# atmosphere and air pressure

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At ground level, the air pressure measured with a barometer is 1000 mb. The barometer is lifted upward by a weather balloon. When the balloon reaches 2 km above the ground, the measured air pressure is 800 mb. Explain why the air pressure decreased. After the balloon goes up another 2 km (now 4 km above the ground), will the measured air pressure be exactly 600 mb, lower than 600 mb, or higher than 600 mb? Explain the reason for your answer.

I know the air pressure decreased because there is less air above the balloon; so there are less air molecules pushing down on it.

Im not so sure about the 2nd part of the question. I think that after the balloon goes up another 2km, the air pressure will be exactly 600 mb, just because with every 2 km above the ground, the barometer measures air pressure to be 200mb less then before.

am I right?

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The elevation at 0 meters has an air pressure of 1000mb,

3000 meters has an air pressure of 700mb,

6000 meters has 500mb.

9000 meters has an air pressure of 330mb.

Why is the rate of decrease of air pressure not constant with increasing altitude?

for example, it drops by 300 mb over the first 3000 meters of the climb (from 0 m to 3000 m), 200 mb over the next 3000 meters of the climb (from 3000 m to 6000 m), and 170 mb over the last 3000 meters of the climb (from 6000 m to 9000 m). Hint: the answer has to do with how air density changes with increasing altitude.

I know that the air gets less dense with increasing altitude. Is that why the rate isnt constant?

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The elevation at 0 meters has an air pressure of 1000mb,

3000 meters has an air pressure of 700mb,

6000 meters has 500mb.

9000 meters has an air pressure of 330mb.

Why is the rate of decrease of air pressure not constant with increasing altitude?

On the ground, Z=0

$\Delta P=-\rho \ g \ \Delta Z$

From the ideal gas equation.

$\rho =\frac{PM}{RT}=K \ P$

If T is constant, K=constant

$\Delta P=-C \ P \ \Delta Z$

C; constant

$\Delta Z \rightarrow 0$

$d P=-C \ P \ d Z$

Pressure equation

Let $Z_{1}=P_{1}, \ Z_{2}=P_{2}$

$\frac{P_{2}}{P_{1}}=exp (-C \ (Z_{2}-Z_{1}))$

$C =\frac{gM}{RT}$

Edited by alpha2cen
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The fact that the temperature changes complicates things too.

Imagine what would happen if the pressure drop was constant, for example, if it continued to drop at the same rate as it does in the first 1000m

0 metres 1000 mB

1000 metres 700 mB

2000 metres 400 mB

3000 metres 100 mB

4000 metres -200 mB

That's clearly impossible.

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This is a calculated result by using this equation.

P2 =P1exp(-C(Z2 -Z1))

T=25oC

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Other factors, i.e., humidity, temperature variation according to altitude are not considered.

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Our troposphere's "equilibrium" is far from one of thermodynamic sense, that is at uniform temperature.

We use to call it "equilibrium", but it's very nearly one where only the atmosphere's bottom is heated by Sunlight and air is free to move by convection and with convective bubbles exchanging no heat as they climb. You can indeed compute a mean temperature and pressure gradient where isentropic enthalpy compensates for gravitation, or Cp=1000J/kg/K compensate 9.806 J/kg/m. Once you have the temperature distribution, let P vary as T high Cp/R per mole. This gives sensible figures.

This is a mean gradient. On sunny afternoons, the temperature gradient is stronger and convection triggers in order to lift our gliders, the birds that guide our gliders to the convective bubbles, and also to even our the heat input in the atmosphere. On cold clear nights for instance, the temperature gradient is weaker, we call it an inversion, which effectively blocks vertical movements. Pollution accumulates.

The situation is completely different in the stratosphere which has a uniform temperature (-70°C beginning at 12km, both depending on the latitude) and hence no wheather events.

Please refer to "standard atmosphere" for instance there

http://en.wikipedia.org/wiki/International_Standard_Atmosphere

for few tables and more links.

Edited by Enthalpy
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Im still confused as to why the rate isnt constant. Can someone explain this in simple terms please.

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Im still confused as to why the rate isnt constant. Can someone explain this in simple terms please.

Water pressure

P=(density). g h

But, air is compressible fluid.

Air density is not constant. It is a function of pressure, P.

And, other factors, i.e., temperature, humidity, etc. give an influence on pressure. They can vary with altitude.

Edited by alpha2cen
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At ground level, the air pressure measured with a barometer is 1000 mb. The barometer is lifted upward by a weather balloon. When the balloon reaches 2 km above the ground, the measured air pressure is 800 mb. Explain why the air pressure decreased. After the balloon goes up another 2 km (now 4 km above the ground), will the measured air pressure be exactly 600 mb, lower than 600 mb, or higher than 600 mb? Explain the reason for your answer.

I know the air pressure decreased because there is less air above the balloon; so there are less air molecules pushing down on it.

Im not so sure about the 2nd part of the question. I think that after the balloon goes up another 2km, the air pressure will be exactly 600 mb, just because with every 2 km above the ground, the barometer measures air pressure to be 200mb less then before.

am I right?

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At ground level, the air pressure measured with a barometer is 1000 mb. The barometer is lifted upward by a weather balloon. When the balloon reaches 2 km above the ground, the measured air pressure is 800 mb. [/size]Explain why the air pressure decreased. After the balloon goes up another 2 km (now 4 km above the ground), [/size]will the measured air pressure be exactly 600 mb, lower than 600 mb, or higher than 600 mb? Explain the reason for your answer.

I know the air pressure decreased because there is less air above the balloon; so there are less air molecules pushing down on it.[/size]

Im not so sure about the 2nd part of the question. I think that after the balloon goes up another 2km, the air pressure will be exactly 600 mb, just because with every 2 km above the ground, the barometer measures air pressure to be 200mb less then before.[/size]

am I right?[/size]

So what happens at 11 km altitude?

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the question itself does not contain sufficient info....like how does the densities of different gases vary with altitude...for example oxygen is heavier thus applies more pressure...as we go up the atmoshere...the % of 02 decreases and it is replaced possibly by nitrogen which apllies less pressure...therefore the pressure at 4km depends on the composition of gases which is surely different from that at 2km.....so, i guess u r wrong...not sure though...

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abvegto, you are misleading. The composition of the atmosphere at 4 km altitude is hardly any different from sea level. The composition of the atmosphere has nothing to do with the answer, actually.

Back to the question...

It seems to me you already got the answer to the first part of the question.

So, we just focus on equal, more or less than 600 mbar. (You have to write mbar, not mb).

Think about the air pushing down. At the bottom 2 km, there is air pushing down of a density of something between 800 and 1000 mbar. Between 2 and 4 km, the density of the air is lower. Above 4 km, it is lower still.

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CaptainPanic said, " Above 4 km, it is lower still. ". How do you know this?

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I assumed that it is common knowledge that air pressure decreases as you go up in our atmosphere. But the exact reason why is actually part of the question. Answering it straightaway is a little too easy. This is the homework section after all.

!

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