Ideal diatomic gas in cylinder with sliding adiabatic plate

[claudiadeluca]

The volume of a cylinder, with adiabatic walls and closed at the ends, has been divided in two parts by an adiabatic plate (with unimportant volume), which can slide without friction inside the cylinder (it is like a movable piston).

The cylinder has been filled with an ideal diatomic gas, and initially the pressures, the temperatures and the volumes are the same in the two parts of the cylinder separated by the plate.

Pin=1.0 atm
Vin=1.14 L
Tin=302 K

We (very slowly) start giving heat to part number 1, using an electric resistance, until the volume of part number 2 has become half than what it was before.

Calculate:

1) the value of the final pressure
2) the final temperatures of both parts
3) the heat "given" to part number 1

NB: when the plate is at equilibrium the pressure at its sides is the same.


Pardon the misuse of physics words, I'm not english.

Thanks, any help is really needed and appreciated.

 

[chaoseverlasting]

Since the process is adiabatic, $$PV^{\gamma}=constant$$.
Therefore, $$P_0V_0^{\gamma}=P_1\frac{V_0}{2}^{\gamma}$$
$$P=P_02^{\gamma}$$

Using the same expression for an adiabatic process and the ideal gas equation (pv=nrt), you can solve for the second part.

For the third part, $$dq=Pdv+\frac{nfR}{2} dt$$. Again, you can find the required paramaters using $$PV^{\gamma}=K$$