0
$\begingroup$

The only formula we learned for work is $W=\int pdV$. In stage 2 of Carnot cycle, there is no heat transfer but the volume increases, pressure decreases, internal energy decreases and temperature too. So even if we plug ideal gas equation, the integral is not solvable because the temperature is changing (we didn't learn what function describes that to plug into the integral).

I have only learned thermodynamics, no heat transfer.

Improve this question
$\endgroup$
9
  • $\begingroup$ Have you learned about the first law of thermodynamics yet? $\endgroup$
    – Chet Miller
    Commented Mar 14 at 11:46
  • $\begingroup$ Indeed. Second law too and how to calculate entropy in isothermal, isobaric and isochoric $\endgroup$
    – Dor
    Commented Mar 14 at 12:10
  • $\begingroup$ So in step 2, $dU=nC_vdT=-PdV=-\frac{nRT}{V}dV$ and $W=-nR\Delta T$You know how to solve for the final temperature, right? $\endgroup$
    – Chet Miller
    Commented Mar 14 at 12:14
  • $\begingroup$ How did u get there? I know that $dU=-W$ because $Q=0$. In isobaric process $dU=mC_pdT$. Did you mean that in ideal gas $C_v=C_p=C$? and what can I write now? I don't know the formula you wrote for $W$ $\endgroup$
    – Dor
    Commented Mar 14 at 12:49
  • $\begingroup$ In an isobaric process, dU is not $nC_PdT$, even for an ideal gas But, for an ideal gas, $cU=nC_VdT$. .Until you understand this better, you will not be able to understand Carnot cycle. I suggest doing more practice problems on 1st law. $\endgroup$
    – Chet Miller
    Commented Mar 14 at 13:03

1 Answer 1

Reset to default
0
$\begingroup$

So, for a reversible expansion of an ideal gas, $$dU=nC_vdT=dQ-dW=dQ-PdV$$If the expansion is also adiabatic, then dQ=0 and, from the ideal gas law, Pv=nRT,so we have $$nC_vdT=-\frac{nRT}{V}dV$$or$$C_v\frac{dT}{T}=-R\frac{dV}{V}$$Integrating this gives: $$\frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$From the 1st law of thermodynamics, the overall change in internal energy must be equal to minus the amount of work: $$\Delta U=-W=nC_v(T_2-T_1)$$or $$W=-nC_v(T_1-T_2)=nRT_1\left[1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}\right]$$

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$
    – Dor
    Commented Mar 14 at 15:17
  • $\begingroup$ I went back to the overall version of the 1st law, from initial state to final state. $\endgroup$
    – Chet Miller
    Commented Mar 14 at 15:38
  • $\begingroup$ 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. $\endgroup$
    – Dor
    Commented Mar 14 at 16:05
  • $\begingroup$ 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. $\endgroup$
    – Philip Wood
    Commented Mar 14 at 17:13
  • $\begingroup$ 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 $$C_V=\frac{1}{n}\left(\frac{\partial U}{\partial T}\right)_V$$ $\endgroup$
    – Chet Miller
    Commented Mar 14 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.