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GR9677 #15
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Thermodynamics$\Rightarrow$}Phase Diagram

Recall that for an ideal gas $U=C_v \Delta T$ and $PV=nRT$. Don't forget the first law of thermodynamics, $Q=W+U$.

For $A\rightarrow B$, $U=0$, since the temperature is constant. Thus, $Q=W=RT_H\ln V_2/V_1$.

For $B\rightarrow C$, $W=P_2(V_1-V_2)=R(T_c-T_h)$. $U=C_v (T_c - T_h)$, and thus $Q=W+U=C_v (T_c - T_h)-R(T_h-T_c)$.

For $C\rightarrow A$, $W=0$, $U=C_v(T_h-T_c)$, thus $Q=U=C_v(T_h-T_c)$.

Add up all the Q's from above, cancel the $C_v$ term, to get $Q_tot=RT_h\ln(V_2/V_1)-R(T_h-T_c)$, as in choice (E).s

Alternate Solutions
 ssp2008-09-08 03:26:43 3-step cycle with isovolumetric part that has W = 0, so we need an added heat for two steps (read two parts). That already eliminates (A) through (C). Then for the last part just dimensional analysis... (D) does not work cause we are missing a mass to convert the $C_P$ to joules.... that only leaves (E) Reply to this comment
nirav_605
2013-10-16 12:44:43
In Process from B to C

Isn't that suppose to be C_p not C_v
The process occurs at constant pressure not volume.
 Giuseppe2015-09-11 15:04:33 For the isobar B-C you have to use C_p for the heat absorbed (Q) and C_v for the change in internal energy (U), since the change in internal energy in B-C has to be equal and opposite to that of C-A.\r\nInfact, for BC:\r\nQ = -Cp(Th-Tl)\r\nU = -Cv(Th-Tl)\r\nW= -P(V2-V1)\r\nwhich also gives P(V2-V1) = R (Th-Tl)
mvgnzls
2011-09-07 18:01:05
why is there no n in the formulas?
pv=(n)RT & W= -nRTln(vf/vi)

why do the n's get left out?
 pam d2011-09-11 18:41:05 n = 1, so you are right in looking for it but make sure you read the question very carefully
jmason86
2009-09-30 17:22:25
I also realized that $\Delta U=0$ therefore Q=W which is the area in the curve. However, the way I actually solved the problem was to look at the answers. (D) and (E) both had a first term that I knew was an isothermal work term. Also, the two answers were essentially identical, so the correct answer was probably one of those.
Looking at the 2nd terms, they obviously had different units. The units in the 2nd term of (D) are not consistent with the units in its first term.. you can't even do this subtraction. Eliminate (D). (E) remains.
 noether2009-11-03 21:04:57 Answer D does have correct units, see the responses to ssp's solution.
mudder
2009-09-28 02:13:31
An alternate way of looking at this is

Work = area of PV diagram
Since delta-U = o, then Q = area of PV diagram

Find the equation of the T_h isotherm using the ideal gas law, and integrate the PV region to find the area. That area is the added heat.
duckduck_85
2008-11-06 16:43:43
radicaltyro's is the best and quickest solution
ssp
2008-09-08 03:26:43
3-step cycle with isovolumetric part that has W = 0, so we need an added heat for two steps (read two parts). That already eliminates (A) through (C). Then for the last part just dimensional analysis... (D) does not work cause we are missing a mass to convert the $C_P$ to joules.... that only leaves (E)
 Walter2009-01-01 08:57:19 Rejecting (D) on the grounds that we are "missing a mass" is not justified. The questions specifies that $C_p$ is a molar heat capacity and also that there is one mole of gas hence any $n C_{p} \Delta T$ term in the result would simplify to $C_{p} \Delta T$. Hence you remain stuck with a 50/50. radicaltyro's solution is best.
 okdisa2009-09-22 18:23:41 Actually, $c_p = \frac{\delta q}{dT}$ so $c_p (T_h - T_c)$ does in fact have the correct units. Don't forget that heat and energy are equivalent dimensionally.
 Giuseppe2015-09-11 15:08:42 You can\'t use dimensional analysis to eliminate choice D, it has the correct units. Cp is heat capacity at constant pressure per mole, and there is a factor of 1mole implicitaly multiplied everywhere in the solutions.
kyros
2007-11-01 11:27:59
Shouldn't the B->C term have a C_p not a C_v?
 Imperate2008-09-04 09:38:50 Yes. B->C: $\delta Q =C_p(T_c-T_h)=-C_p (T_h-T_c)$ C->A: $\delta Q =C_v (T_h-T_c)$ Adding contributions from these parts, one obtains: $-(C_p-C_v)(T_h-T_c)$ and remembering that for an ideal gas $C_p-C_v=nR$ provides(setting n=1mole): $-R(T_h-T_c)$, which is the last term.
2006-11-02 17:21:59
You don't need to calculate $U$ for each step. Just note that this is a cyclic process so $\Delta E=0\rightarrow Q=W$ and add up the $W$'s.
 newton2007-09-29 06:35:00 cool solution
 prismofmoonlight2007-11-01 23:30:12 agreed.
 tonyhong2008-10-04 04:24:55 so there is nothing to do with Cp and Cv, answers B,C,D excluded. A is excluded for simple reason
 wittensdog2009-10-16 15:23:50 Indeed, I think I was almost implicitly doing this without even really thinking about it, but it's definitely a good general principle to remember. The same idea applies for entropy, or any other state variable. As long as we're working with a constant amount of stuff, then things like pressure, volume, entropy, internal energy, and temperature are all specified when you choose a point on the PV diagram. So as long as you move around a closed loop, when you come back to the original point, then you end up with the same value as before. So indeed, when you go around the cycle, since you come back to the same point, you must have the same internal energy as before. This means that the heat added over the course of the cycle (eventually) all went into doing work. Of course this wouldn't be true in general if you just looked at a portion of the plot (though it is true for isotherms of ideal gases).
beibei
2006-06-14 12:40:26
Why I can not see the problems of 9677 15,16...only the solutions. some of the questions have this problem
 godkun12017-03-15 17:29:41 You are right

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$