Number of balloons that can be filled from a gas cylinder

[songoku]

Homework Statement

A gas cylinder contains 4.00 × 10^4 cm^3 of hydrogen at a pressure of 2.50 × 10^7 Pa and a temperature of 290 K. The cylinder is to be used to fill balloons. Each balloon, when filled, contains 7.24 × 10^3 cm^3 of hydrogen at a pressure of 1.85 × 10^5 Pa and a temperature of 290 K. Calculate the number of balloons that can be filled from the cylinder

Relevant Equations

PV = nRT

$\frac{P_{cylinder}.V_{cylinder}}{P_{balloon}.V_{balloon}}=\frac{n_{cylinder}.R.T}{n_{balloon}.R.T}$

$\frac{n_{cylinder}}{n_{balloon}}=\frac{4 \times 10^4 \times 2.5 \times 10^7}{7.24 \times 10^3 \times 1.85 \times 10^5} \approx 746$

But the teacher said I should take the volume of gas cylinder into consideration. If I did, then the answer should be smaller.

I think I have used volume of gas cylinder into consideration in my working. What is my mistake?

Thanks

 

[haruspex]

Homework Statement:: A gas cylinder contains 4.00 × 10^4 cm^3 of hydrogen at a pressure of 2.50 × 10^7 Pa and a temperature of 290 K. The cylinder is to be used to fill balloons. Each balloon, when filled, contains 7.24 × 10^3 cm^3 of hydrogen at a pressure of 1.85 × 10^5 Pa and a temperature of 290 K. Calculate the number of balloons that can be filled from the cylinder
Relevant Equations:: PV = nRT

$\frac{P_{cylinder}.V_{cylinder}}{P_{balloon}.V_{balloon}}=\frac{n_{cylinder}.R.T}{n_{balloon}.R.T}$

$\frac{n_{cylinder}}{n_{balloon}}=\frac{4 \times 10^4 \times 2.5 \times 10^7}{7.24 \times 10^3 \times 1.85 \times 10^5} \approx 746$

But the teacher said I should take the volume of gas cylinder into consideration. If I did, then the answer should be smaller.

I think I have used volume of gas cylinder into consideration in my working. What is my mistake?

Thanks

What are the differences in the state of the cylinder's gas between the initial state and after filling one balloon?

 

[songoku]

What are the differences in the state of the cylinder's gas between the initial state and after filling one balloon?

The pressure and number of moles of the gas?

 

[haruspex]

The pressure and number of moles of the gas?

Right, the pressure is lower. How might that become a problem?
(Are these gauge pressures or absolute?)

 

[songoku]

Right, the pressure is lower. How might that become a problem?

It can make the cylinder "collapse" because the inner pressure is lower than outer pressure (atmospheric pressure)?

(Are these gauge pressures or absolute?)

Absolute pressure?

Thanks

 

[haruspex]

It can make the cylinder "collapse" because the inner pressure is lower than outer pressure (atmospheric pressure)?

Before that happens it will be lower than...?

Absolute pressure?

I'm asking you, but since it gives the balloon pressure as about 1.85 atmospheres I would guess they are absolute.

 

[songoku]

Before that happens it will be lower than...?

Lower than pressure of the balloon. So when the pressure of the gas cylinder is the same as balloon, the gas cylinder can not be used anymore to fill the balloon?

Thanks

 

[haruspex]

Lower than pressure of the balloon. So when the pressure of the gas cylinder is the same as balloon, the gas cylinder can not be used anymore to fill the balloon?

Thanks

Of course.

 

[songoku]

Thank you very much for the help haruspex

 

[songoku]

Sorry I have another question

Why the gas cylinder and balloon do not "explode" even though they have pressure higher than surrounding pressure (atmospheric pressure)?

Thanks

 

[haruspex]

Sorry I have another question

Why the gas cylinder and balloon do not "explode" even though they have pressure higher than surrounding pressure (atmospheric pressure)?

Thanks

Because the container is strong enough to withstand the pressure difference. It will be under tension.

 

[songoku]

Because the container is strong enough to withstand the pressure difference. It will be under tension.

Let's take balloon as example. There will be net pressure directed outwards because the inside pressure is higher than outside pressure. From $P=\frac{F}{A}$, there will be force acting on the balloon, which is also directed outwards.

1. If we consider the balloon is perfect sphere, then the forces on all part of the sphere will cancel each other so the resultant force due to the pressure difference is zero?

2. Pressure is scalar quantity but why it seems that pressure has direction? Pressure inside a container by a gas is "outwards", pressure by atmospheric is "inwards", or pressure on top of a cylinder immersed fully in water is "downwards"?

Thanks

 

[haruspex]

Let's take balloon as example. There will be net pressure directed outwards because the inside pressure is higher than outside pressure. From $P=\frac{F}{A}$, there will be force acting on the balloon, which is also directed outwards.

1. If we consider the balloon is perfect sphere, then the forces on all part of the sphere will cancel each other so the resultant force due to the pressure difference is zero?

2. Pressure is scalar quantity but why it seems that pressure has direction? Pressure inside a container by a gas is "outwards", pressure by atmospheric is "inwards", or pressure on top of a cylinder immersed fully in water is "downwards"?

Thanks

Pressure is scalar, but the force that results on an area element $\vec{dA}$ is $P\vec{dA}$.
The net force on the balloon is zero, unless you leave a small opening and watch the balloon zoom off. But the force on each area element has to be balanced by the tension in the surface. Since each area element is a small cap, there is a net radial force from this tension.
Have you worked with bubbles? Let the surface tension (force per unit length) be $\sigma$ and the internal pressure be P. Consider a hemisphere radius r. The net force from the gas is $P\pi r^2$. This is balanced by the tension around the rim, $2\pi r\sigma$, so $P=\frac{2\sigma}r$. That is for a single-walled bubble.

 

[songoku]

Pressure is scalar, but the force that results on an area element $\vec{dA}$ is $P\vec{dA}$.
The net force on the balloon is zero, unless you leave a small opening and watch the balloon zoom off. But the force on each area element has to be balanced by the tension in the surface. Since each area element is a small cap, there is a net radial force from this tension.
Have you worked with bubbles? Let the surface tension (force per unit length) be $\sigma$ and the internal pressure be P. Consider a hemisphere radius r. The net force from the gas is $P\pi r^2$. This is balanced by the tension around the rim, $2\pi r\sigma$, so $P=\frac{2\sigma}r$. That is for a single-walled bubble.

Thank you very much haruspex