## GR0877 Problem 34

**PROBLEM STATEMENT**: *This problem is still being typed*.

**SOLUTION**: (**C**) The first law of thermodynamics: , . Here we have used to rewrite the second term. Thus, .

*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.*

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.

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.

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

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.

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

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

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.