The isentropic relation between pressure and temperature of a calorically perfect gas is_____(a) \(\frac {p_2}{p_1} =(\frac {T_2}{T_1} )^{\frac {\gamma }{\gamma -1}}\)(b) \(\frac {p_1}{p_2} =(\frac {T_2}{T_1} )^{\frac {\gamma }{\gamma -1}}\)(c) \(\frac {p_2}{p_1} =(\frac {T_2}{T_1} )^{\frac {\gamma -1}{\gamma }}\)(d) \(\frac {p_1}{p_2} =(\frac {T_2}{T_1} )^{\frac {\gamma -1}{\gamma }}\)I had been asked this question by my college professor while I was bunking the class.My query is from A Brief Review of Thermodynamics topic in chapter Compressible Flow – Preliminary Aspects of Aerodynamics

The isentropic relation between pressure and temperature of a calorica

1.	The isentropic relation between pressure and temperature of a calorically perfect gas is_____(a) \(\frac {p_2}{p_1} =(\frac {T_2}{T_1} )^{\frac {\gamma }{\gamma -1}}\)(b) \(\frac {p_1}{p_2} =(\frac {T_2}{T_1} )^{\frac {\gamma }{\gamma -1}}\)(c) \(\frac {p_2}{p_1} =(\frac {T_2}{T_1} )^{\frac {\gamma -1}{\gamma }}\)(d) \(\frac {p_1}{p_2} =(\frac {T_2}{T_1} )^{\frac {\gamma -1}{\gamma }}\)I had been asked this question by my college professor while I was bunking the class.My query is from A Brief Review of Thermodynamics topic in chapter Compressible Flow – Preliminary Aspects of Aerodynamics
Answer» Correct answer is (a) \(\frac {p_2}{p_1} =(\frac {T_2}{T_1} )^{\frac {\gamma }{\gamma -1}}\) To elaborate: The isentropic relation means no heat exchange and no DISSIPATIVE forces. Also, for the calorically perfect gas, specific heats are CONSTANT. When we put these conditions in the entropy EQUATION, we obtain the relation between pressure and TEMPERATURE which is \(\frac {p_2}{p_1} =(\frac {T_2}{T_1} )^{\frac {\gamma }{\gamma -1}}\).

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