Question

Consider an ideal gas undergoing a constant pressure process from state 1 to state

Answer 1

Answer:

$s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})$

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy at constant pressure:

$ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}$

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

$\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,$

We obtain the expression to compute the specific entropy change:

$s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})$

Best regards.