# Thermo homework (HW) problem

I do not believe this question can be solved with the thermodynamics knowledge that I have learned thus far, but someone correct me if I am wrong:

A 0.2-m³ tank containing helium at 15 bar and 22°C will be used to supply 4.5 moles per minute of helium at atmospheric pressure using a controlled adiabatic throttling valve.

If the tank is well insulated, what will be the pressure in the tank and the temperature of the gas stream leaving the throttling valve at any later time $t$?

You may assume helium to be an ideal gas with $C^{*}_{P} = 22 \text{ J / (mol K)}$, and that there is no heat transfer between the tank and the gas.

$$\frac{dS}{dt} = \sum_{k=1}^{K} \dot{N}_{k} \underline{S_k} + \frac{\dot{Q}}{T} + \dot{S}_{gen}$$

No $\dot{Q}$ since it's adiabatic, and no $\dot{S}_{gen}$ since I guess we're assuming it's reversible. One output stream with $\dot{N} = -4.5 \text{ mol/min}$ gives

$$\frac{dS}{dt} = \dot{N}\underline{S}(t) = \dot{N}\frac{S(t)}{N(t)} = \frac{\dot{N}S(t)}{N_0 + \dot{N}t}$$

Solving this gives

$$S(t) = S_0 + \frac{S_0 \dot{N}}{N_0}t$$

For a closed system, my book derives relationships between entropy changes and two of volume, pressure, or temperature changes via the 1st law:

$$d\underline{U} = Td\underline{S} - \underline{P}d\underline{V}$$

$$d\left(\frac{U}{N}\right) = Td\left(\frac{S}{N}\right) - \frac{P}{N}d\left(\frac{V}{N}\right)$$

From here it integrates this to arrive at something like

$$\underline{S}(T_2, \underline{V}_2) - \underline{S}(T_1, \underline{V}_1) = C_{v}^{*}\ \text{ln}\left(\frac{T_2}{T_1}\right) + R\ \text{ln}\left(\frac{\underline{V}_2}{\underline{V}_1}\right)$$

However, this is for constant $N$! With the helium tank, it is an open system and $N$ is not constant! So I don't see how I can proceed further. Without an entropy balance, the mass and energy balances alone are not enough to solve the system.

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• Start with the ideal gas law, $PV=nRT$, and solve for $n_0$ using initial temperature and pressure.
• With a constant mass flow rate, $n(t)=n_0-\dot{n}t$ where $\dot{n}=4.5\text{mol/min}$.
• The drop in energy of the gas in the tank will equal the energy flow out, $H\dot{n}=n(t){C_p}dT$.