Why in the adiabatic equation of an ideal gas do we take $dU= C(v)dT$ , where $C(v)$ is the specific heat capacity at constant volume ? I mean that in an adiabatic expansion or compression the Volume doesn't stay constant so why do we take $C(v)$ ?

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    – Sanya
    Oct 17, 2016 at 13:10

4 Answers 4


Based on Adiabatic Equation Wikipedia

Derivation of $P–V$ relation for adiabatic heating and cooling

The definition of an adiabatic process is that heat transfer to the system is zero, $δQ = 0$. Then, according to the first law of thermodynamics,

$${\displaystyle {\text{(1)}}\qquad dU+\delta W=\delta Q=0,}$$

where $dU$ is the change in the internal energy of the system and δW is work done by the system. Any work (δW) done must be done at the expense of internal energy $U$, since no heat $δQ$ is being supplied from the surroundings. Pressure–volume work $δW$ done by the system is defined as

$${\displaystyle {\text{(2)}}\qquad \delta W=P\,dV.}$$

However, P does not remain constant during an adiabatic process but instead changes along with $V$.

It is desired to know how the values of $dP$ and $dV$ relate to each other as the adiabatic process proceeds. For an ideal gas the internal energy is given by

$${\displaystyle {\text{(3)}}\qquad U=\alpha nRT,}$$

where $α$ is the number of degrees of freedom divided by two, $R$ is the universal gas constant and $n$ is the number of moles in the system (a constant).

Differentiating Equation (3) and use of the ideal gas law, $PV = nRT$, yields

$${\displaystyle {\text{(4)}}\qquad dU=\alpha nR\,dT=\alpha \,d(PV)=\alpha (P\,dV+V\,dP).}$$

Equation (4) is often expressed as $dU = nC_V dT$ because $ C_V = αR$.

  • $\begingroup$ But why can dU be takes as nC(v)dT when volume is not constant in the expansion $\endgroup$
    – user379001
    Oct 17, 2016 at 13:47
  • $\begingroup$ And why can't we take dU=nC(p)dT where C(p) is the specific heat at constant pressure $\endgroup$
    – user379001
    Oct 17, 2016 at 13:49

The subscript v on Cv refers to how Cv can be measured for a given material experimentally, namely, by specifically measuring the amount of heat Q added in a constant volume test. It does not mean that Cv canand not be used determine the effect of temperature on internal energy U for other situations. In fact, for an ideal gas, U is a function only of T, irrespective of v. So, in that case, dU=nCvdT can be used to determine the change in internal energy in all situations.

  • $\begingroup$ So why can't we use dU = nC(p)dT where C(p) is the specific heat capacity at constant pressure $\endgroup$
    – user379001
    Oct 17, 2016 at 13:48
  • $\begingroup$ $C_p$ is defined as $$nC_p=\left(\frac{\partial H}{\partial T}\right)_p$$ $\endgroup$ Oct 17, 2016 at 14:29
  • $\begingroup$ $C_v$ is defined as $$nC_v=\left(\frac{\partial U}{\partial T}\right)_v$$ $\endgroup$ Oct 17, 2016 at 14:38
  • $\begingroup$ @ChesterMiller But what is the point of using C(v) ....why not just use C(p) since you have said that C(v) can be used and it doesn't matter if the volume is constant or not. Same should be applied to C(p). It won't matter is the pressure is constant or not. Why can't we use C(p) and just use C(v) $\endgroup$
    – Shashaank
    Oct 17, 2016 at 15:04
  • 1
    $\begingroup$ If you're solving a problem in terms of internal energy, you use Cv. If you're solving a problem in terms of enthalpy, you use Cp. Actually, U is the fundamental quantity, and H is just a convenient parameter that arises mathematically in many problems. But, all problems that can be solved in terms of H can be solved in terms of U. Don't forget that, for an ideal gas, H=U+PV, so H=U+nRT, and Cp=Cv+R. $\endgroup$ Oct 17, 2016 at 15:18

$U$ is a state function, which means that $\Delta U$ depends solely on initial and final states, and not on which process it went through in between. For ideal gas $U$ depends on $T$ alone. Hence once you know initial and final temperature, $\Delta U$ is thereby completely determined, irrespective of which process it went through in between.


For an ideal gas


is always true regardless of the process.


Consider any reversible process that takes the ideal gas(n=1) from an initial(P1,V1,T1) to final state(P2,V2,T2).

  1. Since the process is reversible we can consider infinitesimal differential changes from (P,V,T) to (P+dP,V+dV,T+dT'') as the system goes from initial to final state.
  2. At each of these steps, since U is a state var,calculate dU from an attached isobar and isochor betweeen same endpoints.
  1. isobar from (P,V,T) to (P,V+dV,T+dT'):
    dQ=$C_{P}$dT' where dT'=PdV/R (from PV=RT-this is why its valid only for ideal gas)

  2. isochor from (P,V+dV,T+dT') to (P+dP,V+dV,T+dT'+dT'')
    dQ=$C_{V}$dT'' where dT''=VdP/R (from PV=RT)

  3. total dU=$C_{P}$PdV/R -PdV + $C_{V}$VdP/R-$0$
    then using $C_{P}$-$C_{V}$=R (this derivation also requires PV=RT) dU simplifies to $C_{V}$dT

  1. Since the process is reversible, we can integrate the differential to get
    ΔU=$C_{V}$ΔT (assuming $C_{V}$ to independent of T-true for an ideal gas)

please see.


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