For an ideal gas with constant properties undergoing a quasi-static process, which one of the following represents the change of entropy (∆s) from state 1 to 2?

The correct relation representing the change in entropy ($\Delta s$) for an ideal gas with constant properties is:
$\Delta s=C_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{P_2}{P_1}\right)$
This corresponds to the expression shown in the first image.

Step-by-step Derivation:

1. We start from the second Tds relation (Gibbs equation) which relates entropy, enthalpy, and pressure:
$Tds=dh-vdP$
where:
• $T$ is the absolute temperature
• $ds$ is the differential change in specific entropy
• $dh$ is the differential change in specific enthalpy
• $v$ is the specific volume
• $dP$ is the differential change in pressure

2. Rearranging the equation to solve for $ds$:
$ds=\frac{dh}{T}-\frac{vdP}{T}$

3. For an ideal gas, the change in specific enthalpy is given by:
$dh=C_{p}dT$
where $C_p$ is the specific heat capacity at constant pressure.

4. Using the ideal gas equation of state, $Pv=RT$, we can express the term $\frac{v}{T}$ as:
$\frac{v}{T}=\frac{R}{P}$
where $R$ is the specific gas constant.

5. Substituting these expressions back into the equation for $ds$:
$ds=C_p\frac{dT}{T}-R\frac{dP}{P}$

6. Integrating both sides from state 1 to state 2, assuming constant specific heat ($C_p$):
${\int }_1^{2}ds=C_p{\int }_1^2\frac{dT}{T}-R{\int }_1^2\frac{dP}{P}$
This yields:
$\Delta s=C_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{P_2}{P_1}\right)$