# Computation of the ratio of barometric pressure

#### Dhamnekar Winod

##### Active member
Hi,

The Homestake gold mine near Lead,South Dakota is excavated to 8000 feet below the surface. Lead is nearly a mile high; the bottom of the Homestake is about 900 m below sea level. Nearby custer peak is about 2100 m above sea level.

What is the ratio of barometric pressure on the top of the custer peak to the barometric pressure at the bottom of Homestake?
(Assume that the entire atmosphere is at 300 K and that it behaves as a single ideal gas whose molar mass is 29.

How to answer this question? The known barometric formula is $P=P_0 e^{-\frac{m_gh}{kT}}$ where T=temperature, k=Boltzmann's constant, $m_g$=mass of an individual atmospheric molecule.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
We can fill in 2100 m respectively -900 m into the formula to find the pressures. And then divide them to find the ratio.

Alternatively, we can first express the ratio as a formula and simplify it, and then evaluate it.
That is, let $h_1=2100\, m$ and $h_2=-900\,m$.
Then the requested ratio is:
$$\frac{P_0 e^{-\frac{m_g h_1}{kT}}}{P_0 e^{-\frac{m_g h_2}{kT}}}=e^{\frac{m_g h_2}{kT} - \frac{m_g h_1}{kT}}=e^{\frac{m_g}{kT}(h_2 - h_1)}$$

#### Dhamnekar Winod

##### Active member
We can fill in 2100 m respectively -900 m into the formula to find the pressures. And then divide them to find the ratio.

Alternatively, we can first express the ratio as a formula and simplify it, and then evaluate it.
That is, let $h_1=2100\, m$ and $h_2=-900\,m$.
Then the requested ratio is:
$$\frac{P_0 e^{-\frac{m_g h_1}{kT}}}{P_0 e^{-\frac{m_g h_2}{kT}}}=e^{\frac{m_g h_2}{kT} - \frac{m_g h_1}{kT}}=e^{\frac{m_g}{kT}(h_2 - h_1)}$$
Hi,
So, $m_g=\frac{29 g}{6.02214e23mol^{-1}}=4.8155e-23 g, e^{\frac{4.8155e-23 g(900 m-2100 m)}{\frac{1.38065e-23J}{K}300 K}}$= $e^{-0.013951577 s^2/m}$

Now, what is the final ratio? What does this answer mean to the readers?
My other conclusions from the information given in this question are

1) South Dakota is about 1538.4 m above sea level.

2)Custer Peak is about 561.6 m above South Dakota.

Are these conclusions correct?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So, $m_g=\frac{29 g}{6.02214e23mol^{-1}}=4.8155e-23 g, e^{\frac{4.8155e-23 g(900 m-2100 m)}{\frac{1.38065e-23J}{K}300 K}}$= $e^{-0.013951577 s^2/m}$
There is something wrong with the units.
The exponent should be dimensionless.

From wiki, it appears the correct formula is $P=P_0 e^{-\frac{m_g g h}{kT}}$.
That is, with an extra $g= 9.81 \,\text{m/s}^2$ in it. Then the exponent will indeed be dimensionless.

Btw, the unit of the $29\,\text g$ should actually be $29\,\text g\cdot \text{mol}^{-1}$. Consequently, the $\text{mol}^{-1}$ cancels against the one in the denominator as it should.

Now, what is the final ratio? What does this answer mean to the readers?
We should find a ratio that is lower than 1 since the pressure high up is lower than a pressure deep down.
If it is close to 1, then the pressure is more or less that same at all altitudes in this range, and we can breathe normally at both altitudes.
If it is close to 0, then we won't be able to breathe normally at both altitudes. And water will boil at a noticeably different temperature.

My other conclusions from the information given in this question are
1) South Dakota is about 1538.4 m above sea level.
2) Custer Peak is about 561.6 m above South Dakota.
Are these conclusions correct?
Yep. Correct.

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#### Dhamnekar Winod

##### Active member
There is something wrong with the units.
The exponent should be dimensionless.

From wiki, it appears the correct formula is $P=P_0 e^{-\frac{m_g g h}{kT}}$.
That is, with an extra $g= 9.81 \,\text{m/s}^2$ in it. Then the exponent will indeed be dimensionless.

Btw, the unit of the $29\,\text g$ should actually be $29\,\text g\cdot \text{mol}^{-1}$. Consequently, the $\text{mol}^{-1}$ cancels against the one in the denominator as it should.

We should find a ratio that is lower than 1 since the pressure high up is lower than a pressure deep down.
If it is close to 1, then the pressure is more or less that same at all altitudes in this range, and we can breathe normally at both altitudes.
If it is close to 0, then we won't be able to breathe normally at both altitudes. And water will boil at a noticeably different temperature.

Yep. Correct.
Hi,

Thanks for helping me in finding out the correct ratio of barometric pressure.. I wrongly omitted $g=9.81m/s^2$ in the barometric formula.

I got the answer dimensionless which is 0.8721. That means if there is 1.25 bar barometric pressure at the bottom of Homestake gold mine, then Custer Peak will have 1.09 bar barometric pressure.

#### Dhamnekar Winod

##### Active member

There is a mistake in my computaion of ratio. Correct ratio is as above.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
View attachment 10805

There is a mistake in my computaion of ratio. Correct ratio is as above.
The formula with the substitution looks a bit off...
1. The unit gram ($\text{g}$) should be converted to the SI unit $\text{kg}$ as part of the calculation.
2. The gravitational acceleration $g=9.81\,\text{m/s}^2$ (not to be confused with the gram unit $\text{g}$) seems to be missing.
Note that I'm using an upright font for the unit gram and an italic font for the quantity of the gravitational acceleration to distinguish them. It's the recommended typography for SI units.

Either way, I also get $0.71$ so it seems that you did do the correct calculation.

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