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We define specific heat as $$C={\frac{\Delta Q}{\Delta T}}$$ In case of isothermal process, if the heat supplied is compensated by P-V work done, then the temperature of the gas will not increase. So, $\Delta T=0$. Then, $C=\infty$.

In case of adiabatic process, however, $\Delta Q=0$, as you cannot supply heat from outside. Therefore, $C=0$.

However, when we want to calculate the amount of work done, in adiabatic process, we write $W=C_v \Delta T$ (for ideal gas). But $C_v$ should be $0$. But that does not make any sense.

How to solve this dialemma?

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    $\begingroup$ The C in isothermal process and the C in adiabatic process are different and should not be written out without differentiating them. For $W=C_v\Delta T$, I doubt it is correct as this is an adiabatic process but not an isochroric process. $\endgroup$ – user115350 Feb 28 '17 at 17:25
  • $\begingroup$ @user115350 For adiabatic process, dU=dq+dw. dq=0. Then, dU=dw. Again, we know that, dU=$C_v$dT. So, dw=$C_v$dT. There is nothing wrong with this equation. $\endgroup$ – Shoubhik Raj Maiti Mar 3 '17 at 17:03
  • $\begingroup$ You first say that $C_{adiabatic}=0$ and then say "$C_v$ should be zero". $\endgroup$ – lucas Mar 3 '17 at 17:22
  • $\begingroup$ Yes, you are right: $W=C_v \delta T$. It is also correct that $C_v \neq 0$ . In your second sentence, you said, for an adiabatic process C=0 which doesn't mean $C_v=0$. $\endgroup$ – user115350 Mar 3 '17 at 17:30
  • $\begingroup$ @user115350 For adiabetic process C=0, since we are applying $C_v$ for adiabetic process, $C_v$ should also be zero. $\endgroup$ – Shoubhik Raj Maiti Mar 5 '17 at 14:18
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The problem is with defining C in terms of the amount of heat Q (which is a function of path). We know that C should be a physical property of the material, so it should not depend on path. All the problems with C go away if, as we do in thermodynamics, we recognize and adopt the more general definitions: $$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$ and $$C_p=\left(\frac{\partial H}{\partial T}\right)_P$$ The subscripts on the C's are how the C's are measured, not how they are used in practice. At constant volume, you can get $\Delta U$ by measuring Q. At constant pressure, you can get $\Delta H$ by measuring Q. So the old definition carries over into these special circumstances where the heat capacities are being measured. But the definitions in terms of U and H are much more general than that, and apply to all processes.

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  • $\begingroup$ Please read the updated question. $\endgroup$ – Shoubhik Raj Maiti Mar 3 '17 at 17:05
  • $\begingroup$ I read the updated version. When we were freshmen, they tricked us by saying that C is the derivative of Q with respect to T. Obviously, you have shown that this definition does not properly carry over to our needs in Thermodynamics, where we need C to be a function of state (i.e., a physical property of the material) rather than a function of path. The two equations I gave are consistent with the freshman version when the volume or the pressure are constant. But, for other situations, the freshman version is not correct. The equations I gave are much more general, and apply to all cases. $\endgroup$ – Chet Miller Mar 3 '17 at 22:08
  • $\begingroup$ So, specific heat is an inherent property of a material, and C cannot be 0 or infinity. Am I right? $\endgroup$ – Shoubhik Raj Maiti Mar 5 '17 at 14:03
  • $\begingroup$ Yes you are right. $\endgroup$ – Chet Miller Mar 5 '17 at 14:57
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There is no dilemma as you have just shown that a gas can have many values of specific heat capacity.
They range from plus infinity through zero to minus infinity.
It all depends on what is happening to the gas as your examples have shown.

Two are chosen and tabulated called the principal specific heat capacities.
They are the specific heat capacity at constant pressure and the specific heat capacity at constant volume.

The first law links the change in internal energy of a system $\Delta U$ to the heat input to the system $\delta Q$ and the work done by the system $\delta W$.

$\Delta U = \delta Q-\delta W$

For an adiabatic process $\delta Q =0$ and for an ideal gas $\Delta U = C_v \Delta T$ and that is where you work done on an ideal gas equation comes from.

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  • $\begingroup$ Please read the updated question. $\endgroup$ – Shoubhik Raj Maiti Mar 3 '17 at 17:04
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In the section 4.7 of the 7th edition of the book Heat and Thermodynamics written by M. Zemansky and R. Dittman, we read

The heat capacity may be negative, zero, positive, or infinite, depending on the process the system undergoes during the heat transfer. Heat capacity has a definite value only for a definite process.

So according to the definition $C=\mathrm{đ}Q/\mathrm{d}T$ and the above extract, I think for an isothermal process $C=\infty$ and for an adiabatic process $C=0$.

Regarding the mentioned $W=C_v \Delta T$ relation for the case of adiabatic process in ideal gas:

According to the 1st law, for changes in internal energy we have

$$\Delta U=Q+W$$

In an adiabatic process $Q=0$, so

$$\Delta U=W(\text{adiabatic})$$

Thus, if we calculate the change in the internal energy, we have calculated the work done. But unlike work and heat, internal energy is a state function; this means its changes are independent of process and depend only on the initial and final thermodynamic states. Moreover, for ideal gases $U$ is a function of $T$ only (see chapter 5 of the same reference). Therefore the change in the internal energy of an ideal gas undergone an adiabatic process from temperature $T_i$ to $T_f$, equals the change in its internal energy when undergone a constant-volume process with initial and final temperatures $T_i$ and $T_f$. In the latter case, where $W(\text{isochoric})=0$, $$\Delta U=Q(\text{isochoric})=\int_{T_i}^{T_f}C_v \mathrm{d}T$$ So assuming $C_v$ is constant, we arrive to $$W(\text{adiabatic})=\Delta U= C_v \Delta T$$

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