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Homework Help: Thermodynamics piston problem.

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data
    An ideal gas at 300 K and 1.5x10^5 Pa is contained in a piston cylinder device. The piston is 0.1m above the base of the cylinder, has a diameter of 0.15 m and is allowed to move freely. Heat is transferred to the cylinder causing the piston to move a distance of 0.05 m. The gas specific heat at constant volume is 718 J/kg·K and the gas constant is 287 J/kg·K.
    (a) What is the work done by the gas and the gas pressure at the end of the process? (b) Calculate the final gas temperature and the amount of heat transferred to the gas during the whole
    process. (Ans: (a) -132.54J, 1.5x10^5 Pa, (b) 450K, 464.26 J )

    2. Relevant equations

    3. The attempt at a solution
    The pressure is constant = 1.5x105 Pa I have no idea why...
    WD by gas = (1.5x10^5)(0.1-0.05)pi(0.075^2)=132.54J <---- It should be positive right ? The answer shows negative.
    T=300(0.15A)/(0.1A)=450K , A is the area.
    OK the main problem starts here. After a few tries i got
    n=PV/(T287) <-------------so this one i should take 287 instead of the usual 8.31? how come ? I know the qn says gas contant is 287 but why did it change ?
    deltaU=(fR/2)(ndeltaT) <---------- Is this eqn right ? f is the degree of freedom since we aren't told what type of gas this is.
    I took fR/2 as the specific heat at constant vol( correct???)
    and obtain deltaU=331.57
    For energy conservation the heat supplied must cause the increase in U by 331.57 and must also provide for the WD by gas. Q=331.57+132.54=464.11 Actual answer is 464.26.
    I cannot grasp this properly. This is a constant pressure process. So why do we use constant vol specific heat?
  2. jcsd
  3. Oct 12, 2011 #2
    Work should be positive.

    Pressure is constant because problem says the piston is allowed to move freely. The portion above the piston is open to the atmosphere.

    One uses constant pressure specific heat when one deals with enthalpy. You are dealing with internal energy so constant volume specific heat is used.
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