Consider an insulating cylinder with a piston There is a sma

Consider an insulating cylinder with a piston. There is a small mass inside the cylinder that can exchange heat with the diatomic gas inside. Assume that the walls are completely insulating and that the gas will undergo compression from Vi to Vf so that no heat is lost to the surroundings. For this problem, assume that the heat capacity of the gas Cg and the heat capacity of the mass Cm are exactly equal. Show that if we made the compression slow enough that the gas will undergo a quasi-static process, the entire process will be “reversible.” Compare the pV curve with an isothermal process and an adiabatic process. The result is known as the polytropic law pV ^ =Constant. Can you figure out what the exponent is? Make graphs and label the axes correctly.

HINT 1: Compress the gas adiabatically by a volume change of dV = and allow the gas and the mass to come to thermal equilibrium. Continue this process n times until we have gone fromVi to V Vn f i = . Then, show that in the limit of large n and small the process can be reversed and the system can be brought back to the initial state. HINT 2: You can do the calculations on a spreadsheet like EXCEL and make plots there too. Bonus if you can do it in MATLAB. HINT 3: Pick say, 10 L at 1 atm, and compress to 9L in 0.01L steps. NOTE: Can you use the results here to analyze the source of systematic uncertainty in the Adiabatic Compression Lab?

Solution

Compress the gas adiabatically by a volume change of dV = and allow the gas and the mass to come to thermal equilibrium. Continue this process n times until we have gone fromVi to V Vn f i = . Then, show that in the limit of large n and small the process can be reversed and the system can be brought back to the initial state.

p1= 150×103 Pa T1=673 K T2=293 K 2 21 1 11 T VP T VP × = × 1. If it is a constant pressure process, V A m T T V 87.02 673 293 1 1 2 2 =××=×= Since it is less than weight of the stops, the piston rests on stops.