state (bosons do not have this restriction)! Imagine that you have an atomic nucleus and you add an electron to it. This electron will want to occupy the lowest energy state allowed, the ground state. It will then either have a spin of +1/2 or -1/2. You can add another electron, and this will also tend to occupy the ground state, with a spin opposite that of the first electron.
What if you try to add more? These will have to occupy a higher energy state! This is the reason for our periodic table being the way it is, and why atoms have different properties, binding together the way that they do!
So, the fact that electrons and neutrons are fermions can keep white dwarf stars and neutron stars from collapsing. If you have a lot of electrons, it turns out that none of them will have the same momentum. Once an electron has claimed a momentum (as a vector, so particles can have the same value of momentum, so long as it is in different directions) no other electron can have this momentum. Even if they want it super super badly.
When the gravitational force from a star works on these electrons, they will exist within a small space, where they all attempt to get the best momentum (the lowest). The momentum is the lowest at the origin, meaning that this is where the particles want to be, but at a point there is just no more room, and in order to get one of the good spots, an electron will need more energy than it has. At this point there is a force pushing electrons out of this radius. This is known as the degenerate pressure.
That's all we've got you guys!
It was a fun ride for our star, and this project has been a fun ride for us! But just like the star we are burnt out, we've thrown away all the energy we has to give, and now there is nothing much left of us- unlike the star we can take a vacation, get some sleep, and start new projects in the future.