- #1
Hobold
- 82
- 1
This is a very simple problem, though I have arrived at two different results and I would like some help to procceed (I'm not sure if this is the right section, though this is not a homework):
"2kg are confined in an adiabatic cilinder through a frictionless piston activated by an elastic spring, as in the picture. The piston is initially locked by an inner pin and keeps the air at 300 K and 400 kPa. The piston is then unlocked and moves freely, allowing the air to expand until the final volume is equal to double the initial volume. Determine both final pressure and temperature."
This problem is to be solved using only the first law of thermodynamics. Here is what I have done:
Considering \delta Q = dU + \delta W = mc_{vo}dT + pdV = 0 and air as a ideal gas, differentiating T = pV/(mR) you get dT = \frac{1}{mR}(pdV + Vdp) and, with some algebraic manipulation and taking c_{p0} = R + c_{v0}, you get \frac{dp}{p} + \frac{dV}{V} \gamma = 0 \rightarrow pV^{\gamma} = \text{constant}, \ \gamma = c_{p0}/c_{v0}
We are allowed to use a table from which we can take c_p0 and c_v0.
The initial volume is easily found by doing V_1 = \frac{MRT_1}{p_1}
then, taking the expression found before, we get p_1 V_1^\gamma = p_2 V_2^\gamma \rightarrow p_2 = p_1(V_1/V_2)^\gamma
Finally, to find final temperature, we use Clapeyron's equation for ideal gas.
Though it is clear that T_2 < T_1. The solution my professor's PhD student proposed is much simpler: you consider the spring as a part of the system and this is nothing but a case of free expansion, then T_2 = T_1. I talked to him later and he told me that my solution is also right, depending on how you interpret the problem, so that's why I posted exactly the same question here. I want some clue whether my solution is right or not.
Thanks.
"2kg are confined in an adiabatic cilinder through a frictionless piston activated by an elastic spring, as in the picture. The piston is initially locked by an inner pin and keeps the air at 300 K and 400 kPa. The piston is then unlocked and moves freely, allowing the air to expand until the final volume is equal to double the initial volume. Determine both final pressure and temperature."
This problem is to be solved using only the first law of thermodynamics. Here is what I have done:
Considering \delta Q = dU + \delta W = mc_{vo}dT + pdV = 0 and air as a ideal gas, differentiating T = pV/(mR) you get dT = \frac{1}{mR}(pdV + Vdp) and, with some algebraic manipulation and taking c_{p0} = R + c_{v0}, you get \frac{dp}{p} + \frac{dV}{V} \gamma = 0 \rightarrow pV^{\gamma} = \text{constant}, \ \gamma = c_{p0}/c_{v0}
We are allowed to use a table from which we can take c_p0 and c_v0.
The initial volume is easily found by doing V_1 = \frac{MRT_1}{p_1}
then, taking the expression found before, we get p_1 V_1^\gamma = p_2 V_2^\gamma \rightarrow p_2 = p_1(V_1/V_2)^\gamma
Finally, to find final temperature, we use Clapeyron's equation for ideal gas.
Though it is clear that T_2 < T_1. The solution my professor's PhD student proposed is much simpler: you consider the spring as a part of the system and this is nothing but a case of free expansion, then T_2 = T_1. I talked to him later and he told me that my solution is also right, depending on how you interpret the problem, so that's why I posted exactly the same question here. I want some clue whether my solution is right or not.
Thanks.
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