Thermodynamics: Entropy and specific heat

AI Thread Summary
The discussion revolves around the confusion regarding entropy change equations in thermodynamics, specifically the differences in formulations based on moles versus mass. The first equation uses moles of gas and incorporates specific heat at constant volume (C_V) and the gas constant (R), while the second equation uses mass and includes isobaric specific heat (c_p). The user seeks clarification on how these specific heats relate to entropy change and acknowledges the relationship between C_p, C_V, and the gas constant (R). Ultimately, the user realizes that the equations are consistent when considering the different measures of gas amount and the definitions of specific heats. Understanding these relationships clarifies the application of specific heats in entropy calculations.
Pandris
Messages
3
Reaction score
0
Hi here!

(before scriptum. Sorry for my lousy English and LaTex.)

I am a bit confused about entropy change statements:

In most textbooks is given:
\Delta S = n (C_V ln \frac{ \ T_{2}}{T_1} + R ln \frac{ \ V_{2}}{V_1})

where n - moles of gas. And dimension of entropy is [J/K]

But now I have one book there entropy change is defined using specific heats:

<br /> \Delta S = m (c_v ln \frac{ \ p_{2}}{p_{1}} + c_p ln \frac{ \ V_{2}}{V_{1}})<br />

where m - mass of gass, and dimension is [J/(kmol*K)]

I can't understand where this isobaric specific heat gets here!

Ok, I understand that these entropies essentially are given for different measures of amount of gass. First is given for moles, but second for kilograms. (am I right??)

But how to involve specific heats there is obscure for me.

Thanks!
 
Last edited:
Physics news on Phys.org
Oh my!

Am I blind or what?!

Everything works out if take in mind that
<br /> C_p - C_V = R<br />
and

<br /> c_v = \frac { \ C_{V}}{ \mu}<br />

That's it! Sorry for buzzer!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top