/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Chemicals are stored in a labora... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Chemicals are stored in a laboratory with volume \(V\left(\mathrm{m}^{3}\right) .\) As a consequence of poor laboratory practices, a hazardous species, A, enters the room air (from inside the room) at a constant rate \(\dot{m}_{\mathrm{A}}(\mathrm{g} \mathrm{A} / \mathrm{h})\) The room is ventilated with clean air flowing at a constant rate \(\dot{V}_{\text {air }}\left(\mathrm{m}^{3} / \mathrm{h}\right) .\) The average concentration of A in the room air builds up until it reaches a steady-state value \(C_{\mathrm{A}, \mathrm{r}}\left(\mathrm{g} \mathrm{A} / \mathrm{m}^{3}\right)\) (a) List at least four situations that could lead to A getting into the room air. (b) Assume that the A is perfectly mixed with the room air and derive the formula $$\dot{m}_{\mathrm{A}}=\dot{V}_{\mathrm{air}} C_{\mathrm{A}}$$ (c) The assumption of perfect mixing is never justificd when the enclosed space is a room (as opposed to, say, a stirred reactor). In practice, the concentration of A varies from one point in the room to another: it is relatively high near the point where A enters the room air and relatively low in regions far from that point, including the ventilator outlet duct. If we say that \(C_{\mathrm{A}, \text { duct }}=k C_{\mathrm{A}}\) where \(k < 1\) is a nonideal mixing factor (generally between 0.1 and \(0.5,\) with the lowest value corresponding to the poorest mixing), then the equation of Part (b) becomes $$\dot{m}_{\mathrm{A}}=k \dot{V}_{\mathrm{air}} C_{\mathrm{A}}$$ Use this equation and the ideal-gas equation of state to derive the following expression for the average mole fraction of \(A\) in the room air: $$y_{\mathrm{A}}=\frac{\dot{m}_{\mathrm{A}}}{k \dot{V}_{\mathrm{air}}} \frac{R T}{M_{\mathrm{A}} P}$$ where \(M_{\mathrm{A}}\) is the molecular weight of \(\mathrm{A}\) (d) The permissible exposure level (PEL) for styrene \((M=104.14\) ) defined by the U.S. Occupational Safcty and Health Administration is 50 ppm (molar basis). \(^{21}\) An open storage tank in a polymerization laboratory contains styrene. The evaporation rate from this tank is estimated to be \(9.0 \mathrm{g} / \mathrm{h}\). Room temperature is \(20^{\circ} \mathrm{C}\). Assuming that the laboratory air is reasonably well mixed (so that \(k=0.5\) ), calculate the minimum ventilation rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) required to keep the average styrene concentration at or below the PEL. Then give several reasons why working in the laboratory might still be hazardous if the calculated minimum ventilation rate is used. (e) Would the hazard level in the situation described in Part (d) increase or decrease if the temperature in the room were to increase? (Increase, decrease, no way to tell.) Explain your answer, citing at least two effects of temperature in your explanation.

Short Answer

Expert verified
In part (a), various situations are listed where hazardous A can get into the room air, mainly related to the mishandling of chemicals. In (b), the formula \(\dot{m}_{A} = \dot{V}_{air} C_{A}\) is derived based on the steady-state assumption. In (c), for nonideal mixing, the equation is modified to \(\dot{m}_{A} = k \dot{V}_{air} C_{A}\), and the mole fraction \(y_{A}\) is derived using the ideal gas law. In (d), the minimum ventilation rate is computed ensuring the styrene concentration below PEL, and potential hazards of working in such a lab are discussed. In (e), it is reasoned that increasing the room temperature would increase the hazard level due to increased volatilization and diffusion rates of styrene.

Step by step solution

01

Part (a) - Situations introducing hazardous species A

1. Improper handling or storage of chemical substances known to off gas hazardous species A.\n2. Accidental spillage of chemicals containing species A.\n3. Lack of proper sealing on containers holding species A.\n4. Chemical reactions taking place in the room producing species A as a byproduct.
02

Part (b) - Derivation of the formula

The rate of accumulation of species A in the room is equal to the rate of its entry \(\dot{m}_{A}\) minus its rate of removal which is given by \(\dot{V}_{air} C_{A}\). In a steady state condition, accumulation is zero, hence, \(\dot{m}_{A} = \dot{V}_{air} C_{A}\).
03

Part (c) - Modification for nonideal mixing and deriving average mole fraction

In case of nonideal mixing, the rate of removal of species A becomes \(k \dot{V}_{air} C_{A}\). Hence, the equation becomes \(\dot{m}_{A} = k \dot{V}_{air} C_{A}\). Applying the ideal gas law, express the concentration \(C_{A}\) as \(n_{A}/V_{air}\), where \(n_{A}\) is the number of moles, given as \(\dot{m}_{A}/M_{A}\). Substitute this into the equation to get \(y_{A} = \frac{\dot{m}_{A}}{k \dot{V}_{air}} \frac{R T}{M_{A} P}\), where \(y_{A}\) is the mole fraction of A.
04

Part (d) - Calculating Minimum Ventilation Rate

Given that the permissible exposure level for styrene is 50 ppm, this has to be converted into mole fraction, \(y_{styrene} = 50 parts per million = 50 x 10^{-6}\). Substitute the given values into the equation \(y_{A} = \frac{\dot{m}_{A}}{k \dot{V}_{air}} \frac{R T}{M_{A} P}\) and solve for \(\dot{V}_{air}\), the minimum ventilation rate. Working might still be hazardous due to: Variations in the concentration of styrene in different locations within the room, Other potential sources of styrene or other harmful substances, Poor assumption of perfect mixing.
05

Part (e) - Temperature Effects on Hazard Level

Increasing the temperature in the room would increase the hazard level. This is because higher temperature would: Increase the volatilization rate of styrene, hence introducing more of it into the room air. Increase the diffusion rate of styrene, potentially leading to higher exposure in regions of the room that were previously considered safe.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hazardous Species Ventilation
When handling chemicals in a lab setting, the matter of safety is paramount. Understanding the concept of hazardous species ventilation is essential for maintaining air quality and reducing potential risks. This involves the process of removing airborne contaminants through continuous flow of fresh air. Imagine this as a dilution process where 'bad' air is replaced by 'good', clean air at a rate sufficient to keep the concentration of hazardous substances at acceptable levels.

Key to this is ensuring that there are no stagnant zones where contaminants might accumulate. Situations that can preempt the need for vigilant ventilation can include accidental spills, reactions that produce airborne byproducts, or even the off-gassing of certain materials due to improper storage. It's like having a busy and slightly messy 'roommate' in the form of a chemical compound that you have to continuously clean after to maintain a livable environment. In essence, understanding the ventilation rate hinges on balancing the introduction of hazardous species with their removal to sustain a safe indoor air quality.
Steady-State Concentration Derivation
The concept of steady-state concentration derivation can seem quite complex, but let’s break it down. A steady state is like a tug-of-war match where both teams are equally strong—there’s a lot of activity, but no movement. In our chemical 'match', the rate at which a contaminant is added to the lab air is countered by the rate at which the ventilation system removes it.

The derived steady-state equation, \(\dot{m}_{A} = \dot{V}_{\text{air}} C_{A}\), succinctly shows this balance. This formula is a snapshot of a dynamic equilibrium, meaning that even though the substances are still being emitted and ventilated, their overall concentration in the air remains constant. It’s a pivotal concept in maintaining safe lab environments, ensuring that the exposure to hazardous substances remains consistently within safe limits.
Ideal Gas Law Application
When faced with the question of how to relate the amount of a gas to volume, temperature, and pressure, the ideal gas law is the go-to tool. By applying this law, we’re equipped to solve a variety of problems, such as gases mixing in ventilation or reactions scaling from test tubes to industrial reactors. The longevity of this law lies in its simplicity and adaptability.

In the context of our lab scenario, the ideal gas law recalibrates our understanding of concentration and ventilation. Specifically, it allows us to express the concentration of a contaminant as part of the room's air mixture and defines the relationship between mass, volume, and mole fraction. Through the equation \(y_A = \frac{\dot{m}_A}{k \dot{V}_{\text{air}}} \frac{RT}{M_A P}\), the mole fraction—an indication of how 'diluted' our problematic 'roommate' is—gives a clearer picture of air quality. Always remember, this equation stands as a testament to the power of understanding basic physical relationships and applying them to complex systems.
Permissible Exposure Level Calculation
Determining the permissible exposure level (PEL) is like setting a strict curfew for the concentration of hazardous substances in the workplace: it's the maximum amount allowed, no excuses. Calculating the correct ventilation rate to stay within this 'curfew' involves not just math but a careful consideration of working conditions, like the rate at which chemicals evaporate into the air and the mixing effectiveness of the ventilation system.

In cases where you have a value in parts per million (ppm), like the PEL for styrene, you can convert it to a mole fraction. Use this mole fraction to back-calculate the minimum amount of fresh air needed per hour to 'escort out' enough of the hazardous 'guest'. Yet this number isn't the be all and end all. There could still be 'hot spots' of high concentrations, or changes in room temperature could invite our problematic 'guest' to be more 'active'. Thus, understanding the PEL and how to calculate the required ventilation provides not just a figure for engineers to aim at but also highlights the dynamic nature of maintaining safety in chemical environments.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Phosgene (CCl, O) is a colorless gas that was used as an agent of chemical warfare in World War I. It has the odor of new-mown hay (which is a good warning if you know the smell of new-mown hay). Pete Brouillette, an innovative chemical engincering student, came up with what he believed was an effective new process that utilized phosgene as a starting material. He immediately set up a reactor and a system for analyzing the reaction mixture with a gas chromatograph. To calibrate the chromatograph (i.e., to determine its response to a known quantity of phosgene), he evacuated a 15.0 cm length of tubing with an outside diameter of \(0.635 \mathrm{cm}\) and a wall thickness of \(0.559 \mathrm{mm}\), and then connected the tube to the outlet valve of a cylinder containing pure phosgene. The idea was to crack the valve, fill the tube with phosgene, close the valve, feed the tube contents into the chromatograph, and observe the instrument response. What Pete hadn't thought about (among other things) was that the phosgene was stored in the cylinder at a pressure high enough for it to be a liquid. When he opened the cylinder valve, the liquid rapidly flowed into the tube and filled it. Now he was stuck with a tube full of liquid phosgene at a pressure the tube was not designed to support. Within a minute he was reminded of a tractor ride his father had once given him through a hayfield, and he knew that the phosgene was leaking. He quickly ran out of the lab, called campus security, and told them that a toxic leak had occurred, that the building had to be evacuated, and the tube removed and disposed of properly. Personnel in air masks shortly appeared, took care of the problem, and then began an investigation that is still continuing. (a) Show why one of the reasons phosgene was an effective weapon is that it would collect in low spots soldiers often mistakenly entered for protection. (b) Pete's intention was to let the tube equilibrate at room temperature ( \(23^{\circ} \mathrm{C}\) ) and atmospheric pressure. How many gram-moles of phosgene would have been contained in the sample fed to the chromatograph if his plan had worked? (c) The laboratory in which Pete was working had a volume of \(2200 \mathrm{ft}^{3}\), the specific gravity of liquid phosgene is \(1.37,\) and Pete had read somewhere that the maximum "safe" concentration of phosgene in air is \(0.1 \mathrm{ppm}\) \(\left(0.1 \times 10^{-6} \mathrm{mol} \mathrm{CCl}_{2} \mathrm{O} / \mathrm{mol}\) air) \right. Would the "safe" concentration have been exceeded if all the liquid phosgene in the tube had evaporated into the room? Even if the limit would not have been exceeded, give several reasons why the lab would still have been unsafe. (d) List several things Pete did (or failed to do) that made his experiment unnecessarily hazardous.

An ideal-gas mixture contains \(35 \%\) helium, \(20 \%\) methane, and \(45 \%\) nitrogen by volume at 2.00 atm absolute and \(90^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of each component, (b) the mass fraction of methane, (c) the average molecular weight of the gas, and (d) the density of the gas in \(\mathrm{kg} / \mathrm{m}^{3}\).

A gas consists of 20.0 mole \(\% \mathrm{CH}_{4}, 30.0 \% \mathrm{C}_{2} \mathrm{H}_{6},\) and \(50.0 \% \mathrm{C}_{2} \mathrm{H}_{4} .\) Ten kilograms of this gas is to be compressed to a pressure of 200 bar at \(90^{\circ} \mathrm{C}\). Using Kay's rule, estimate the final volume of the gas.

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?

A balloon \(20 \mathrm{m}\) in diameter is filled with helium at a gauge pressure of 2.0 atm. A man is standing in a basket suspended from the bottom of the balloon. A restraining cable attached to the basket kecps the balloon from rising. The balloon (not including the gas it contains), the basket, and the man have a combined mass of \(150 \mathrm{kg}\). The temperature is \(24^{\circ} \mathrm{C}\) that day, and the barometer reads \(760 \mathrm{mm} \mathrm{Hg}\) (a) Calculate the mass (kg) and weight (N) of the helium in the balloon. (b) How much force is exerted on the balloon by the restraining cable? (Recall: The buoyant force on a submerged object equals the weight of the fluid- -in this case, the air- -displaced by the object. Neglect the volume of the basket and its contents.) (c) Calculate the initial acceleration of the balloon when the restraining cable is released. (d) Why does the balloon eventually stop rising? What would you need to know to calculate the altitude at which it stops? (e) Suppose at its point of suspension in midair the balloon is heated, raising the temperature of the helium. What happens and why?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.