Ideal gas law problem with two cylinders

  • Chemistry
  • Thread starter DottZakapa
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  • #1
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Homework Statement:
There are two cylinders A and B with the same capacity and at the same temperature. If the mass of hydrogen contained in cylinder A is double the mass of helium contained in cylinder B. How does the pressure vary? (Consider the ideal gases)
Relevant Equations:
ideal gas law
my answer will be ##P_1=2 P_2## but I have some doubts, if that is correct or not
 

Answers and Replies

  • #2
Borek
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That would be correct if both cylinders were filled with the same gas.
 
  • #3
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ok so the reasoning will be:
Concerning H, in the first container, the moles will be given by ## \frac {2 *He (g)} {\text{molar mass of} H_2 \text{(because comes as diatomic molecule} } ##
that is :

## \frac {2*He (g)}{2\frac {g} {mol} } = He\text{ mol } \rightarrow P_1=\frac {He \text{ mol} * R*T}{V} ##

Concerning the second, which contains He :

## \frac {He (g)}{4\frac {g} {mol} } = \frac {He}{4} mol \rightarrow P_2=\frac {He \text{ mol} * R*T}{4V} \rightarrow 4P_2=\frac {He \text{ mol} * R*T}{V}##

concluding, merging everything :

## P_1=4*P_2##

is it how it should be managed ?
 
  • #4
mjc123
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The answer is correct, but your working is a bit confusing. Is He a variable equal to the number of grams of helium in cylinder 2? Then it is correct that there are He moles of hydrogen in cylinder 1; but the notation is confusing. "He mol" could easily be understood as "the number of moles of He" rather than, as I think is meant, " a number of moles equal to the number of grams of He". Then the pressures would be out by a factor of 4, but the ratio would still be correct. It would be better to use e.g. mHe for the mass of helium - or a variable like x; mass of helium = x g. Then P1 = xRT/V and P2 = xRT/4V.
 
  • #5
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The answer is correct, but your working is a bit confusing. Is He a variable equal to the number of grams of helium in cylinder 2? Then it is correct that there are He moles of hydrogen in cylinder 1; but the notation is confusing. "He mol" could easily be understood as "the number of moles of He" rather than, as I think is meant, " a number of moles equal to the number of grams of He". Then the pressures would be out by a factor of 4, but the ratio would still be correct. It would be better to use e.g. mHe for the mass of helium - or a variable like x; mass of helium = x g. Then P1 = xRT/V and P2 = xRT/4V.
The problem says:
"the mass of hydrogen contained in cylinder A is double the mass of helium contained in cylinder B"
so, if :
PV=nRT
where n are the moles of gas.
If n is equal to the mass of the gas divided by the Molecular Mass of the gas, then, knowing that in cylinder A the mass of Hydrogen is twice the mass of Helium, in the above mentioned mass of the gas I should write twice the mass of Helium (g)divided by the Molecular mass of Hydrogen (## \frac g {mol}##).
that is :
##n_a= \frac {H (g)}{H \frac g {mol}}= \frac {2He (g)}{H\frac {g} {mol} } = \frac {2He(g)}{2 \frac g {mol}} ##
This is for cylinder A.
For cylinder B always mass of gas (Helium) divided by molecular mass of Helium.
##n_b= \frac {He (g)}{He \frac g {mol}}= \frac {He (g)}{4\frac {g} {mol} } = \frac {He(g)}{4\frac g {mol}} ##
correct?
 
  • #6
mjc123
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Your method is correct. It is your notation that I'm concerned about. I don't like "He (g)" for "the mass of helium in grams", because it's potentially confusing, especially when you go on to write "He mol", not meaning "the number of moles of helium". Further confusion is caused by writing H when you should write H2. The molar mass of H is 1 g/mol, not 2. But you don't have H, you have H2. So your calculation is right, but the way you write it is confusing.
 
  • #7
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Your method is correct. It is your notation that I'm concerned about. I don't like "He (g)" for "the mass of helium in grams", because it's potentially confusing, especially when you go on to write "He mol", not meaning "the number of moles of helium". Further confusion is caused by writing H when you should write H2. The molar mass of H is 1 g/mol, not 2. But you don't have H, you have H2. So your calculation is right, but the way you write it is confusing.
it was an MCQ, but, just in case I will have it in a written test, would you Kindly explain how should I write? Yes in H it was a typo.
 
  • #8
mjc123
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Let mass of helium be x g; MW of He = 4. No. of moles helium = x/4. PB = nBRT/V = xRT/4V.
Mass of hydrogen is twice mass of helium = 2x g. MW of H2 = 2. No. of moles hydrogen = 2x/2 =x.
PA = nART/V = xRT/V = 4PB.
 
  • #9
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Let mass of helium be x g; MW of He = 4. No. of moles helium = x/4. PB = nBRT/V = xRT/4V.
Mass of hydrogen is twice mass of helium = 2x g. MW of H2 = 2. No. of moles hydrogen = 2x/2 =x.
PA = nART/V = xRT/V = 4PB.
thank you :)
 

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