## 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!*

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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.