Another thermodinamics exercise,

[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 $$PV^{\gamma}=K$$

 

[m2003]

1) The final pressure will be 2.0 atm. This can be calculated using the ideal gas law, where P1V1/T1 = P2V2/T2. Since the initial pressures and temperatures are the same in both parts of the cylinder, we can set them equal to each other and solve for P2. This will give us P2 = 2P1 = 2 x 1.0 atm = 2.0 atm.

2) The final temperature of part number 1 can be calculated using the equation PV = nRT, where n is the number of moles of gas and R is the gas constant. Since the volume and pressure of part number 1 have changed, we can set up the equation as P1V1 = nR(T1+T2), where T1 is the initial temperature of part number 1. We can then solve for T2, which will give us T2 = 2T1 = 2 x 302 K = 604 K. The final temperature of part number 2 can be calculated in a similar way, using the fact that the volume and temperature of part number 2 have remained constant. This will give us T2 = T1/2 = 302 K/2 = 151 K.

3) The heat "given" to part number 1 can be calculated using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. Since the walls are adiabatic, there is no heat transfer between the two parts of the cylinder. Therefore, the change in internal energy of part number 1 is equal to the work done by the system. This can be calculated using the equation W = PΔV, where P is the pressure and ΔV is the change in volume. In this case, ΔV = (1.14 L - 0.57 L) = 0.57 L. Therefore, the heat "given" to part number 1 can be calculated as Q = PΔV = (1.0 atm)(0.57 L) = 0.57 L atm.