Topic: Thermodynamics
In a reversible adiabatic process, the ratio of $T_{1}/T_{2}$ is equal to
- A. $(\frac{p_{1}}{p_{2}})^{\frac{\gamma-1}{\gamma}}$
- B. $(\frac{p_{2}}{p_{1}})^{\frac{\gamma-1}{\gamma}}$
- C. $(\frac{v_{1}}{v_{2}})^{\frac{\gamma-1}{\gamma}}$
- D. $(\frac{v_{2}}{v_{1}})^{\frac{\gamma-1}{\gamma}}$
Answer: A. $(\frac{p_{1}}{p_{2}})^{\frac{\gamma-1}{\gamma}}$
Explanation: A reversible adiabatic process is known as an isentropic process. In an adiabatic process, the change in heat energy is zero $(Q=0)$. For a reversible adiabatic process, the temperature and pressure are related by the expression $\frac{T_{1}}{T_{2}}=(\frac{p_{1}}{p_{2}})^{\frac{\gamma-1}{\gamma}}$ where $\gamma = \frac{C_p}{C_v}$.