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Several Old Exam Questions. (Mechanics and thermodynamics)

  1. Dec 13, 2008 #1
    Hello everyone,

    I am studying the old exams to prepare for my physics final, and I have gotten sort of confused on a few of them. Help on any of the following is much appreciated.
    Question 1:
    1. The problem statement, all variables and given/known data
    Assume 3 moles of diatomic gas has internal energy of 10kJ. Determine the temperature of the gas after it has reached equilibrium. [The gas constant R is 8.314 J/(mol * K)).]

    I also know that each diatomic molecule has 5 degrees of freedom.

    2. Relevant equations
    This is where I am sketchy. By guess, I ventured that:

    Temp (K) = (Internal energy(J))/((degrees of freedom/2)(R (J/(mol * K)))(number of moles)

    3. The attempt at a solution
    Given that 10kJ(1000J/1kJ) = 10000J
    10000(J)/((5/2)(8.314(J/(mol*K)))(3mol) = 160.372K, which is correct.

    I guess my real question is, is my process correct, and will it hold for any other question of the same sort?

    Question 2:
    1. The problem statement, all variables and given/known data
    A sealed 75-m3 tank is filled with 9000 moles of oxygen gas (diatomic) at an initial temperature of 270 K. The gas is heated to a final temperature of 320 K. The atomic mass for oxygen is 16.0 g/mol. The final pressure of the gas, in MPa, is closest to:

    I went under the assumption that the ideal gas equations would produce a close enough answer.
    2. Relevant equations
    pV = nRT
    and thus:
    (p1)/(T1) = (p2)/(T2)
    ((p1)(T2))/(T1) = (p2)
    3. The attempt at a solution
    (p1 (Pa)) = (9000 mol)(8.314 J/(mol * K))(270 K)/(75 (m^3 = 269373.6 Pa = .269373 mPa
    (.269373 (mPa))(320K)/(270K) = .3192576 mPa
    Again this is correct, but is it fair to use ideal gas laws for something that doesn't appear to be ideal?

    Question 3:
    1. The problem statement, all variables and given/known data
    An ice cube at 0°C is placed in a very large bathtub filled with water at 30°C and allowed to melt, causing no
    appreciable change in the temperature of the bath water. Which one of the following statements is true?
    A. The entropy gained by the ice cube is equal to the entropy lost by the water.
    B. The entropy of the water does not change because its temperature did not change.
    C. The net entropy change of the system (ice plus water) is zero because no heat was added to the system.
    D. The entropy of the system (ice plus water) increases because the process is irreversible.
    E. The entropy lost by the ice cube is equal to the entropy gained by the water.

    2. Relevant equations
    dS = dQ / T

    3. The attempt at a solution
    I figured that A would be the most likely answer, if the system is closed. However
    the answer sheet says it is D which also sort of makes sense, may be it is an open system?
    Can someone explain this to me?

    Question 4:
    1. The problem statement, all variables and given/known data
    25. You may have noticed that when you get out of a swimming pool and stand dripping wet in a light breeze, you feel much colder than you feel after you dry off. Why is this?
    A. Water has a relatively large heat capacity.
    B. The water on your skin is colder than the surrounding air.
    C. The moisture on your skin has good thermal conductivity.
    D. 540 calories of heat are required to evaporate each gram of water from your skin, and most of this heat
    flows out of your body.
    E. This is a purely psychological effect resulting from the way in which sensory nerves in the skin are

    2. The attempt at a solution
    As far as I remember water has a higher specific heat and higher rate of heat transfer than air, so I kind of assumed since heat transferred quicker through water, that as the water was dripped off the body, more heat would be transferred, thus allowing the body to cool quicker than if it were only surrounded by air. So I guessed it was B, but it turns out it is D. Can someone please explain this?

    Thanks in advance for your help.

    Best Regards,
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 15, 2008 #2
    Sorry. Bump.
  4. Dec 15, 2008 #3

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Your equation is equivalent to:

    [tex]U = nC_vT = n\frac{5}{2}RT[/tex]

    The water and the ice are the system.

    The heat flow from the water to the ice cube is Q. The change in entropy of the water is negative:

    [tex]-Q/T_h = -Q/303[/tex]

    The change in entropy of the icecube is more complicated to calculate but you can see that it is positive and denominator starts out smaller than 303 (T = 273). So the increase in entropy of the ice is greater than the decrease in entropy of the water.

    The answer is provided in D.

    In order for water to turn to vapour, it must acquire the energy of vaporisation to break the bonds between water molecules. Where does it get this from?

  5. Dec 15, 2008 #4
    Thanks a lot! Makes good sense now..
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