Timeline for How to calculate work done by ideal gas in Carnot cycle stage 2?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 14 at 18:41 | comment | added | Chet Miller | For an adiabatic reversible change $nC_VdT=-pdV$ , but this is not the defining equation for $C_V$. The definition equation for molar $C_V$ is $$C_V=\frac{1}{n}\left(\frac{\partial U}{\partial T}\right)_V$$ | |
| Mar 14 at 17:13 | comment | added | Philip Wood | In $dU=nc_v\ dT$, $n$ is the number of moles and $c_v$ is the molar heat capacity. In $dU=mC_v\ dT$, $m$ is the mass of gas and $C_v$ is the specific heat capacity (that is heat capacity per unit mass). One advantage of the molar version is that $c_v$ is the same for all gases with the same atomicity, for example $c_v=\tfrac 32 R$ for all monatomic gases, whereas the specific heat capacity, $C_v$ depends on the mass of the molecule. Note: the use of $c_v$ for molar and $C_v$ for specific is not, I think, standardised notation. | |
| Mar 14 at 16:05 | comment | added | Dor | I still don't understand. $dU=nC_vdT=-dW=-pdV$ so $C_v=-\frac{pdV}{ndT}$ but $R=\frac{pV}{ndT}$, why is that same? and why it's correct to write $n$ instead of $m$ in $nC_vdT$? which $C_v$ do you mean when you write that? I'm very confused because we simply didn't use moles not even one time this semester. | |
| Mar 14 at 15:38 | comment | added | Chet Miller | I went back to the overall version of the 1st law, from initial state to final state. | |
| Mar 14 at 15:17 | comment | added | Dor | After the integration it's $e^{C_v}\frac{T_2}{T_1}=e^R\frac{V_1}{V_2}$, assuming $C_v$ is constant. how did you proceed? | |
| Mar 14 at 14:49 | history | answered | Chet Miller | CC BY-SA 4.0 |