Home > 0877 > GR0877 Problem 34

GR0877 Problem 34

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (C) The first law of thermodynamics: Q = \frac{3}{2} \nu R \Delta T, Q' = \frac{3}{2} \nu R \Delta T + p \Delta V = \frac{3}{2} \nu R \Delta T + \nu R \Delta T = \frac{5}{2} \nu R \Delta T. Here we have used PV = \nu RT to rewrite the second term. Thus, Q' = 5Q/3.

Found a typo? Comment!

Jump to the problem

1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100

Want to get a full PDF with solutions? Read THIS.

The comments appending particular solutions are due to the respective users. Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 2008 by ETS.

Categories: 0877 Tags: , , , ,
  1. betelguese05
    07.10.2012 at 09:52

    This solution confuses me. You say that Q = 3/2 n R \Delta T follows from the first Law. This is of course true, but requires the use of the fact that \Delta V = 0 (so that work W=P \Delta V = 0, so U=3/2 n R T = Q). Now, you say that Q’ = 3/2 n R \Delta T + p \Delta V. But we just used the assumption that p \Delta V = 0! How could this be simply replaced by a nonzero quantity you got from the ideal gas law? It is also impossible that the work is so simple as p \Delta V, since the pressure will change in this thermodynamic process. Then we really need that W = p dV, so we’d integrate instead of saying p \Delta V. This gets the right answer but your reasoning seems unclear.

  2. betelguese05
    07.10.2012 at 10:03

    Sorry, I see now that I misread the question. I didn’t read that it said to use constant pressure for the calculation at hand (last 5 words or so). Ignore the above comment.

  3. 09.09.2015 at 07:48

    Can someone explain how the formula for Q’ was obtained?

  4. 20.10.2015 at 17:02

    There is an easier way to solve this sum. Please correct me if I am wrong.

    Q1 = Cv * dT and Q2 = Cp * dT.

    Taking Cv = 3/2 and Cp = 5/2 for monoatomic gas.

    dT = 2/3 Q1 = 2/5 Q2.

    From this we get Q2 = 5/3 Q1.

    • 19.07.2016 at 01:08

      where did you get those constants? Otherwise, looks good to me

  5. 19.07.2016 at 01:08

    are we using v for little n in this formula? v means #moles, not velocity, correct?

  6. Hanna
    23.08.2016 at 19:24

    I did this problem quickly by remembering that Cp/Cv = (alpha + 1)/alpha where 2*alpha is the degrees of freedom of a gas. For an ideal, monoatomic gas there are 3 degrees of freedom (movement in x, y, and z) so alpha = 3/2 — which means that the ratio of Cp/Cv is equal to 5/3.It might just be a good idea to remember that the ratio Cp/Cv for an ideal monoatomic gas is 5/3.

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: