233 reputation
315
bio website physics.stackexchange.com/…
location Waterloo, Canada
age 23
visits member for 3 years, 11 months
seen Apr 20 at 17:38

Mathematical Physics Major at the University of Waterloo


Dec
30
awarded  Benefactor
Dec
29
accepted Independence of thermodynamic variables
Dec
23
comment Independence of thermodynamic variables
@Christoph Thank you Christoph. If you convert your comments to an answer, I'd be happy to award you the bounty.
Dec
23
comment Independence of thermodynamic variables
cont.. This is because for any partition of extensive/intensive variables that we choose, there is some thermodynamic potential for which the variables serve as natural variables and hence is enough to determine the entire thermodynamic state. So I guess this is a sufficient requirement. Am I correct in assuming that this is also sufficient? I.e. the inclusion of a conjugate pair of variables means that the state will not be uniquely determined. So for a simple system with variables $\{U,S,V,T,p\}$, we cannot use $\{S,T,p\}$ or $\{V,p,T\}$ as coordinates for example.
Dec
23
comment Independence of thermodynamic variables
@Christoph Thank you for your response. My current understanding is this: Let $\{X_i\}_{i=1}^k$ denote the extensive parameters of the system and let $\{Y_j\}$ denote the conjugate intensive parameters. Let $I\cup J = {1,2,\cdots,k}$ denote a partition of the indices. Then $\{X_i\}_{i\in I} \cup \{Y_j\}_{j\in J}$ will serve as a valid coordinate system for the thermodynamic states so long as we include at least one extensive parameter.
Dec
23
awarded  Promoter
Dec
21
asked Independence of thermodynamic variables
Oct
24
accepted A vector function of a vector $\mathbf{S}$ must be given by a multiple of $\mathbf{S}$?
Oct
15
awarded  Curious
Oct
14
asked A vector function of a vector $\mathbf{S}$ must be given by a multiple of $\mathbf{S}$?
Sep
18
awarded  Critic
Apr
23
awarded  Popular Question
Mar
21
awarded  Popular Question
Feb
4
awarded  Notable Question
Feb
3
awarded  Nice Question
Feb
3
accepted How do we know that heat is a differential form?
Feb
1
comment How do we know that heat is a differential form?
Thank you for your answer. Just a small question of mine. I had always assumed (and been told) that the only restriction on $\delta Q = T\ dS$ was that the process needs to be quasi-static, regardless of whether its reversible or irreversible. Can you give me an example of an irreversible quasi-static process for which $\delta Q = T\ dS$ does not hold?
Jan
31
awarded  Nice Question
Jan
31
comment How do we know that heat is a differential form?
I have little experience with thermodynamics of non-quasi-static processes, but it seems from your addendum that the differential formulation of the first law $dU = \delta Q - \delta W$ is practically limited to quasi-static thermodynamics. This has been a deeper question than I had anticipated. I appreciate everyone's help with this.
Jan
31
awarded  Commentator