Expansion of an ideal gas in an evacuated bottle having nonconducting walls In the free expansion of ideal gas in a vacuum $dW = 0$ as $p_{ext} = 0$ and $dU = 0$ therefore there is no change in temperature of the ideal gas also. what i want to know that can the following situation also classify as a free expansion of a ideal gas and if no why ? the situation is as follows-
" An evacuated bottle with nonconducting walls is connected through a valve to a large supply of gas, where the pressure is $P_0$ and the temperature is $T_0$. The valve is open slightly, and helium flows into the bottle until the pressure inside the bottle is $P_0$. "
is the process also quasi static. As the bottle is a vacuum it seems that the process is free expansion but i dont think it is. 
 A: The question whether or not it is a free expansion relates to what you choose as your thermodynamic system. Here are two choices, both applicable to the situation you describe.


*

*We suppose the 'large supply of gas' is itself large but finite, and we suppose it is contained in a large vessel with rigid walls. Let the thermodymamic system under study be this gas. No heat enters or leaves, and since the walls are rigid no work is done on the system, so this system undergoes a process of free expansion as the gas expands into the bottle. 

*Alternatively, take as your thermodynamic system just that part of the gas which ends up in the bottle. You could imagine a thin flexible bag around it if you like. Now the process is not a free expansion, because the rest of the gas does work on your system as it pushes against the bag and this bag (or boundary) moves.
A: As @Andrew Steene pointed out, it depends on how you define the "system". If you define the system as the combination of the two rigid insulated vessels and their contents (Andrew's choice 1) then the system is isolated (can exchange no mass or energy with its surroundings).Then you have a free expansion.

is the process also quasi static?

I would add a free expansion is not considered quasi-static. Even though the helium may slowly move into the evacuated chamber, it nonetheless expands across a finite pressure difference until it approaches equilibrium. That makes the process irreversible. You know it is irreversible because you would not expect the helium to spontaneously return to its original.  Work would be required.
UPDATE:
This update is given in light of the following additional comments you have given.

if i take both the container as my system then the process is a free
  expansion right. If the gas expanding from the container has
  temperature 0 then after expanding into the evacuated bottle the
  temperature should remain the same ($_0$) but in my textbook it is
  written that the temperature of the gas will change to $\frac{C_p}{C_v}T_0$.
  Where is that i am wrong?

For a free expansion of an ideal gas (aka Joule expansion) there is no change in internal energy. Also, for an ideal gas, its internal energy is only a function of the temperature of the gas according to
$$\Delta U=c_{v}\Delta T$$
Consequently if the book says there is a temperature change, we do not, by definition have a free expansion of an ideal gas. (It should be noted that real gases can experience a temperature change)

what if the gas coming from the container is infinitely large as
  compared to the bottle wont the pressure be constant in that
  situation?

Such an assumption would be inconsistent with the temperature change given in the book, and here's why:
Given the specific heats of helium, the book says the final temperature of the helium is about 1.67 times the original temperature, or a 67% increase in temperature. For an ideal gas that means the increase in volume of the helium gas is also increase 67% because the idea gas law gives us 
$$\frac{P_{i}V_{i}}{T_{i}}=\frac{P_{f}V_{f}}{T_{f}}$$. 
If the initial and final pressures are the same, the initial and final ratios of volume to temperature are also the same. Finally, that means the volume of the bottle, into which helium expanded, is 67% of the initial infinite volume of the helium! The container can’t be infinitely larger compared to the bottle. 
The above being said, I must admit I do not know what kind of process can produce the final temperature given in the book. I will reach out to a trusted colleague of mine to see if he knows.
Hope this helps. 
A: This is in response to @BobD 's inquiry as to how to obtain the final temperature using the first law of thermodynamics.  There are two equivalent ways of doing this, using either (1) the usual closed system version of the first law, or (2) the open system (control volume) version of the first law.  Both methods give exactly the same results.


*

*Closed system version


The system here is chosen as all the gas that will eventually end up in the bottle.  There is work done on this gas as it is pushed into the valveby the gas immediately behind it in the gas supply.  This work is given by $P_0v_0n$, where $v_0$ is the specific volume in the gas supply and n is the number of moles of gas that enter the tank.  So, from the closed system version of the first law, we have:$$\Delta U=n(u_f-u_0)=nC_v(T_f-T_0)=nP_0v_0$$where $u_0$ is the molar specific internal energy of the gas in the supply.  Now, from the ideal gas law, $$P_0v_0=RT_0$$Therefore we have $$C_v(T_f-T_0)=RT_0$$From this, it follows that $$T_f=\frac{C_p}{C_v}T_0$$


*Open system version


In applying the open system version of the first law, we take as our control volume the combination of the valve and the bottle.  For this system, there is no shaft work done, and the open system version of the first law is expressed as $$U_f=nu_f=nh_0$$where $h_0$ is the enthalpy per mole in the supply (and thus equal to the enthalpy per mole of the inlet stream to the control volume), with $$h_0=u_0+P_0v_0$$and$$u_f=u_0+C_v(T_f-T_0)$$ Combining these equations, we obtain $$u_f=u_0+C_v(T-T_0)=u_0+P_0v_0$$or again:$$C_v(T_f-T_0)=RT_0$$
