Question

Can two states of an ideal gas be connected by an isothermal process as well as an adiabatic process?

Answer 1

Answer:

I was thinking that the statement in the problem is wrong. Let the initial state be (P1,V1)(P1,V1) and final state be (P2,V2)(P2,V2). Let the isothermal curve passes through the initial state and final state. Also there is a relation between the slope of adiabatic curve and the isothermal curve.

slope of adiabatic curve = γdPdVγdPdV where dPdVdPdV is the slope of the isothermal curve.

Now if the adiabatic curve passes through the initial state (P1,V1)(P1,V1) then it won't pass through the final state.

Am I right or wrong? Is the statement in the problem means something else?

If both processes are reversible, then, as you said, they can't pass through the same initial and final states. If they are irreversible, I'm not so sure.

In any event, the intent of the problem seems to have been to get you to use the first law of thermodynamics. Both processes have the same ΔUΔU because the initial and final states are the same. So, WI−WA=QI−QAWI−WA=QI−QA. But, QA=0QA=0. Therefore,

WI=WA+QIWI=WA+QI

QIQI can be either positive of negative. So, it seems to me, none of the answers are right.

Answer 2

If the $V_{1}=V_{2}$  condition is met then an isothermal as well as adiabatic process will connect the two states.

Explanation:

To connect two states through an isothermal process,

$P_{1} V_{1}=P_{2} V_{2} \ldots \text { eq }^{n}(1)$

To connect the same two States by an adiabatic mechanism,

$P_{1} V_{1}^{y}=P_{2} V_{2}^{y} \dots e q^{n}(\text { ii })$

If both equations hold together then we get equation (ii) divided by equation (i)

$V_{1}^{r-1}=V_{2}^{r-1}$

Let the gas is considered as mono atomic. Then,

$\begin{aligned}&\gamma=\frac{5}{3}\\&V_{1}^{\frac{2}{3}}=V_{2}^{\frac{2}{8}}\end{aligned}$

$V_{1}=V_{2}$  

If this condition is met then an isothermal as well as adiabatic process will connect the two states.