Use of Mass and Energy Balances to Solve an Ideal Gas Problem^8 A gas cylinder of 1 m^3 volume containing nitrogen initially at a pressure of 40 bar and a temperature of 200 K is connected to another cylinder of 1 m^3 volume that is evacuated. A valve between the two cylinders is opened until the

Question

Use of Mass and Energy Balances to Solve an Ideal Gas Proble[latex]m^{8}[/latex]

A gas cylinder of 1 [latex]m ^{3}[/latex] volume containing nitrogen initially at a pressure of 40 bar and a temperature of 200 K is connected to another cylinder of 1 [latex]m ^{3}[/latex] volume that is evacuated. A valve between the two cylinders is opened until the pressures in the cylinders equalize. Find the final temperature and pressure in each cylinder if there is no heat flow into or out of the cylinders or between the gas and the cylinder. You may assume that the gas is ideal with a constant-pressure heat capacity of 29.3 J/(mol K).

[latex]^{8}[/latex]An alternative solution to this problem giving the same answer is given in the next chapter.

Accepted Answer

This problem is more complicated than the previous ones because we are interested in changes that occur in two separate cylinders. We can try to obtain a solution to this problem in two different ways. First, we could consider each tank to be a separate system, and so obtain two mass balance equations and two energy balance equations, which are coupled by the fact that the mass flow rate and enthalpy of the gas leaving the first cylinder are equal to the like quantities entering the second cylinde[latex] ext { r. }^{9}[/latex] Alternatively, we could obtain an equivalent set of equations by choosing a composite system of the two interconnected gas cylinders to be the first system and the second system to be either one of the cylinders. In this way the first (composite) system is closed and the second system is open. We will use the second system choice here; you are encouraged to explore the first system choice independently and to verify that the same solution is obtained.
The difference form of the mass and energy balance equations (on a molar basis) for the twocylinder composite system are

[latex]N_{1}^{i}=N_{1}^{f}+N_{2}^{f}[/latex] (a)

and

[latex]N_{1}^{i} \underline{U}_{1}^{i}=N_{1}^{f} \underline{U}_{1}^{f}+N_{2}^{f} \underline{U}_{2}^{f}[/latex] (b)

Here the subscripts 1 and 2 refer to the cylinders, and the superscripts i and f refer to the initial and final states. In writing the energy balance equation we have recognized that for the system consisting of both cylinders there is no mass flow, heat flow, or change in volume.
Now using, in Eq. a, the ideal gas equation of state written as N = P V/RT and the fact that the volumes of both cylinders are equal yields

[latex]\frac{P_{1}^{i}}{T_{1}^{i}}=\frac{P_{1}^{f}}{T_{1}^{f}}+\frac{P_{2}^{f}}{T_{2}^{f}}[/latex] ([latex]a ^{\prime}[/latex])

Using the same observations in Eq. b and further recognizing that for a constant heat capacity gas we have, from Eq. 3.3-8, that

[latex]\begin{aligned}\underline{U}^{ IG }(T) &=\underline{U}^{ IG }\left(T_{R} ight)+\int_{T_{R}}^{T} C_{ V }^{*}(T) d T=\left\{\underline{H}^{ IG }\left(T_{R} ight)-R T_{R} ight\}+\int_{T_{R}}^{T} C_{ V }^{*}(T) d T \&=\int_{T_{R}}^{T} C_{ V }^{*}(T) d T-R T_{R}\end{aligned}[/latex] (3.3-8)

[latex]\underline{U}(T)=C_{ V }^{*} T-C_{ P }^{*} T_{R}[/latex]

yields

[latex]\frac{P_{1}^{i}}{T_{1}^{i}}\left\{C_{ V }^{*} T_{1}^{i}-C_{ P }^{*} T_{R} ight\}=\frac{P_{1}^{f}}{T_{1}^{f}}\left\{C_{ V }^{*} T_{1}^{f}-C_{ P }^{*} T_{R} ight\}+\frac{P_{2}^{f}}{T_{2}^{f}}\left\{C_{ V }^{*} T_{2}^{f}-C_{ P }^{*} T_{R} ight\}[/latex]

which, on rearrangement, gives

[latex]-\left\{\frac{P_{1}^{i}}{T_{1}^{i}}-\frac{P_{1}^{f}}{T_{1}^{f}}-\frac{P_{2}^{f}}{T_{2}^{f}} ight\} C_{ P }^{*} T_{R}+C_{ V }^{*}\left\{P_{1}^{i}-P_{1}^{f}-P_{2}^{f} ight\}=0[/latex]

Since the bracketed quantity in the first term is identically zero (see Eq. [latex]a ^{\prime}[/latex]), we obtain

[latex]P_{1}^{i}=P_{1}^{f}+P_{2}^{f}[/latex] (c)

(Note that the properties of the reference state have canceled. This is to be expected, since the solution to a change-of-state problem must be independent of the arbitrarily chosen reference state. This is an important point. In nature, the process will result in the same final state independent of our arbitrary choice of reference temperature, TR. Therefore, if our analysis is correct, TR must not appear in the final answer.)
Next, we observe that from the problem statement [latex]P_{1}^{f}=P_{2}^{f}[/latex] ; thus

[latex]P_{1}^{f}=P_{2}^{f}=\frac{1}{2} P_{1}^{i}=20 bar[/latex]

and from Eq. a′

[latex]\frac{1}{T_{1}^{f}}+\frac{1}{T_{2}^{f}}=\frac{2}{T_{1}^{i}}[/latex] (c′)

Thus we have one equation for the two unknowns, the two final temperatures. We cannot assume that the final gas temperatures in the two cylinders are the same because nothing in the problem statement indicates that a transfer of heat between the cylinders necessary to equalize the gas temperatures has occurred.
To get the additional information necessary to solve this problem, we write the mass and energy balance equations for the initially filled cylinder. The rate-of-change form of these equations for this system are

[latex]\frac{d N_{1}}{d t}=\dot{N}[/latex] (d)

and

[latex]\frac{d\left(N_{1} \underline{U}_{1} ight)}{d t}=\dot{N} \underline{H}_{1}[/latex]

In writing the energy balance equation, we have made use of the fact that [latex]\dot{Q}, \dot{W}_{s}[/latex], and dV/dt are all zero. Also, we have assumed that while the gas temperature is changing with time, it is spatially uniform within the cylinder, so that at any instant the temperature and pressure of the gas leaving the cylinder are identical with those properties of the gas in the cylinder. Thus, the molar enthalpy of the gas leaving the cylinder is

[latex]\underline{H}=\underline{H}\left(T_{1}, P_{1} ight)=\underline{H}_{1}[/latex]

Since our interest is in the change in temperature of the gas that occurs as its pressure drops from 40 bar to 20 bar due to the escaping gas, you may ask why the balance equations here have been written in the rate-of-change form rather than in terms of the change over a time interval.
The answer is that since the properties of the gas within the cylinder (i.e., its temperature and pressure) are changing with time, so is [latex]\underline{H}_{1}[/latex], the enthalpy of the exiting gas. Thus, if we were to use the form of Eq. e integrated over a time interval (i.e., the difference form of the energy balance equation),

[latex]N_{1}^{f} \underline{U}_{1}^{f}-N_{1}^{i} \underline{U}_{1}^{i}=\int \dot{N} \underline{H}_{1} d t[/latex]

we would have no way of evaluating the integral on the right side. Consequently, the difference equation provides no useful information for the solution of the problem. However, by starting with Eqs. d and e, it is possible to obtain a solution, as will be evident shortly.
To proceed with the solution, we first combine and rearrange the mass and energy balances to obtain

[latex]\frac{d\left(N_{1} \underline{U}_{1} ight)}{d t} \equiv N_{1} \frac{d \underline{U}_{1}}{d t}+\underline{U}_{1} \frac{d N_{1}}{d t}=\dot{N} \underline{H}_{1}=\underline{H}_{1} \frac{d N_{1}}{d t}[/latex]

so that we have

[latex]N_{1} \frac{d \underline{U}_{1}}{d t}=\left(\underline{H}_{1}-\underline{U}_{1} ight) \frac{d N_{1}}{d t}[/latex]

Now we use the following properties of the ideal gas (see Eqs. 3.3-7 and 3.3-8)

[latex]\underline{H}^{ IG }(T)=\int_{T_{R}}^{T} C_{ P }^{*}(T) d T[/latex] (3.3-7)

[latex]N=P V / R T \quad \underline{H}=C_{ P }^{*}\left(T-T_{R} ight)[/latex]

and

[latex]\underline{U}=C_{ V }^{*}\left(T-T_{R} ight)-R T_{R}[/latex]

to obtain

[latex]\frac{P_{1} V}{R T_{1}} C_{ V }^{*} \frac{d T_{1}}{d t}=R T_{1} \frac{d}{d t}\left(\frac{P_{1} V}{R T_{1}} ight)[/latex]

Simplifying this equation yields

[latex]\frac{C_{ V }^{*}}{R} \frac{1}{T_{1}} \frac{d T_{1}}{d t}=\frac{T_{1}}{P_{1}} \frac{d}{d t}\left(\frac{P_{1}}{T_{1}} ight)[/latex]

or

[latex]\frac{C_{ V }^{*}}{R} \frac{d \ln T_{1}}{d t}=\frac{d}{d t} \ln \left(\frac{P_{1}}{T_{1}} ight)[/latex]

Now integrating between the initial and final states, we obtain

[latex]\left(\frac{T_{1}^{f}}{T_{1}^{i}} ight)^{C_{ V }^{*} / R}=\left(\frac{P_{1}^{f}}{P_{1}^{i}} ight)\left(\frac{T_{1}^{i}}{T_{1}^{f}} ight)[/latex]

or

[latex]\left(\frac{T_{1}^{f}}{T_{1}^{i}} ight)^{C_{ P }^{*} / R}=\left(\frac{P_{1}^{f}}{P_{1}^{i}} ight)[/latex] (f)

where we have used the fact that for the ideal gas [latex]C_{ P }^{*}=C_{ V }^{*}+R[/latex]. Equation f provides the means to compute [latex]T_{1}^{f}, ext { and } T_{2}^{f}[/latex] can then be found from Eq. c′. Finally, using the ideal gas equation of state we can compute the final number of moles of gas in each cylinder using the relation

[latex]N_{ J }^{f}=\frac{V_{ cyl J } P_{ J }^{f}}{R T_{ J }^{f}}[/latex] (g)

where the subscript J refers to the cylinder number. The answers are

[latex]\begin{array}{ll}T_{1}^{f}=164.3 K & N_{1}^{f}=1.464 kmol \T_{2}^{f}=255.6 K & N_{2}^{f}=0.941 kmol\end{array}[/latex]

Comments
The solution of this problem for real fluids is considerably more complicated than for the ideal gas. The starting points are again

[latex]P_{1}^{f}=P_{2}^{f}[/latex] (h)

and l

[latex]N_{1}^{f}+N_{2}^{f}=N_{1}^{i}[/latex] (i)
and Eqs. d and e. However, instead of Eq. g we now have
[latex]N_{1}^{f}=\frac{V_{c y 11}}{\underline{V}_{1}^{f}}[/latex] (j)

and

[latex]N_{2}^{f}=\frac{V_{c y 12}}{\underline{V}_{2}^{f}}[/latex] (k)

[latex] ext { where } \underline{V}_{1}^{f} ext { and } \underline{V}_{2}^{f} ext { are related to }\left(T_{1}^{f}, P_{1}^{f} ight) ext { and }\left(T_{2}^{f}, P_{2}^{f} ight)[/latex] , respectively, through the equation of state or tabular P V T data of the form

[latex]\underline{V}_{1}^{f}=\underline{V}_{1}^{f}\left(T_{1}^{f}, P_{1}^{f} ight)[/latex] (l)

[latex]\underline{V}_{2}^{f}=\underline{V}_{2}^{f}\left(T_{2}^{f}, P_{2}^{f} ight)[/latex] (m)

The energy balance for the two-cylinder composite system is

[latex]N_{1}^{i} \underline{U}_{1}^{i}=N_{1}^{f} \underline{U}_{1}^{f}+N_{2}^{f} \underline{U}_{2}^{f}[/latex] (n)

Since a thermal equation of state or tabular data of the form [latex]\underline{U}=\underline{U}(T, \underline{V})[/latex] are presumed available, Eq. n introduces no new variables.

Thus we have seven equations among eight unknowns [latex]\left(N_{1}^{f}, N_{2}^{f}, T_{1}^{f}, T_{2}^{f}, P_{1}^{f}, P_{2}^{f}, \underline{V}_{1}^{f} ight.[/latex] , and [latex]\left.\underline{V}_{2}^{f} ight)[/latex]. The final equation needed to solve this problem can, in principle, be obtained by the manipulation and integration of Eq. e, as in the ideal gas case, but now using the real fluid equation of state or tabular data and numerical integration techniques. Since this analysis is difficult, and a simpler method of solution (discussed in Chapter 6) is available, the solution of this problem for the real fluid case is postponed until Sec. 6.5.

Question 3.4-5: Use of Mass and Energy Balances to Solve an Ideal Gas Proble...

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