Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Win,odd Dhamnekar on November 05, 2020, 10:19:37 AM
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Hello,
Here are the questions with unknown answers.
1) At 1 bar, the boiling point of water is 372.78 K. At this temperature and pressure, the density of liquid water is 958.66 kg/m3 and that of gaseous water is 0.59021 kg/m3. What are the molar volumes. in m3 mol-1 of liquid and gaseous water at this temperature and pressure? in Liters/mol?
2) Refer to the answer to 1) question. Assuming that a water molecule excludes the other water molecules from a cubic region centered on itself, estimate the average distance between nearest-neighbor water molecules in the liquid and in the gas.
3)Calculate the molar volume and gaseous water at 1 bar and 372.78 K from ideal gas equation. What is the error, expressed as a percentage of the value, you calculated in question 1).
4) At 372.78 K, the virial coefficient B* for water is -1.487 × 10-7 Pa-1. Calculate the molar volume of gaseous water at 1 bar and 372.78 K from the virial equation: Z=PV̅/RT= 1+B*P. What is the error, expressed as a percentage of the value, you computed in question 1).
5) Compute the molar volume of gaseous water at 1 bar and 372.78 K from van der Waals' equation. The van der Waals' parameters for water are a=5.537 bar L2 mol-1 and b=0.0305 L mol-1. What is the error, expressed as the percentage of the value, you computed in question 1)?
6)What are your comments on the results in questions 3,4 and 5? At this temperature, would you expect the accuracy to increase or decrease at lower pressure?
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Please read the forum rules. You must show work to receive help.
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Answer to 1)Molar volume of liquid water is 0.000017275 m³/mol at 1 bar and 372.78 K.
Molar volume of gaseous water is 0.000013229 m3/mol.
Answer to 3) Ideal gas equation is PV=nRT. So, V=(8.314472J/(mol.K) ⨉ 372.78 K)/1 bar=30.9947 litres/mol. By the way, are these answers correct?
I am working on other questions. If i found answers to other questions, I shall post them here.
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Answer to 1)Molar volume of liquid water is 0.000017275 m³/mol at 1 bar and 372.78 K.
Molar volume of gaseous water is 0.000013229 m3/mol.
That would mean molar volume of the gas is substantially lower than the molar volume of the liquid, doesn't make much sense.
Answer to 3) Ideal gas equation is PV=nRT. So, V=(8.314472J/(mol.K) ⨉ 372.78 K)/1 bar=30.9947 litres/mol.
That looks reasonably close to what I got. Note how this result is orders of magnitude different from the one you calculated in 1.
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Answer to 3) The computed value of the molar volume of gaseous water is 0.000013229 m3/mol in question 1) and the computed value of the molar volume of gaseous water is 0.0309947 m3/mol in question 3) So the error is 234294% of 0.000013229 m3/mol.
Answer to 4) Z= 0.98513. and V̅=30.5338 litres /mol. So, the error is 230809% of 0.000013229 m3/mol.
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The computed value of the molar volume of gaseous water is 0.000013229 m3/mol in question
It is not, you are making some error in the way you are approaching the problem. Show how you got your result.
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My revised answer to question 1) is 0.0309946887216 m3/mol.
My revised answer to question 3). We don't find any error in the computed value in question 3)
My revised answer to question 4). The error, expressed as a percentage of the value. computed in question 1) is 1.487%.
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My revised answer to question 1) is 0.0309946887216 m3/mol.
It is still wrong. Please show how you are getting your values from the data given.
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1m3=1000 l. Molar volume of gaseous water is V=(1 mol* 8.314472J/(mol*K)*372.78 K)/1 bar=3099.4688722 J/bar [Recall 1J/bar=0.01 litre] So, V=30.994688722 Litres
Now, density of a gaseous water is 0.00059021 kg/0.001m3. For 1 liter of gaseous water, density is 0.001m3. So, for 30.994688722 litres, density is 0.030994688722m3/mol
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1m3=1000 l. Molar volume of gaseous water is V=(1 mol* 8.314472J/(mol*K)*372.78 K)/1 bar=3099.4688722 J/bar [Recall 1J/bar=0.01 litre] So, V=30.994688722 Litres
That's OK, but this is answer to 3, not to 1.
Now, density of a gaseous water is 0.00059021 kg/0.001m3
A bit strange use of units/numbers, but OK.
For 1 liter of gaseous water, density is 0.001m3
No, that's a volume, not density.
So, for 30.994688722 litres, density is 0.030994688722m3/mol
Sorry, it doesn't make any sense. Density is not measured in m3/mol, that's a molar volume unit. Plus, density doesn't depend on the volume which you seem to be suggesting, density is an intensive property.
In 1. you are asked to calculate the molar volume from the experimental data, using given density of 0.59021 kg/m3. How is the molar volume defined?
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Molar volume[tex]=\frac{ Automic weight \times molar mass}{density(\rho)}[/tex]
I don't know the automic weight of H2O(liquid and vapour). But i know the automic mass of hydrogen[itex](1.6736e-24 \text{g})[/itex] and oxygen[itex](2.65673e-23\text{g})[/itex]
So, how to compute the automic weight of H2O(liquid and vapour)?
And also how to compute molar volumes of liquid water and gaseous water at 1 bar pressure and 372.78 K temperature?
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No idea what is automic weight, where did you got this definition from?
https://en.wikipedia.org/wiki/Molar_volume
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Automic weights of H2O (liquid and vapour) is not given in your link.
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It was not intended to contain information on made up or misread/misnamed quantity, it was intended to help you calculate the molar volume of the water from the given, experimental data.
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Molar volume of liquid water is 0.00001879 m3/mol.
Molar volume of gaseous water is 0.030589 m3/mol using Charles' law for temperature- volume relationship.
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Van der waals' equation is [tex]\big( 1 bar+\frac{5.537 bar*l^2*mol}{V^2}\big)*(V-0.0305 l)=3099.4688722 J[/tex]
We can express [itex]V= Xlitres[/itex] So, how to answer question 5)?
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Molar volume of gaseous water is 0.030589 m3/mol using Charles' law for temperature- volume relationship.
You are not asked to use Charles' law for 1.
I pointed you in the right direction. You know the mass, you know the density, that's enough to calculate molar volume using a correct definition. If you prefer to ignore the advice and waste time - so be it.
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I didn't know that the molar mass of liquid water and gaseous water are one and the same that is 18.02 g/mol.
that's why i used charles' law.
Any way, how to answer question 2) and 5) ?
Regarding question 2), i don't have any idea. But while answering question 5), I faced difficulty in inputing the equation into equation solver of HP 50g calculator. It said " Invalid object type". How to solve it manually for V in easy way?
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My attempt to answer question 5)
Final Van der Waals' equation is [tex]100x_J+ \frac{553.7_{J·Mol}}{x}-3.05_J -\frac{16.88785_{J·Mol}}{x^2}=3099.4688722_J[/tex]
Where x= Volume in liters. [itex]1_{bar·liter}=100_J.[/itex] Now, how to find x?
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Answers 1) The molar volume of liquid water is [itex]0.0000188m^3/mol[/itex] viz. 0.0188 liters/mol.
The molar volume of gaseous water is [itex]0.03053150574 m^3/mol[/itex] viz. 30.53150574 liters/mol.
2) The average distance between nearest-neighbor water molecules in the liquid phase is 315 pm and in the gaseous phase 3701 pm.
3) From ideal gas equation, the computed value of molar volume of gaseous water is [itex]0.030994688722 m^3/mol [/itex] The error, expressed as a percentage of the computed value in question 1 is [itex]\frac{0.030994688722 m^3/mol}{0.03053150574 m^3/mol}=101.517[/itex]%
4) The molar volume of gaseous water at 1 bar and 372,78 K is [itex]0.0305338 m^3/mol[/itex] viz.30.5338 liters/mol. The error, expressed as a percentage of the computed value in question 1 is [itex]\frac{0.0305338m^3/mol}{0.03053150574m^3/mol}=100.075144[/itex]%
5) The molar volume of gaseous water at 1 bar and 372.78 K from van der waals' equation is [itex]0.000140306968475 m^3/mol[/itex] viz.0.140306968475 liters/mol.
The error, expressed as a percentage of the computed value in question 1 is [itex]\frac{0.000140306968475m^3/mol}{0.03053150574 m^3/mol}=0.459548[/itex]%
6)What are your comments on the results of questions 3,4 and 5? At this temperature, would you expect the accuracy to increase or decrease at lower pressure?