Ideal gas law problem with two cylinders

In summary,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.
  • #1
DottZakapa
239
17
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
 
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  • #2
That would be correct if both cylinders were filled with the same gas.
 
  • #3
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
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
mjc123 said:
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
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
mjc123 said:
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
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.
 
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  • #9
mjc123 said:
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 :)
 

1. What is the Ideal Gas Law?

The Ideal Gas Law is a mathematical equation that describes the relationship between the pressure, volume, temperature, and amount of gas in a system. It is written as PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature.

2. How do you solve an Ideal Gas Law problem with two cylinders?

To solve an Ideal Gas Law problem with two cylinders, you need to apply the law separately to each cylinder. This means that you will have two equations, one for each cylinder, and you can solve for any of the variables (pressure, volume, temperature, or number of moles) as long as you have values for the other three variables in each equation.

3. What are the assumptions of the Ideal Gas Law?

The Ideal Gas Law assumes that the gas particles have no volume, there are no intermolecular forces between the particles, and the particles are in constant, random motion. It also assumes that the gas is in a closed system and that the temperature is measured in Kelvin.

4. Can the Ideal Gas Law be applied to real gases?

The Ideal Gas Law is an approximation and is most accurate for gases at low pressures and high temperatures. Real gases deviate from ideal behavior at high pressures and low temperatures due to the presence of intermolecular forces and the volume of the gas particles. However, the Ideal Gas Law can still be used as an estimate for real gases in many situations.

5. How does the Ideal Gas Law relate to the Kinetic Molecular Theory?

The Ideal Gas Law is based on the Kinetic Molecular Theory, which states that gas particles are in constant, random motion and that their kinetic energy is directly proportional to the absolute temperature of the gas. This theory helps to explain the relationships between pressure, volume, temperature, and number of moles in the Ideal Gas Law.

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