Why is the equation for change in Entropy for a reversible change in $T$ at constant $P$ described as
$$\Delta S = n C_p \ln\frac{T_f}{T_i}$$
[Physics] Why is the equation for Entropy of an ideal gas that undergoes reversible change in T at constant Pressure like this
entropythermodynamics
Related Solutions
$$H = U + PV \Rightarrow$$
$$dH = dU +PdV + VdP\tag{6}$$
In other words, equation 6 is missing the $VdP$ term.
$$dH = dU + nRT \frac{dV}{V} + nRT \frac{dP}{P}\tag{7}$$
$$ \Delta H = \Delta U + nRT \ln\frac{V_2}{V_1} + nRT \ln\frac{P_2}{P_1}\tag{8}$$
$$P_1 V_1 = P_2 V_2 \text{ (isothermal)}$$
$$\Delta H = \Delta U + nRT \left( \ln\frac{V_2}{V_1} + \ln\frac{V_1}{V_2} \right) = \Delta U = 0\tag{9}$$
If you can take the environment to be one with constant specific heats $c_p$ and $c_V$, then you can write $$dS_{\textrm{env}}=\frac{\delta Q_{\textrm{env}}}{T_{\textrm{env}}} = \begin{cases} \frac{nc_VdT}{T} & \textrm{if isochoric} \\ \frac{nc_pdT}{T} & \textrm{if isobaric} \end{cases}, $$ where the temperatures all refer to the environment. Then, when integrated between final and initial temperatures, we arrive at $$\Delta S_{\textrm{env}} = \begin{cases} n c_V \ln\left(\frac{T_f}{T_i}\right) & \textrm{if isochoric} \\ n c_p \ln\left(\frac{T_f}{T_i}\right) & \textrm{if isobaric} \end{cases}, $$ where, again, all quantities refer to the environment.
I assumed above that the processes undergone by the environment were quasi-static (or quasi-equilibrium). If the process was not quasi-static, we can only use the above expressions to compute the change in entropy provided the initial and final state variables of the environment match up correctly.
For instance, suppose that during the quasi-static, isobaric process assumed above, the environment moves from the state described by $(T_i,V_i,p_i)$ to the state described by $(T_f,V_f,p_f = p_i)$. Then the change in entropy of the environment during a non-quasi-static process taking place between the same two states would again be given by $$\Delta S_{\textrm{env}} = n c_p \ln\left(\frac{T_f}{T_i}\right). $$ However, this process would most likely not be isobaric, since intensive state variables like pressure tend to not be well-defined during non-quasi-static processes.
For further reference: supposing that the system and the environment are exchanging energy with each other via heat but are otherwise isolated from their environments, we can write $$\delta Q_{\textrm{env}} = -\delta Q_{\textrm{sys}},$$ in which case $$dS_{\textrm{env}} = \frac{\delta Q_{\textrm{env}}}{T_{\textrm{env}}} = -\frac{\delta Q_{\textrm{sys}}}{T_{\textrm{env}}} \geq -\frac{\delta Q_{\textrm{sys}}}{T_{\textrm{sys}}}. $$ Let's prove the inequality:
If $T_{\textrm{sys}} \geq T_{\textrm{env}}$, then the system loses energy to the environment via heat, and so $\delta Q_{\textrm{sys}} < 0$. In that case, $-\delta Q_{\textrm{sys}} = |\delta Q_{\textrm{sys}}|$, and so $$-\frac{\delta Q_{\textrm{sys}}}{T_{\textrm{env}}} =\frac{|\delta Q_{\textrm{sys}}|}{T_{\textrm{env}}} \geq \frac{|\delta Q_{\textrm{sys}}|}{T_{\textrm{sys}}} = -\frac{\delta Q_{\textrm{sys}}}{T_{\textrm{sys}}}, $$ where the inequality comes in because we have replaced the smaller environment temperature with the larger system temperature.
If $T_{\textrm{sys}} \leq T_{\textrm{env}}$, then the system gains energy from the environment via heat, and so $\delta Q_{\textrm{sys}} > 0$. In that case, $-\delta Q_{\textrm{sys}} = -|\delta Q_{\textrm{sys}}|$, and so $$-\frac{\delta Q_{\textrm{sys}}}{T_{\textrm{env}}} =-\frac{|\delta Q_{\textrm{sys}}|}{T_{\textrm{env}}} \geq \frac{|\delta Q_{\textrm{sys}}|}{T_{\textrm{sys}}} = -\frac{\delta Q_{\textrm{sys}}}{T_{\textrm{sys}}}, $$ where here, the replacement of a smaller denominator makes a "larger negative" number, in which case we get the same inequality.
Then, in the case that the system undergoes either an isochoric or isobaric process, we can write $$\Delta S_{\textrm{env}} \geq \begin{cases} -n_{\textrm{sys}} c_{V,\textrm{sys}} \ln\left(\frac{T_{f,\textrm{sys}}}{T_{i,\textrm{sys}}}\right) & \textrm{if isochoric} \\ -n_{\textrm{sys}} c_{p,\textrm{sys}} \ln\left(\frac{T_{f,\textrm{sys}}}{T_{i,\textrm{sys}}}\right) & \textrm{if isobaric} \end{cases}. $$ Of course, we knew this already, because if we go back to $$dS_{\textrm{env}}\geq -\frac{\delta Q_{\textrm{sys}}}{T_{\textrm{sys}}}, $$ we can identify the right-hand side as minus the change in entropy of the system, and re-arranging, we get---of course---the Second Law of Thermodynamics: \begin{align*} dS_{\textrm{env}}&\geq -\frac{\delta Q_{\textrm{sys}}}{T_{\textrm{sys}}} = -dS_{\textrm{sys}} \Longrightarrow\\ 0 &\leq dS_{\textrm{env}} + dS_{\textrm{sys}}. \end{align*}
Best Answer
Begin with $$ \delta Q = T dS $$ and recall that the heat transfer for an ideal gas at constant pressure is $$ \delta Q = nC_p dT $$ Combining these gives $$ dS = n C_p \frac{dT}{T} $$ and integrating both sides with respect to temperature from $T_i$ to $T_f$ gives $$ S_f - S_i = nC_p \ln\frac{T_f}{T_i} $$ as desired.