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Chapter 5: Problem 13

The volume of a dry box (a closed chamber with dry nitrogen flowing through it) is \(2.0 \mathrm{m}^{3}\). The dry box is maintained at a slight positive gauge pressure of \(10 \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\) and room temperature \(\left(25^{\circ} \mathrm{C}\right) .\) If the contents of the box are to be replaced every five minutes, calculate the required mass flow rate of nitrogen in \(g / \min\) by (a) direct solution of the ideal-gas equation of state and (b) conversion from standard conditions. You may assume the gas in the dry box is well mixed.

Short Answer

Expert verified
Obtain the required mass flow rate by directly applying the ideal gas law and converting the obtained value to standard conditions. The resulting value should be converted from kilograms per second (kg / s) to grams per minute (g / min) using appropriate conversion factors. The detailed steps illustrate the method for doing this.

Step by step solution

01

Determining the absolute pressure

Firstly, convert the gauge pressure from \(cmH_{2}O \) to \(\ kPa \) (kilopascal). The conversion factor is 1 \(\ cmH_{2}O \) = 0.0980665 \(\ kPa \). Hence the absolute pressure \(P\) at room temperature in \(kPa\) is \(P = 101.325 + (10\times0.0980665)\) that gives us \(P=102.256665 kPa\).
02

Using the ideal gas law for direct solution

The ideal gas law equation is \(PV = nRT\), but we want it in terms of mass flow rate \((\dot{m})\) which is \(\frac{m}{t} \), where \(t\) is the time taken to replace the gas and \(V \) the volume. We rewrite the equation in the form required: \(PV = \frac{\dot{m}}{R'}T\), where \(\dot{m}= \frac{PV}{R'T} \), \(R'\) is the specific gas constant for nitrogen (\(R'=0.2968 \, kPa.m^3/(kg.K)\)). Now, we can substitute the values of \(P\), \(V\), \(R'\) and \(T\)(temperature given) into the equation to find out \(\dot{m}\).
03

Converting to standard conditions

To calculate mass flow rate under standard conditions, we use the Standard Temperature and Pressure (STP) conditions. STP temperature \(T_{STP}=273 K\) and STP pressure \(P_{STP}=101.325 kPa\). So for STP, using \(\dot{m_{STP}}= \frac{PV}{R'T} \) equation, we substitute \(T_{STP}\) and \(P_{STP}\) while keeping the volume the same. This will provide the mass flow rate in the standard conditions.
04

Converting the units

Since the problem asked for answer in \(g/min\), the result obtained in \(Kg/sec\) has to be converted to the required units. To do so, one multiplies the results by \(60 sec / min\) to convert \(sec\) to \(min\), and by \(1000 g/Kg\) to convert \(Kg\) to \(g\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The ideal gas law is a foundational equation in thermodynamics and physical chemistry that relates the pressure (P), volume (V), temperature (T), and amount (n) of an ideal gas, where the amount can be in terms of moles. The equation is given by:
\[ PV = nRT \]
In this context, 'R' represents the universal gas constant, which has a value dependent on the units used for pressure, volume, and temperature. When solving for the mass flow rate, or the rate at which mass enters or exits a system, the ideal gas law can be rearranged to include the mass and specific gas constant (\( R' \)).
For a given mass (\( m \)) and specific gas constant (\( R' \)), the ideal gas law becomes:
\( PV = (m/t)R'T \)
Where \( \dot{m} \) (mass flow rate) is equivalent to \( m/t \). For the calculation of mass flow rate in a particular system, such as a dry box with nitrogen gas, this form of the equation allows us to find the rate at which nitrogen must flow to maintain the specified conditions, using the pressure, volume, and temperature of the system along with the specific gas constant for nitrogen.
Standard Temperature and Pressure
Standard Temperature and Pressure (STP) are reference conditions used in the field of chemistry and physics to enable comparisons between different sets of data. STP is typically defined as a temperature of 0掳C (273.15K) and an absolute pressure of 1 atmosphere (101.325 kPa). These conditions are essential when using the ideal gas law to calculate properties such as density or volume that can vary with changes in temperature and pressure.

Calculations involving gases often require adjustments to account for deviations from STP due to the fact that real-world conditions usually differ. In the case of our mass flow rate calculation for nitrogen, converting to and from STP allows us to comprehend the changes in gas behavior under different conditions, and it is instrumental in solving for the mass flow rate under the specific conditions mentioned in the exercise.
Absolute Pressure
Pressure is essential in understanding how gases behave under different conditions, and it is a critical variable in the ideal gas law. Absolute pressure is the total pressure exerted by a gas, including the atmospheric pressure and any additional pressure applied by the gas itself. It is a sum of the gauge pressure, which can be positive or negative, and the atmospheric pressure.

For instance, in our example with the dry box, gauge pressure is given, but for accurate calculations using the ideal gas law, absolute pressure is needed. This is why the first step in the solution involves converting the gauge pressure to absolute pressure by adding it to the atmospheric pressure. The atmospheric pressure is equivalent to the pressure exerted by the Earth's atmosphere at sea level (101.325 kPa at STP). The inclusion of absolute pressure in calculations ensures accuracy when applying the ideal gas law in practical scenarios.
Gas Constant
The gas constant, often symbolized as 'R', is a physical constant that appears in several fundamental equations in the physical sciences, such as the ideal gas law. Its value depends on the units chosen for pressure, volume, and temperature, with a common value being 8.314 J/(mol路K) when pressure is in pascals and volume in cubic meters.

However, each gas also has a specific gas constant (\( R' \)), which relates the universal gas constant to the molar mass of the gas. For nitrogen, which is used in the exercise, the specific gas constant is 0.2968 kPa路m鲁/(kg路K). This particular value permits the calculation of the mass flow rate without first finding the number of moles, simplifying the process. Understanding the role of the gas constant is vital for working with gas laws and carrying out calculations related to the behavior of gases.

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Most popular questions from this chapter

The label has come off a cylinder of gas in your laboratory. You know only that one species of gas is contained in the cylinder, but you do not know whether it is hydrogen, oxygen, or nitrogen. To find out, you evacuate a 5 -liter flask, seal it and weigh it, then let gas from the cylinder flow into it until the gauge pressure equals 1.00 atm. The flask is reweighed, and the mass of the added gas is found to be 13.0g. Room temperature is \(27^{\circ} \mathrm{C}\), and barometric pressure is 1.00 atm. What is the gas?

Determining the value of newly located natural gas sites involves estimating the gas composition. quantity, and ease of access. For example, one report described a find of 2 trillion cubic feet of natural gas that is significantly offshore, in 20 feet of water, and at a drilled depth of 25,000 ft. (In North America and the OPEC countries, reported volumes are determined at 14.73 psia and \(60^{\circ} \mathrm{F}\).) The pressure in this find is estimated to be 750 atm, and the gas is 94 mole \(\%\) methane, \(3.5 \%\) ethane, and the balance \(\mathrm{CO}_{2}\) (a) Estimate the total Ib-moles of gas in the find. (b) Use the compressibility-factor equation of state to estimate the specific volume (ft \(^{3} /\) /b-mole) in the well. The temperature of such wells can vary depending upon a number of factors; for the purposes of this problem, assume that it is \(200^{\circ} \mathrm{C}\).

The concentration of oxygen in a 5000 -liter tank containing air at 1 atm is to be reduced by pressure purging prior to charging a fuel into the tank. The tank is charged with nitrogen up to a high pressure and then vented back down to atmospheric pressure. The process is repeated as many times as required to bring the oxygen concentration below 10 ppm (i.c., to bring the mole fraction of \(\mathrm{O}_{2}\) below \(10.0 \times 10^{-6}\) ). Assume that the temperature is \(25^{\circ} \mathrm{C}\) at the beginning and end of each charging cycle. When doing \(P V T\) calculations in Parts (b) and (c), use the generalized compressibility chart if possible for the fully charged tank and assume that the tank contains pure nitrogen. (a) Speculate on why the tank is being purged. (b) Estimate the gauge pressure (atm) to which the tank must be charged if the purge is to be done in one charge-vent cycle. Then estimate the mass of nitrogen (kg) used in the process. (For this part, if you can't find the tank condition on the compressibility chart, assume ideal-gas behavior and state whether the resulting estimate of the pressure is too high or too low.) (c) Suppose nitrogen at 700 kPa gauge is used for the charging. Calculate the number of charge-vent cycles required and the total mass of nitrogen used. (d) Use your results to explain why multiple cycles at a lower gas pressure are preferable to a single cycle. What is a probable disadvantage of multiple cycles?

Most of the concrete used in the construction of buildings, roads, dams, and bridges is made from Portland cement, a substance obtained by pulverizing the hard, granular residue (clinker) from the roasting of a mixture of clay and limestone and adding other materials to modify the setting properties of the cement and the mechanical properties of the concrete. The charge to a Portland cement rotary kiln contains \(17 \%\) of a dried building clay \(\left(72 \mathrm{wt} \% \mathrm{SiO}_{2}\right.\) \(\left.16 \% \mathrm{Al}_{2} \mathrm{O}_{3}, 7 \% \mathrm{Fe}_{2} \mathrm{O}_{3}, 1.7 \% \mathrm{K}_{2} \mathrm{O}, 3.3 \% \mathrm{Na}_{2} \mathrm{O}\right)\) and \(83 \%\) limestone \(\left(95 \mathrm{wt} \% \mathrm{CaCO}_{3}, 5 \% \text { impuritics }\right)\) When the solid temperature reaches about \(900^{\circ} \mathrm{C},\) calcination of the limestone to lime (CaO) and carbon dioxide occurs. As the temperature continues to rise to about \(1450^{\circ} \mathrm{C},\) the lime reacts with the minerals in the clay to form such compounds as \(3 \mathrm{CaO} \cdot \mathrm{SiO}_{2}, 3 \mathrm{CaO} \cdot \mathrm{Al}_{2} \mathrm{O}_{3},\) and \(4 \mathrm{CaO} \cdot \mathrm{Al}_{2} \mathrm{O}_{3} \cdot \mathrm{Fe}_{2} \mathrm{O}_{3} .\) The flow rate of \(\mathrm{CO}_{2}\) from the kiln is \(1350 \mathrm{m}^{3} / \mathrm{h}\) at \(1000^{\circ} \mathrm{C}\) and 1 atm. Calculate the feed rates of clay and limestone ( \(\mathrm{kg} / \mathrm{h}\) ) and the weight percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the final cement.

Methanol is produced by reacting carbon monoxide and hydrogen at \(644 \mathrm{K}\) over a \(\mathrm{ZnO}-\mathrm{Cr}_{2} \mathrm{O}_{3}\) catalyst. A mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) in a ratio \(2 \mathrm{mol} \mathrm{H}_{2} / \mathrm{mol}\) CO is compressed and fed to the catalyst bed at \(644 \mathrm{K}\) and 34.5 MPa absolute. A single-pass conversion of 25\% is obtained. The space velocity, or ratio of the volumetric flow rate of the feed gas to the volume of the catalyst bed, is The product gases are passed through a condenser, in which the methanol is liquefied. (a) You are designing a reactor to produce \(54.5 \mathrm{kmol} \mathrm{CH}_{3} \mathrm{OH} / \mathrm{h}\). Estimate (i) the volumetric flow rate that the compressor must be capable of delivering if no gases are recycled, and (ii) the required volume of the catalyst bed. (Use Kay's rule for pressure-volume calculations.) (b) If (as is done in practice) the gases from the condenser are recycled to the reactor, the compressor is then required to deliver only the fresh feed. What volumetric flow rate must it deliver assuming that the methanol produced is completely recovered in the condenser? (In practice it is not; moreover, a purge stream must be taken off to prevent the buildup of impurities in the system.)
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