Question
By combining the ideal gas equation of state with (7.43), show that during an adiabatic expansion $T V^{\gamma-1}$ is constant and $T^\gamma p^{1-\gamma}$ is constant, when $\gamma$ is independent of temperature.
Step 1
First, let's start with the ideal gas equation of state: \[PV = nRT\] where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of gas, $R$ is the ideal gas constant, and $T$ is the temperature. Show more…
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For an adiabatic process involving an ideal gas (A) $\mathrm{P}^{\gamma-1}=\mathrm{T}^{\gamma-1}=$ constant (B) $\mathrm{P}^{1-\gamma}=\mathrm{T}^{\gamma}=$ constant (C) $\mathrm{PT}^{\gamma-1}=$ constant (D) $\mathrm{P}^{\gamma-1}=\mathrm{T}^{\gamma}=$ constant
The ideal gas equation for an adiabatic process is: (a) $P V^{\gamma}=$ constant (b) $P V^{\gamma+1}=$ constant (c) $P V^{\gamma}=0$ (d) $P V^{\gamma-1}=$ constant
THERMODYNAMICS
Thermodynamics
Show that for a quasistatic adiabatic process in a perfect gas, with constant specific heats, $$ \left(\gamma \equiv C_{p} / C_{y}\right) $$ $$ P V^{\prime}=\text { const. } $$
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