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I can’t seem to understand this part regarding the concept of temperature as mentioned in the book Heat and Thermodynamics by Zemansky. The text is as follows:

A scientific understanding of the concept of temperature builds upon thermal equilibrium, established in the zeroth law of thermodynamics. Consider a system A in the state $X_1,Y_1$ in thermal equilibrium with another system B in the state $X_1',Y_1'$. If system A is removed and its state changed, there will be found a second state $X_2,Y_2$ that’s in thermal equilibrium with the original state $X_1',Y_1'$ of system B.

I’m stuck at this part and can’t figure out why the second state $X_2,Y_2$ would be in thermal equilibrium with the original state of B.

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  • $\begingroup$ What edition, page of Zemansky is this on? $\endgroup$
    – Bob D
    Commented Sep 7, 2021 at 17:49
  • $\begingroup$ For the benefit of other contributors, the quote is from page 10 of the 7th edition of Heat and Thermodynamics, Zemansky & Dittman $\endgroup$
    – Bob D
    Commented Sep 7, 2021 at 21:35

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To put the quote in context (pp 7-10 of Zemansky), he states that it is applicable to systems of constant mass (a closed system) and composition, each requiring only one pair of independent coordinates for its description, such as pressure and volume for a gas.

In that context, let's break down the quote for the case of an ideal gas for a closed system (n=constant). The equation of state for an ideal gas for one mole of gas is

$$pV=RT$$

Now, the beginning of the quote:

Consider a system A in the state $X_1,Y_1$ in thermal equilibrium with another system B in the state $X_1',Y_1'$.

If $X$ and $X'$ are the pressures of systems A and B and $Y$ and $Y'$ are the volumes of systems A and B, and the two system are in thermal equilibrium, $T_{A}=T_{B}$, then it follows for an ideal gas that the product of the pressure and volume of system A will equal the product of the pressure and volume of system B, i.e.,

$$X_{1}Y_{1}=X'_{1}Y'_1$$

If system A is removed and its state changed, there will be found a second state $X_2,Y_2$ that’s in thermal equilibrium with the original state $X_1',Y_1'$ of system B.

For our ideal gas, if the state of system A changes to state 2 with a different pressure and volume it will still be in thermal equilibrium with the original state of system B as long as the product of the pressure and volume in the new state is the same as the original, or

$$X_{2}Y_{2}=X_{1}Y_{1}=X'_{1}Y'_{1}$$

Hope this helps.

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