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## 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, $Q' = 5Q/3$.

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1. 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. 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. 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.

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