Tin Friction
Another thermodinamics exercise, please help?
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.
Note that it specifies a DIATOMIC ideal gas. This is because the specific heats c_p and c_v will be needed, and they depend on the size and shape of the molecule.
(Some background information for you: c_v depends on the number of “degrees of freedom” of the molecule. A degree of freedom is a way the molecule can move. Monatomic ideal gas molecules can move in three dimensions; therefore they have 3 degrees of freedom. Diatomic ideal gas molecules can move in three dimensions and rotate in two directions; therefore, they have 5 degrees of freedom. The Equipartition Law states that each degree of freedom gets an energy kT/2, where k is Boltzmann’s constant and T is the absolute temperature. Therefore, one molecule of a diatomic ideal gas will have an energy 5kT/2. Boltzmann’s constant * Avogadro’s Number = Gas constant R; therefore one mole of diatomic ideal gas molecules will have an energy 5kT/2 * N_A = 5RT/2. Therefore, since by definition U = c_v*T, you see that c_v = 5R/2. Comparing c_p and c_v, it is always true that c_p = c_v + R; therefore for a diatomic ideal gas, c_p = 7R/2.)
Let us start the problem by looking at the gas in chamber 2. Work is done on this gas, but no heat flows into it; therefore it will follow an “adiabat curve” if you plot its temperature or pressure versus volume. I do not know whether you know calculus, so I will tell you the result instead of showing the derivation. If you want to see where these formulas come from, I could show you that in a separate post.
For an adiabatic process, the pressure, temperature, and volume are related by PV^x = constant, where x = c_p / c_v. In the case of a diatomic ideal gas, x = (7R/2) / (5R/2) = 7/5 = 1.4. Since the volume of chamber 2 is reduced to half of its initial value, you can calculate that P_final * (V_final)^1.4 = P_initial * (V_initial)^1.4 —> P_final / P_initial = [(V_initial)/(V_final)]^1.4 = 2^1.4. Therefore P_final = 2^1.4 * 1.0 atm = 2.639 atm.
Now that the final pressure is known, you can find the final temperature in chamber 2, by using PV = nRT. It is easiest to write a ratio of initial an final equations: (P_f / P_i)(V_f / V_i) = (n_f / n_i)(R/R)(T_f / T_i), which reduces to 2.639*0.5 = 1*1*(T_f / T_i). Therefore, T_f = 1.3195*T_i = 398.5K.
To study chamber 1, you must remember that the TOTAL volume of both chambers is constant, but the division into two volumes changes. Since the final volume of chamber 2 is half of its initial volume, the lost volume must now be part of chamber 1. Therefore, chamber 1 has a new volume V_f = 1.5*V_i. The pressures are balanced in the two chambers, so chamber 1 also has pressure 2.639 atm. Therefore, the temperature of chamber 1 can be found ty a ratio in the same way as for chamber 2: (P_f / P_i)(V_f / V_i) = (n_f / n_i)(R/R)(T_f / T_i), which reduces to 2.639*1.5 = 1*1*(T_f / T_i). Theerfore, T_f = 3.9585*T_i = 1195.5K. This is exactly 3 times the temperature of chamber 2, because chamber 1 has the same pressure and three times the volume of chamber 2.
To find the heat given to chamber 1, you must use U_f – U_i = Q – W. You could look at each chamber separately, but then you would need to account for the work done by the moving plate. It is easier to consider the WHOLE system. In that case, W = 0, because no OUTSIDE force does work on the system. The only place where heat enters the system is through chamber 1, so the heat for the total system is the same as the heat given to chamber 1. Therefore, Q = U_f – U_i. U_f = c_v * n * T_1f + c_v * n * T_2f = 5R/2 * n * (1195.5 + 398.5)K = 3985K * nR. U_i = c_v * 2n * T_i = 5R/2 * 2n * 302K = 755K * nR. Therefore, Q = 3985K * nR – 755K * nR = 3230K * nR.
Now, what is nR? From the ideal gas law, it is nR = PV/T = 1atm * 1.14L / 302K = .00037748L*atm/K. Therefore, Q = 3230K * .00037748L*atm/K = 12.1927L*atm. since 1L*atm = 101.3J, this gives a work of 12.1927L*atm * 101.3J/L/atm = 1235J.
Hope this helps!
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