In his lecture (26:30-38:40), Shankar derives the adiabatic pressure-volume relationship $P_1V_1^\gamma = P_2V_2^\gamma$, where $\gamma = C_P / C_V$, from the First Law of Thermodynamics $\Delta U =Q - W$.

His first step in doing is is to make the substitution $\Delta U = n C_V \Delta T$ into the First Law. In adiabatic processes, volumes are not held constant, so why is using the specific heat at constant volume $C_V$ valid?

Glossary of Notation

$P$ - pressure

$V$ - volume

$T$ - temperature

$U$ - internal energy

$Q$ - heat added to system

$W$ - work done by system

$C_P$ - specific heat at constant pressure

$C_V$ - specific heat a constant volume

$n$ - number of moles

  • $\begingroup$ As the process is adiabatic, the first law reads $0 = q = \Delta U + p \Delta V$. Also $dU = q_V$ if only work of expansion is considered and volume is constant. And $q_V = C_V dT$. Does it solve your problem? $\endgroup$ – user1420303 Nov 24 '18 at 22:29
  • $\begingroup$ Notice that if only p-V work is allowed: $dU = q - pdV$, and if V is constant $dU = q_V$. This is valid even if you have a process that is not performed at constant volume. $\endgroup$ – user1420303 Nov 24 '18 at 22:33
  • $\begingroup$ @user1420303 Volume is NOT constant, so why is that substitution valid? $\endgroup$ – Trevor Kafka Nov 24 '18 at 22:40
  • $\begingroup$ Interesting! Thank you for your spot-on answer. Does this imply $C_V = \frac 32 N_A k_B$ (in per-mole units) for an ideal gas? $\endgroup$ – Trevor Kafka Nov 24 '18 at 22:48
  • $\begingroup$ Not sure why so many answers were downvoted, even if they say the same as the accepted answer $\endgroup$ – Andrei Nov 24 '18 at 22:54

$C_v$ is defined for a gas by$$Q=nC_v\Delta T$$ in which $Q$ is the heat inflow needed to raise the temperature of the gas by $\Delta T$ at constant volume.

At constant volume, no work is done, so the First Law collapses to $$\Delta U = Q.$$ Therfore$$\Delta U = nC_v\Delta T.$$ But for an ideal gas, $U$ is proportional to $n$ and $T$, whatever sort of process the gas is undergoing. There is therefore a process-independent proportionality constant. But we know from the constant volume process that this constant is equal to $C_v$ ! We name the constant after just one of the roles that it plays.

[This is a perennial cause of confusion to students learning thermodynamics. I find the following comparable case instructive… The fundamental role of the space-time constant, $c,$ is as a scale factor between times and displacements, so, for example we can write the invariant interval as$$(c\Delta \tau)^2=(c\Delta t)^2-[(\Delta x)^2+(\Delta y)^2+(\Delta z)^2]$$Yet we call $c$ "the speed of light", after one of the roles that it plays.]

| cite | improve this answer | |
  • $\begingroup$ This does not answer the question. You can't derive the ideal gas adiabatic isentropic relations for a constant volume process, and $\Delta U = nC_v \Delta T$ even if the process is not constant volume. $\endgroup$ – Drew Nov 25 '18 at 15:23
  • $\begingroup$ @Drew (a) "This does not answer the question." Clearly we disagree. The key paragraph in my answer is the one that starts "But for an ideal gas…". (b) "You can't derive the ideal gas adiabatic isentropic relations for a constant volume process" Who is claiming that you can? (c) "ΔU=nCvΔT even if the process is not constant volume." Exactly so, as I try to explain in my answer. $\endgroup$ – Philip Wood Nov 25 '18 at 19:09

He started with the first law

$\Delta U = \Delta Q - P \Delta V$

and then noted that $\Delta Q = 0$ for the adiabatic process, therefore we have

$0 = \Delta U + P \Delta V$

Now, for an ideal gas, we know that $\Delta U = nC_v \Delta T$. Therefore

$0 = nC_v \Delta T + P \Delta V$

and this was the basis of the rest of his derivation.

| cite | improve this answer | |
  • $\begingroup$ Typo fixed; question still stands. $\endgroup$ – Trevor Kafka Nov 24 '18 at 22:43
  • $\begingroup$ @TrevorKafka This still answers your question... For an ideal gas, $\Delta U = nC_v \Delta T$. That is why it was used. It is valid even if the process is not constant volume. $\endgroup$ – Drew Nov 25 '18 at 0:49

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