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I have a vessel of fixed volume with adiabatic boundary which is initially filled with air at pressure of 1 atm and temperature of 300 K. Now, I inject air from an external source at a fixed flow rate for a certain duration. The injection flow rate and duration are known to me. The injected air is also at 300 K. After the duration of injection, I leave the system to achieve equilibrium.

My question is: at the final state, will the temperature be higher than 300 K? Being an adiabatic boundary, there is no heat transfer; and moreover, the initial air and injected air are both at the same temperature of 300 K.

But, the kinetic energy of the injected gas should get dissipated and increase the internal energy of the system. How do I calculate this and get the final temperature?

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  • $\begingroup$ Point is the temperature of both gases is same this implies their internal energy is also the same. $\endgroup$
    – Qwerty
    Jan 19 at 7:51

2 Answers 2

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Adiabatic means no heat is conducted, but not that temperature does not rise. The gas fill is compressed, which takes work, and raises the pressure of the filled container.

Adiabatic equation $$P \times {(V + {dV\over dt} \times t)^\gamma} = constant = P_{final} \times V^\gamma $$

applies. For air at normal temperatures, gamma is about 1.4' it's known as the ratio of specific heats.

Since the volume flow rate and time are known, that's an equation that can solve for the final pressure, and the ideal gas equation (or van der Waals variant) can convert to temperature.

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Yes, you are correct that the final temperature will be higher.

Assume that the gas is ideal, and let the subscript 0 represent the initial steady state parameters for the tank and the subscript 1 represent the parameters when the pressure in the tank reaches $P_1$. Let $m_0$ represent the steady sate moles of gas in the tank and $\delta m=m_1-m_0$. Then the ideal gas law gives: $$P_0V=m_0RT_0$$ and $$P_1V=(m_0+\delta m)RT_1$$ The gas being fed to the tank is at the same temperature as the steady state temperature $T_0$. Taking as our closed adiabatic system the initial steady state moles in the tank $m_0$ plus the number of moles $\delta m$ injected until the tank pressure is $P_1$, we have $$\Delta U=(m_0+\delta m)C_v(T-T_0)=P_{ext}\delta v=P_{ext}\frac{\delta m RT_0}{P_{ext}}=\delta m RT_0$$where $\delta v$ is the volume injected at the injectate pressure $P_{ext}$. Solving this for the final temperature gives: $$\frac{T_1}{T_0}=1+(\gamma-1)\frac{\delta m}{m_0+\delta m}$$. $\delta v$ is related to $\delta m$ by the ideal gas law: $$P_{ext}\delta v=\delta m RT_0$$

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