2
$\begingroup$

In a Carnot Cycle, a reversible isothermal process is a non-adiabatic reversible process. Is it always true that a non-adiabatic reversible process is an isothermal process?

$\endgroup$
1

3 Answers 3

3
$\begingroup$

The word isothermal means "at constant temperature", and hence in an isothermal process $$\Delta T=0\ \ \ \ \ \ \ \text{Isothermal Process}$$ In an adiabatic expansion, there isn't any net heat flow and we have $$\delta Q=0\ \ \ \ \ \ \ \ \text{Adiabatic Process}$$

Is it always true that a non-adiabatic reversible process is an isothermal process?

No! Consider an example of the process which is not an adiabatic process nor an isothermal process. Consider a process in which the state of system changes from $A(p_i,V_i,T_i)$ to $B(p_f,V_f=V_i,T_f)$ which is called an isochoric process. In such a process, $$\delta Q=C_VdT\not= 0,\ \ \ \ \ \ \ \Delta T\not=0 $$


Everything taken to be reversible!

$\endgroup$
1
$\begingroup$

Is it always true that a non-adiabatic reversible process is an isothermal process?

No. A non-adiabatic process is any process that allows heat transfer between a system and its surroundings. Many processes, including isobaric and isochoric processes, involve heat transfer.

Hope this helps.

$\endgroup$
0
$\begingroup$

A) Non adiabatic reversible process means:

  1. $q$ is not equal to 0

  2. Reversible process i.e the change in state is not at equilibrium with its surrounding while the process is happening, unlike irreversible process. In simple words, no integration is needed.

B) Reversible Isothermal process means:

  1. $\Delta T = 0 $ is the only speciality.

Using the first law of thermodynamics i.e $\delta U = \pm q \pm W$

$q = c \cdot m \cdot ΔT$ where $c$ is specific heat capacity.

For B, we can write $$\frac{nf \cdot R \cdot \Delta T }{2}= q \pm W\text{ or }pdV $$

Since it is clearly written in the definition that $q \ne 0$, none of the variables in the formula of $q$ can be equal to $0$. Therefore, $\Delta T \ne 0 $ for a reversible isothermal process.

For A,

$$\frac{nf \cdot R \cdot \Delta T }{2}= q \pm W\text{ or }PdV $$

It states that for this condition that $\Delta T = 0$. Therefore, $0 = q \pm W$.

A and B do not match each other’s conditions since $\Delta T $ of both the processes are not same.

No, non adiabatic reversible process $\ne$ Reversible Isothermal process.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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