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Chapter 5: Problem 24

Many references give the specific gravity of gases with reference to air. For example, the specific gravity of carbon dioxide is 1.53 relative to air at the same temperature and pressure. Show that this value is correct as long as the ideal-gas equation of state applies.

Short Answer

Expert verified
The calculated specific gravity of Carbon Dioxide relative to air under ideal conditions is approximately 1.52, which is close to the given figure of 1.53. Thus, within the assumptions of the ideal gas law, the specific gravity of 1.53 is substantially correct.

Step by step solution

01

Understanding Molar Mass

First, we need to calculate the molar masses of both CO2 and air. As per the periodic table, the molar mass of Carbon (C) is roughly 12 g/mole and Oxygen (O) is about 16 g/mole. Therefore, as CO2 has one atom of Carbon and two of Oxygen, the molar mass of CO2 is \(12g/mole + 2 * 16g/mole = 44g/mole\). Air is primarily composed of Nitrogen (N2) about 78%, Oxygen (O2) about 21%, and Argon (Ar) about 1%. The molar mass of these gases are approximately 28, 32 and 40 g/mole respectively, weighted by their percentages, we find the average molar mass of air to be roughly \( (0.78*28g/mole) + (0.21*32g/mole) + (0.01*40g/mole) = 28.97g/mole \).
02

Calculating Specific Gravity

Knowing that the specific gravity is the ratio of the two densities, but under ideal gas conditions, density is proportional to the molar mass, we could thus compare the molar masses instead since the volume, temperature, and pressure are constants. Therefore, we have the specific gravity as the ratio of the molar mass of gas to the molar mass of air. Our equation is thus: Specific Gravity \(= \frac{Molar mass of CO2}{Molar mass of air} = \frac{44g/mole}{28.97g/mole}\).
03

Checking the Result

Now, it's just a simple division problem. Our answer after calculation approximates close to 1.52, which is close to 1.53. You may point out that our value does not exactly match the provided specific gravity (1.53). This small discrepancy can be attributed to approximations in both our calculation and the provided value. The processes a gas undertakes in real world might not strictly adhere to the ideal gas law, and air is not an ideal gas. Therefore the answer provided may be a rounded-off laboratory value. We can thus say however, that the statement that the specific gravity of CO2 is 1.53 relative to air with reference to the ideal gas law, is substantially correct.

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Key Concepts

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

Molar Mass Calculation
To grasp the concept of molar mass calculation, think of it as finding the weight of one mole of a substance, similar to counting pennies in a dollar. A mole is simply a way of expressing a quantity of atoms or molecules in a specific volume. Let's break down how we calculate it using carbon dioxide (CO2) as our example.
  • Carbon (C) has a molar mass of approximately 12 grams per mole (g/mol).
  • Oxygen (O) has a molar mass of roughly 16 g/mol.
Carbon dioxide contains one carbon atom and two oxygen atoms, hence its total molar mass is calculated as follows:
\[ 12 \ g/mol + 2 \times 16 \ g/mol = 44 \ g/mol \]
This means every mole of CO2 weighs about 44 grams. Similarly, air is composed mainly of Nitrogen (N2), Oxygen (O2), and Argon (Ar). To find air's average molar mass, we weigh the contribution of each gas:
  • Nitrogen, 78% present, has a molar mass of 28 g/mol.
  • Oxygen, 21% present, has a molar mass of 32 g/mol.
  • Argon, 1% present, has a molar mass of 40 g/mol.
The average molar mass of air is approximately:
\[ (0.78 \times 28) + (0.21 \times 32) + (0.01 \times 40) = 28.97 \ g/mol \]
This calculation helps us when comparing the gas densities.
Ideal Gas Law
The Ideal Gas Law is central to understanding how gases behave under various conditions. It combines several gas laws to create a simple equation that provides an insight into the state of a gas. The formula is represented as:
\[ PV = nRT \]
where:
  • \( P \) is the pressure of the gas,
  • \( V \) is the volume,
  • \( n \) is the amount of substance in moles,
  • \( R \) is the ideal gas constant,
  • \( T \) is the temperature in Kelvin.
This equation tells us how changing one aspect, like pressure or temperature, affects the others when the gas behaves ideally. Understanding this is crucial when comparing the specific gravity of gases because it implies that under ideal conditions, the volume, temperature, and pressure are consistent. This consistency allows us to use the ratio of molar masses as a proxy for comparing densities, simplifying the calculation of specific gravity.
Density Comparison
In the realm of gases, density plays a key role in determining how substances will interact with one another under given conditions. Specific gravity is essentially a density comparison between a gas of interest and a reference gas, which is often air.
Since the state of a gas is influenced by temperature and pressure, these elements must remain constant when performing comparisons. What's fascinating is that under ideal gas conditions, a gas's density is directly proportional to its molar mass. This brings us to a neat trick: instead of measuring density, we can simply use the molar masses when these conditions are stable.
Thus, the specific gravity of carbon dioxide relative to air can be calculated by:
  • Taking the molar mass of CO2: 44 g/mol,
  • Dividing by the molar mass of air: 28.97 g/mol.
This gives us a specific gravity value close to 1.52. The closeness of this number to 1.53 demonstrates the practicality of this approach with slight deviations due to real-world approximations. By understanding these concepts, you can confidently compare the behaviors of various gases under similar conditions.

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Most popular questions from this chapter

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?

Steam reforming is an important technology for converting refined natural gas, which we take here to be methane, into a synthesis gas that can be used to produce a varicty of other chemical compounds. For example, consider a reformer to which natural gas and steam are fed in a ratio of 3.5 moles of steam per mole of methane. The reformer operates at 18 atm, and the reaction products leave the reformer in chemical equilibrium at \(875^{\circ} \mathrm{C}\). The steam reforming reaction is $$\mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CO}+3 \mathrm{H}_{2}$$ and the water-gas shift reaction also occurs in the reformer. $$\mathrm{CO}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CO}_{2}+\mathrm{H}_{2}$$ The equilibrium constants for these two reactions are given by the expressions At \(875^{\circ} \mathrm{C}, K_{\mathrm{R}}=872.9 \mathrm{atm}^{2}\) and \(K \mathrm{w} \mathrm{G}=0.2482 .\) The process is to produce \(100.0 \mathrm{kmol} / \mathrm{h}\) of hydrogen. Calculate the feed rates (kmol/h) of methane and steam and the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{min}\right)\) of gas leaving the reformer.

A stream of liquid \(n\) -pentane flows at a rate of \(50.4 \mathrm{L} / \mathrm{min}\) into a heating chamber, where it evaporates into a stream of air \(15 \%\) in excess of the amount needed to burn the pentane completely. The temperature and gauge pressure of the entering air are \(336 \mathrm{K}\) and \(208.6 \mathrm{kPa}\). The pentane-laden heated gas flows into a combustion furnace in which a fraction of the pentane is burned. The product gas, which contains all of the unreacted pentane and no \(\mathrm{CO},\) goes to a condenser in which both the water formed in the furnace and the unreacted pentane are liquefied. The uncondensed gas leaves the condenser at \(275 \mathrm{K}\) and 1 atm absolute. The liquid condensate is separated into its components, and the flow rate of the pentane is measured and found to be \(3.175 \mathrm{kg} / \mathrm{min}\). (a) Calculate the fractional conversion of pentane achieved in the furnace and the volumetric flow rates ( \(\mathrm{L} / \mathrm{min}\) ) of the feed air, the gas leaving the condenser, and the liquid condensate before its components are separated. (b) Sketch the apparatus that could have been used to separate the pentane and water in the condensate. Hint: Remember that pentane is a hydrocarbon and recall what is said about oil (hydrocarbons) and water.

A stream of oxygen enters a compressor at \(298 \mathrm{K}\) and 1.00 atm at a rate of \(127 \mathrm{m}^{3} / \mathrm{h}\) and is compressed to \(358 \mathrm{K}\) and 1000 atm. Estimate the volumetric flow rate of compressed \(\mathrm{O}_{2},\) using the compressibility-factor equation of state.

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.
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