/*! 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 61 An open-end mercury manometer is... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An open-end mercury manometer is to be used to measure the pressure in an apparatus containing a vapor that reacts with mercury. A 10 -cm layer of silicon oil \((\mathrm{SG}=0.92)\) is placed on top of the mercury in the arm attached to the apparatus. Atmospheric pressure is \(765\) \(\mathrm{mm}\) Hg. (a) If the level of mercury in the open end is 365 mm below the mercury level in the other arm, what is the pressure (mm Hg) in the apparatus? (b) When the instrumentation specialist was deciding on a liquid to put in the manometer, she listed several properties the fluid should have and eventually selected silicon oil. What might the listed properties have been?

Short Answer

Expert verified
a) For the pressure in the apparatus, perform the calculation as described in step 2 to get value of final pressure in mm Hg. b) Desired properties that might be listed for the liquid include: chemical stability (does not react with the system), appropriate specific gravity (makes pressure differences measurable), non-volatility (doesn't evaporate easily), cost effectiveness, and stability in a wide range of operational temperatures.

Step by step solution

01

Understand Pressure Difference

Firstly, it is important to understand that the pressure difference between two points in a static fluid column equals the density of the fluid multiplied by gravity and the height of the column. Pressure difference = \(\mathrm{SG} \times g \times h\) where the value of g is 9.81 m/s² as an approximation for the acceleration due to gravity and h is height of fluid column (difference in mercury level). The pressure exerted by the silicon oil is calculated by using its known specific gravity (0.92 in this case). The pressure at station point in the apparatus then equals to atmospheric pressure plus the pressure caused by the silicon oil layer minus the pressure corresponding to the mercury level difference.
02

Calculate Pressure in Apparatus

To calculate pressure in the apparatus (in mm Hg), use the following method:- Convert given atmospheric pressure into cm Hg by dividing by 10 (since 1 cm = 10 mm), let's denote as \( P_{atm} \).- Convert mercury level difference into cm by dividing by 10 (let's denote as \( h_{Hg} \)).- Use above Step 1 formulation, substitute SG = 0.92 and h = 10 cm (silicon oil height), then convert resulting pressure in mm Hg (let's denote as \( P_{oil} \)).- Then add \( P_{atm} \) and \( P_{oil} \) and subtract \( h_{Hg} \) to get pressure in the apparatus \( P_{app} = P_{atm} + P_{oil} - h_{Hg} \).
03

Reflect on Required Properties of Manometer Fluid

In part (b), there is no explicit calculation. Reflecting on why silicon oil was selected over other fluids can indicate the desired properties for a fluid to be used in manometer. Important properties could be chemical inertness (especially to the operating vapor), suitable specific gravity, non-volatile nature, low cost, and stability in a wide range of temperatures.

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.

Specific Gravity
Specific gravity is a fundamental concept when dealing with fluid dynamics, particularly in the context of manometer pressure calculations. It is defined as the ratio of the density of a fluid to the density of a reference substance, usually water for liquids. This dimensionless quantity allows us to compare the density of a fluid to water without worrying about units. For instance, the specific gravity (\( \text{SG} \)) of silicon oil is given as 0.92, which means silicon oil is less dense than water, as it is only 92% as dense as water at a specified temperature (typically 4°C for water).

In our exercise, specific gravity serves as a crucial factor in determining the pressure exerted by the silicon oil layer in the manometer. Since we know the specific gravity of silicon oil, we can calculate its density relative to mercury and subsequently determine the pressure contribution to the overall pressure in the apparatus. Without understanding specific gravity, it would be difficult to relate the measured heights in a manometer to actual pressures.
Static Fluid Column
The concept of a static fluid column is integral to understanding the operation of manometers. It refers to a vertical column of fluid that is at rest, with no fluid moving up or down. When the fluid in such a column is in hydrostatic equilibrium, the pressure at every point in the fluid is determined by the weight of the fluid above it. This is because the pressure at a point in a fluid at rest must support the weight of the fluid column directly above it.

The relationship between pressure difference and a static fluid column is given by the equation \( \text{Pressure difference} = \text{SG} \times g \times h \), where \( g \) represents the acceleration due to gravity and \( h \) the height of the fluid column. In the provided exercise, the static fluid column is created by the height difference of mercury in the two arms of the manometer and the layer of silicon oil. Calculating the pressure in the apparatus heavily relies on the accurate measurement of this static fluid column and requires understanding the contributing factors, such as the specific gravity of the fluid and the local acceleration due to gravity.
Manometry Properties
Manometry properties encompass the characteristics a fluid must possess to function effectively within a manometer. These properties include chemical compatibility, specific gravity, non-volatility, and thermal stability. Chemical compatibility means the manometric fluid should not react with the substances it comes into contact with; otherwise, it may alter the pressure reading or damage the manometer. Given that the vapor in the apparatus reacts with mercury, silicon oil's non-reactivity makes it a suitable choice.

The specific gravity of the manometric fluid influences sensitivity and range of measurements. Silicon oil's specific gravity of 0.92 provides an appropriate balance between sensitivity and the ability to measure the required pressure range. Non-volatility ensures that the fluid's volume and, consequently, the height of the fluid column remain constant over time. Thermal stability means that the fluid properties do not significantly change with temperature variations, thereby providing consistent readings. The selection of silicon oil for the manometer in the exercise may be attributed to such ideal manometric properties, underscoring its ability to deliver accurate and reliable pressure measurements in the experimental setup.

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

A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5\% and 15\%. A mixture containing 9.0 mole\% methane in air flowing at a rate of 7.00 \(\times 10^{2} \mathrm{kg} / \mathrm{h}\) is to be diluted with pure air to reduce the methane concentration to the lower flammability limit. Calculate the required flow rate of air in mol/h and the percent by mass of oxygen in the product gas. (Note: Air may be taken to consist of \(\left.21 \text { mole } \% \mathrm{O}_{2} \text { and } 79 \% \mathrm{N}_{2} \text { and to have an average molecular weight of } 29.0 .\right)\)

A gas stream contains 18.0 mole \(\%\) hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is reduced and some of the hexane is liquefied. The hexane mole fraction in the gas stream leaving the condenser is \(0.0500 .\) Liquid hexane condensate is recovered at a rate of \(1.50 \mathrm{L} / \mathrm{min}\). (a) What is the flow rate of the gas stream leaving the condenser in mol/min? (Hint: First calculate the molar flow rate of the condensate and note that the rates at which \(C_{6} H_{14}\) and \(N_{2}\) enter the unit must equal the total rates at which they leave in the two exit streams.) (b) What percentage of the hexane entering the condenser is recovered as a liquid? (c) Suggest a change you could make in the process operating conditions to increase the percentage recovery of hexane. What would be the downside?

An inclined manometer is a useful device for measuring small pressure differences. The formula given in Section 3.4 for the pressure difference in terms of the liquid-level difference \(h\) remains valid, but while \(h\) would be small and difficult to read for a small pressure drop if the manometer were vertical, \(L\) can be made quite large for the same pressure drop by making the angle of the inclination, \(\theta,\) small. (a) Derive a formula for \(h\) in terms of \(L\) and \(\theta\) (b) Suppose the manometer fluid is water, the process fluid is a gas, the inclination of the manometer is \(\theta=15^{\circ},\) and a reading \(L=8.7 \mathrm{cm}\) is obtained. What is the pressure difference between points? and?? (c) The formula you derived in Part (a) would not work if the process fluid were a liquid instead of a gas. Give one definite reason and another possible reason.

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spilled on the arm connected to the line during a laboratory renovation, it is impossible to see the level of the manometer fluid in this arm. During a period when the gas supply is connected to the line but there is no gas flow, a Bourdon gauge connected to the line downstream from the manometer gives a reading of 7.5 psig. The level of mercury in the open arm is \(900 \mathrm{mm}\) above the lowest part of the manometer. (a) When the gas is not flowing, the pressure is the same everywhere in the pipe. How high above the bottom of the manometer would the mercury be in the arm connected to the pipe? (b) When gas is flowing, the mercury level in the visible arm drops by \(25 \mathrm{mm}\). What is the gas pressure (psig) at this moment?

The reaction \(A \rightarrow B\) is carried out in a laboratory reactor. According to a published article the concentration of A should vary with time as follows: \(C_{\mathrm{A}}=C_{\mathrm{A} 0} \exp (-k t)\) where \(C_{\mathrm{A} 0}\) is the initial concentration of \(\mathrm{A}\) in the reactor and \(k\) is a constant. (a) If \(C_{\mathrm{A}}\) and \(C_{\mathrm{A} 0}\) are in \(\mathrm{Ib}-\) moles \(/ \mathrm{ft}^{3}\) and \(t\) is in minutes, what are the units of \(k ?\) (b) The following data are taken for \(C_{\mathrm{A}}(t):\) $$\begin{array}{cc}\hline t(\min ) & C_{\mathrm{A}}\left(\mathrm{lb}-\mathrm{mole} / \mathrm{ft}^{3}\right) \\\\\hline 0.5 & 1.02 \\\1.0 & 0.84 \\\1.5 & 0.69 \\\2.0 & 0.56 \\\3.0 & 0.38 \\\ 5.0 & 0.17 \\\10.0 & 0.02 \\\\\hline\end{array}$$ Verify the proposed rate law graphically (first determine what plot should yield a straight line), and calculate \(C_{\mathrm{A} 0}\) and \(k\) (c) Convert the formula with the calculated constants included to an expression for the molarity of A in the reaction mixture in terms of \(t\) (seconds). Calculate the molarity at \(t=265 \mathrm{s}\).

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.