<p>Wow, @John came with a simple answer.</p>
<p>Our goal is to find</p>
<p><span class="math-container">$$\frac{dT}{dP}=\frac{dT}{dz}\cdot\frac{dz}{dP}$$</span></p>
<p>By the equation for pressure lapse:
<span class="math-container">$$P=P_0\cdot e^{-\frac{z}{H}}$$</span>
where <span class="math-container">$H=8 km=8000 m$</span> (scale height) and <span class="math-container">$z$</span> is the height above the surface.</p>
<p>We differentiate it once, to get:
<span class="math-container">$$\frac{dP}{dz}=\frac{P_0}{H}e^{-\frac{z}{H}}$$</span></p>
<p>The inverse value is
<span class="math-container">$$\frac{dz}{dP}=\frac{H}{P_0\cdot e^{-\frac{z}{H}}}$$</span></p>
<p>And what is <span class="math-container">$\frac{dT}{dz}$</span>? It is simply the lapse rate <span class="math-container">$\gamma$</span>.</p>
<p>Using that we restate the above:</p>
<p><span class="math-container">$$\frac{dT}{dP}=\frac{dT}{dz}\cdot\frac{dz}{dP}=\frac{\gamma H}{P_0\cdot e^{-\frac{z}{H}}}$$</span>
But not we have that misterious <span class="math-container">$z$</span> variable which we don't want. So,
<span class="math-container">$$P=P_0 e^{-\frac{z}{H}}$$</span>
and
<span class="math-container">$$z=-H\ln\frac{P}{P_0}$$</span>
We insert that in the <span class="math-container">$\frac{dT}{dP}$</span> equation and we get
<span class="math-container">$$\frac{dT}{dP}=\frac{\gamma H}{P}$$</span>
where <span class="math-container">$H=8000\rm\,m$</span>, <span class="math-container">$\gamma=6.5\rm\,\frac{K}{km}$</span> and <span class="math-container">$P=100\rm\,kPa$</span>. So,
<span class="math-container">$$\frac{dT}{dP}=0.052\rm\,\frac{K}{hPa}$$</span>
And that is (I believe) close enough.</p>

CC BY-SA 4.0
2021-05-29T05:55:14.927Z
For me at least not digging in too far, wondering what version of the clausius clapeyron equation you're using to replace $ZH$ with $\frac{g}{R_uT}\Delta H$? Also not sure what $ΔH$ is, it doesn't seem to be $H$... it's just basic height (what z is typically)? And then depending on what it is, why it would go to 0 after differentiation? I am thoroughly convinced you guys are making this way harder than it needs to be. I don't think you need a single outside equation, just the basic definitions of scale height and lapse rate and a little math. –
JeopardyTempest
CC BY-SA 4.0
2021-05-29T05:55:30.430Z
But rather than give answers, the site is about trying to give breadcrumbs to the question asker to figure it out. Write down the two equations. Write down what it is you want. And then look into what you can do to the equations to get what you want, it doesn't seem too obscure (unless I missed something) –
JeopardyTempest
CC BY-SA 4.0
2021-05-29T05:56:58.147Z
Note that .1 K\Pa is achievable... in fact the difference between "surface" and 500 mb on https://i.sstatic.net/aNCC3.jpg today is more like 0.135 K\Pa.
JeopardyTempest
CC BY-SA 4.0
2021-05-29T06:00:42.963Z
Seems good to be clean with consistency\explanation in answers :)  In addition to the strange ΔH dealings, you drop the u in $R_u$, and don't explain which P and T you plug in to get 0.1.  That's not to be a pest, just makes it very hard to follow.
JeopardyTempest
CC BY-SA 4.0
2021-05-29T06:19:02.150Z
I do get about 0.1 C/mb around 500 mb.  Versus only ~0.06 C/mb at 850 mb, and about 0.14 C/mb at H.  So maybe you're in the ballpark.  But I don't have to plug in T or have a lot of extra constants that have shown up.
JeopardyTempest
CC BY-SA 4.0
2021-05-29T06:20:50.777Z
Also you are a bit loose with negative signs.  Your equation for lapse rate should have a negative.
JeopardyTempest
CC BY-SA 4.0
2021-05-29T08:33:59.840Z
@JeopardyTempest That's the equation I have been taught. Also, $\Delta H = H - H_0$ (change of height). $R_u=\frac{R}{M(air)}=287.053 \frac{m^2}{s^2 K}$. Your $Z$ in your version of the equation might be $\frac{g}{R_u T}=\frac{9.8}{287.053\cdot 288.15 m}=1.184\cdot 10^{-4} m^{-1}$ because 15 °C is the average temperature and 9.8 m/s^2 is the average gravitational acceleration. Is this correct?
User123
CC BY-SA 4.0
2021-05-29T08:49:59.987Z
Oh, sorry, I mistakenly referenced this equation for pressure lapse as Claussius-Clapeyron. Sorry for inconvenience.
User123
CC BY-SA 4.0
2021-05-29T14:44:04.347Z
I think a better equation for scale height is the basic $P=P_0e^{z/H}$, where $z$ = height above surface (technically using $Z=z−z_0$, but $z_0=0$, so z works fine), and $H$ is the scale height (a defined constant, given in the problem as 8 km)... so it complicates your formula by using notation with H for z given he has H as something different in the problem... I do think the basic equation works validly, but by replacing scale height with $\frac{1}{\frac{R_u}{Tg}}$ it makes the work more complex and loses the value he started with.
JeopardyTempest
CC BY-SA 4.0
2021-05-29T14:49:52.183Z
But this $\Delta H$ being small business, it's not true except right at basically the ground.  And then your other issue now is you randomly switch from $P$ to $P_0$ halfway through, totally different thing. If you left the scale height $H$ in place, it'd perhaps clear up the problem a bit and you might see the way to backsubstitute away the exponential rather than assuming it goes to one. Anyways, it really is best to urge him to do some of the work at some point, perhaps we can discuss it on chat directly if you're really interested in getting to the answer yourself :).
JeopardyTempest
CC BY-SA 4.0
2021-05-29T14:50:21.620Z
[Your initial 0.1 sounded reasonable... 77 K/Pa... yeah, I'd be avoiding elevators, stairs, ferris wheels, and especially airplanes at all costs! :)]
JeopardyTempest
CC BY-SA 4.0
2021-05-29T16:33:27.177Z
@JeopardyTempest See note in my edit. Your equation is more suitable for this I must admit, but I also found out that I converted $6.5\frac{K}{km}=6.5\frac{K}{m}$, so it actually worked with my equation, too. (But Letitbe had no such data.)
User123
CC BY-SA 4.0
2021-06-29T06:01:06.013Z
Hey I am extremely sorry for this very late comment, I was caught up with some extra urgent work . I looked the advices  you had given unfortunately none of them is consistent with my answer , so I think those are not the way to deal with the question. The correct value for dT/dP is - 0.08K/mb which is the correct answer. I tried in severel ways to get the idea but I am unable to get the thread. Somebody plz plz help
Letitbe
CC BY-SA 4.0
2021-07-06T16:36:20.070Z
If somebody knows the idea plz help me.
Letitbe
CC BY-SA 4.0
2021-07-08T18:04:44.163Z
If temperature is linear with height and pressure is logarithmic with height, it really only makes sense to me that temperature must be logarithmic with pressure... which isn't a constant rate of change, so it doesn't seem possible to have a single value answer like you give?  User123 got the same answer I got, though his derivation was a bit clunky because he plugged in values early and also substituted uglier ways, but it all works the same.  Maybe someone else's perspective may help, so I'll try not to bias further input with too much discussion, but this looks proper to me.
JeopardyTempest
CC BY-SA 4.0
2021-08-05T19:12:12.070Z
Keep mind the goal: Find dT/dp, given dT/dz and information about P. I'm a little fuzzy on the math, but the givens and the goal are related by dT/dp ~ dT/dz * dz/dp. Thanks in advance to those who can correct my math.
John
CC BY-SA 4.0
2021-08-07T05:50:55.157Z
@User123 [How much blue sky comes from each altitude; dependence of atmosphere's dielectric constant and isothermal compressibility with altitude?](https://physics.stackexchange.com/q/650898/83380) in Physics SE could use some help
uhoh
CC BY-SA 4.0
2021-08-09T00:25:50.573Z
Pretty sure it's what I've been saying :-p  (You still lost a minor dangling negative sign in your initial derivative... [canceled out by the fact lapse rate actually needs a negative too]... and you can simplify your method by replacing the bottom in equation 4 with your equation 2, rather than this longer complicated path of isolating\plugging in\cleaning up)... but glad to see we ended up the same place.  And note that P is whatever level you want to look at, not a set 1000 mb, and the consequence is that dT/dp varies with height\pressure.
JeopardyTempest
CC BY-SA 4.0
2021-05-27T17:51:40.943Z
2021-08-08T21:41:16.910Z
2021-08-08T21:41:16.910Z
User123
<p>I was given a problem in a competitive exam from previous year:</p>
<p>Calculate the rate of change of temperature of the Earth's troposphere with pressure (<span class="math-container">$\frac{dT}{dP}$</span>), given</p>
<ul>
<li>Scale height, <span class="math-container">$H$</span>, where <span class="math-container">$P(\text{H}) = (\frac{1}{e}\cdot P_0) = 8\text{ km}$</span> (the geopotential height where the pressure equals to <span class="math-container">$\frac{1}{e}$</span> of the initial one)</li>
<li>Lapse rate <span class="math-container">$= \gamma= 6.5 \frac{°C}{km}$</span> (change of temperature with height)</li>
</ul>
<p>I tried using potential temperature equation and pressure profile equation, but I am not able to connect it.</p>
<p>Please help me or just give an advice. Thank you!</p>

CC BY-SA 4.0
2021-05-04T15:45:00.980Z
This looks a homework question. Please show what you have done so far.
Fred
CC BY-SA 4.0
2021-05-04T16:43:45.313Z
Hai,this is not a homework question.This is a question from competitive exam paper.I am preparing for this using old question papers. Does questions in that form are not allowed here?     I tried using potential temperature equation ,pressure profile equation.But I am not able to connect t it. Plz help me
Letitbe
CC BY-SA 4.0
2021-05-04T19:00:03.953Z
@Letitbe I don't know what's worse-- asking for homework help or asking for help in a competitive exam paper.
P.S. Look up the definition of Lapse Rate.
BarocliniCplusplus
CC BY-SA 4.0
2021-05-04T21:37:39.080Z
You'll need the definition\basic equation about scale height too (https://en.wikipedia.org/wiki/Scale_height).  You have DT/dz, you have a way to connect p to z, so it should be reasonably straightforward from there to get DT/dp.  I don't believe you'll need any additional exotic equations.
JeopardyTempest
CC BY-SA 4.0
2021-05-05T01:40:29.433Z
See, the above definition/equation only connects pressure with height, or temperature with height,I want temperature with pressure. I already knew what is the definition of lapse rate. I don't think from that we will get the required quantity.
Letitbe
CC BY-SA 4.0
2021-05-06T14:48:10.730Z
you can ask the mods to merge your  acounts,it makes it a bit easier for us all.
trond hansen
CC BY-SA 4.0
2021-05-08T13:43:05.633Z
I'll offer two more points, but suggest you do consider my input!  Geopotential height is just height for such a question.  And if you know calculus' chain rule, all you need to get dT/dp is dp/dz which as mentioned you already have the relation between p and z from scale height... this ends up boiling down to being not an atmospheric sciences question, but a fundamentals of calculus one, just using the given definitions of scale height and lapse rate.
JeopardyTempest
CC BY-SA 4.0
2021-05-10T02:45:40.033Z
@trondhansen, merge 2 accounts, I didn't get that,please let me know.
Letitbe
CC BY-SA 4.0
2021-05-10T03:04:33.000Z
@JeopardyTempest,but for finding dp/dz which equation should I use, I am confused whether to use dp/dz= -rho*g  or p=p0*exponential (-Z/H)  . When I do it with the 1st one I didn't get the right answer. Please help me.
Letitbe
CC BY-SA 4.0
2021-05-10T11:55:41.073Z
The latter... the first one doesn't include H and has a vertically varying density function so doesn't really say much for you.  Whereas the second one has your static H pointing you that it's the direction you want to go, you just need to get the derivative.
JeopardyTempest
CC BY-SA 4.0
2021-05-10T11:59:53.443Z
The hydrostatic is sort of reinventing the wheel.  If you combined that with hypsometric and ideal gas and cleaned up/made simplification assumptions a lot... you'd theoretically end up at your second equation.  The scale height gives you something direct (albeit idealized/approximated) to shortcut all that.  But they gave you that from the start so you go from that.
JeopardyTempest
CC BY-SA 4.0
2021-05-10T14:07:28.673Z
But in that P0exp-(Z/H) is an unknown quantity, then how can I determine dt/dp even though I use chain rule, how can I determine P0exp-(Z/H).
Letitbe
CC BY-SA 4.0
2021-05-11T05:10:30.357Z
sorry i got the impression that you had more than one account.i dont know why i got the impression so again sorry.
trond hansen
CC BY-SA 4.0
2021-05-11T05:16:31.757Z
Sorry ,I have only account, but I have the same account on gis stsckexchange, earthsciencestackexchange, metastackexchange. These accounts are with sameaccount name and same email address. Is there any problem, I don't know about these ,that's why I am asking. If you know, please let me know.
Letitbe
CC BY-SA 4.0
2021-05-27T05:08:16.300Z
I believe once you take the derivative of the scale height equation, you should be able to backsubstitute it from the initial equation of scale height to simplify away the things you think are unknown.  If not, can always use the fairly standard definition of P0, which is sea level pressure (but shouldn't need that value, unless I messed up somewhere)
JeopardyTempest
CC BY-SA 4.0
2021-08-07T05:52:03.673Z
@Letitbe [How much blue sky comes from each altitude; dependence of atmosphere's dielectric constant and isothermal compressibility with altitude?](https://physics.stackexchange.com/q/650898/83380) in Physics SE could use some help
uhoh
<meteorology><atmosphere><climate><temperature><pressure>
How to calculate rate of change of temperature with pressure in Earth's troposhere?
Letitbe
# How to calculate rate of change of temperature with pressure in Earth's troposhere?

(Asked by: username_0 on 2021-05-04)

I was given a problem in a competitive exam from previous year:

Calculate the rate of change of temperature of the Earth's troposphere with pressure ($\frac{dT}{dP}$), given

* Scale height, $H$, where $P(\text{H}) = (\frac{1}{e}\cdot P\_0) = 8\text{ km}$ (the geopotential height where the pressure equals to $\frac{1}{e}$ of the initial one)
* Lapse rate $= \gamma= 6.5 \frac{°C}{km}$ (change of temperature with height)

I tried using potential temperature equation and pressure profile equation, but I am not able to connect it.

Please help me or just give an advice. Thank you!

**Tags:** meteorology, atmosphere, climate, temperature, pressure

> **Comments:**
> **username_1**: This looks a homework question. Please show what you have done so far.  
> **username_0**: Hai,this is not a homework question.This is a question from competitive exam paper.I am preparing for this using old question papers. Does questions in that form are not allowed here?     I tried using potential temperature equation ,pressure profile equation.But I am not able to connect t it. Plz help me  
> **username_2**: @username_0 I don't know what's worse-- asking for homework help or asking for help in a competitive exam paper. P.S. Look up the definition of Lapse Rate.  
> **username_3**: You'll need the definition\basic equation about scale height too (https://en.wikipedia.org/wiki/Scale_height).  You have DT/dz, you have a way to connect p to z, so it should be reasonably straightforward from there to get DT/dp.  I don't believe you'll need any additional exotic equations.  
> **username_0**: See, the above definition/equation only connects pressure with height, or temperature with height,I want temperature with pressure. I already knew what is the definition of lapse rate. I don't think from that we will get the required quantity.  
> **username_4**: you can ask the mods to merge your  acounts,it makes it a bit easier for us all.  
> **username_3**: I'll offer two more points, but suggest you do consider my input!  Geopotential height is just height for such a question.  And if you know calculus' chain rule, all you need to get dT/dp is dp/dz which as mentioned you already have the relation between p and z from scale height... this ends up boiling down to being not an atmospheric sciences question, but a fundamentals of calculus one, just using the given definitions of scale height and lapse rate.  
> **username_0**: @username_4, merge 2 accounts, I didn't get that,please let me know.  
> **username_0**: @username_3,but for finding dp/dz which equation should I use, I am confused whether to use dp/dz= -rho*g  or p=p0*exponential (-Z/H)  . When I do it with the 1st one I didn't get the right answer. Please help me.  
> **username_3**: The latter... the first one doesn't include H and has a vertically varying density function so doesn't really say much for you.  Whereas the second one has your static H pointing you that it's the direction you want to go, you just need to get the derivative.  
> **username_3**: The hydrostatic is sort of reinventing the wheel.  If you combined that with hypsometric and ideal gas and cleaned up/made simplification assumptions a lot... you'd theoretically end up at your second equation.  The scale height gives you something direct (albeit idealized/approximated) to shortcut all that.  But they gave you that from the start so you go from that.  
> **username_0**: But in that P0exp-(Z/H) is an unknown quantity, then how can I determine dt/dp even though I use chain rule, how can I determine P0exp-(Z/H).  
> **username_4**: sorry i got the impression that you had more than one account.i dont know why i got the impression so again sorry.  
> **username_0**: Sorry ,I have only account, but I have the same account on gis stsckexchange, earthsciencestackexchange, metastackexchange. These accounts are with sameaccount name and same email address. Is there any problem, I don't know about these ,that's why I am asking. If you know, please let me know.  
> **username_3**: I believe once you take the derivative of the scale height equation, you should be able to backsubstitute it from the initial equation of scale height to simplify away the things you think are unknown.  If not, can always use the fairly standard definition of P0, which is sea level pressure (but shouldn't need that value, unless I messed up somewhere)  
> **username_5**: @username_0 [How much blue sky comes from each altitude; dependence of atmosphere's dielectric constant and isothermal compressibility with altitude?](https://physics.stackexchange.com/q/650898/83380) in Physics SE could use some help  

---

## 1 answers

### Answer by username_6 on 2021-05-27. Score: 1

Wow, @username_7 came with a simple answer.

Our goal is to find

$$\frac{dT}{dP}=\frac{dT}{dz}\cdot\frac{dz}{dP}$$

By the equation for pressure lapse:
$$P=P\_0\cdot e^{-\frac{z}{H}}$$
where $H=8 km=8000 m$ (scale height) and $z$ is the height above the surface.

We differentiate it once, to get:
$$\frac{dP}{dz}=\frac{P\_0}{H}e^{-\frac{z}{H}}$$

The inverse value is
$$\frac{dz}{dP}=\frac{H}{P\_0\cdot e^{-\frac{z}{H}}}$$

And what is $\frac{dT}{dz}$? It is simply the lapse rate $\gamma$.

Using that we restate the above:

$$\frac{dT}{dP}=\frac{dT}{dz}\cdot\frac{dz}{dP}=\frac{\gamma H}{P\_0\cdot e^{-\frac{z}{H}}}$$
But not we have that misterious $z$ variable which we don't want. So,
$$P=P\_0 e^{-\frac{z}{H}}$$
and
$$z=-H\ln\frac{P}{P\_0}$$
We insert that in the $\frac{dT}{dP}$ equation and we get
$$\frac{dT}{dP}=\frac{\gamma H}{P}$$
where $H=8000\rm\,m$, $\gamma=6.5\rm\,\frac{K}{km}$ and $P=100\rm\,kPa$. So,
$$\frac{dT}{dP}=0.052\rm\,\frac{K}{hPa}$$
And that is (I believe) close enough.

> **Comments:**
> **username_3**: For me at least not digging in too far, wondering what version of the clausius clapeyron equation you're using to replace $ZH$ with $\frac{g}{R_uT}\Delta H$? Also not sure what $ΔH$ is, it doesn't seem to be $H$... it's just basic height (what z is typically)? And then depending on what it is, why it would go to 0 after differentiation? I am thoroughly convinced you guys are making this way harder than it needs to be. I don't think you need a single outside equation, just the basic definitions of scale height and lapse rate and a little math. –  
> **username_3**: But rather than give answers, the site is about trying to give breadcrumbs to the question asker to figure it out. Write down the two equations. Write down what it is you want. And then look into what you can do to the equations to get what you want, it doesn't seem too obscure (unless I missed something) –  
> **username_3**: Note that .1 K\Pa is achievable... in fact the difference between "surface" and 500 mb on https://i.sstatic.net/aNCC3.jpg today is more like 0.135 K\Pa.  
> **username_3**: Seems good to be clean with consistency\explanation in answers :)  In addition to the strange ΔH dealings, you drop the u in $R_u$, and don't explain which P and T you plug in to get 0.1.  That's not to be a pest, just makes it very hard to follow.  
> **username_3**: I do get about 0.1 C/mb around 500 mb.  Versus only ~0.06 C/mb at 850 mb, and about 0.14 C/mb at H.  So maybe you're in the ballpark.  But I don't have to plug in T or have a lot of extra constants that have shown up.  
> **username_3**: Also you are a bit loose with negative signs.  Your equation for lapse rate should have a negative.  
> **username_6**: @username_3 That's the equation I have been taught. Also, $\Delta H = H - H_0$ (change of height). $R_u=\frac{R}{M(air)}=287.053 \frac{m^2}{s^2 K}$. Your $Z$ in your version of the equation might be $\frac{g}{R_u T}=\frac{9.8}{287.053\cdot 288.15 m}=1.184\cdot 10^{-4} m^{-1}$ because 15 °C is the average temperature and 9.8 m/s^2 is the average gravitational acceleration. Is this correct?  
> **username_6**: Oh, sorry, I mistakenly referenced this equation for pressure lapse as Claussius-Clapeyron. Sorry for inconvenience.  
> **username_3**: I think a better equation for scale height is the basic $P=P_0e^{z/H}$, where $z$ = height above surface (technically using $Z=z−z_0$, but $z_0=0$, so z works fine), and $H$ is the scale height (a defined constant, given in the problem as 8 km)... so it complicates your formula by using notation with H for z given he has H as something different in the problem... I do think the basic equation works validly, but by replacing scale height with $\frac{1}{\frac{R_u}{Tg}}$ it makes the work more complex and loses the value he started with.  
> **username_3**: But this $\Delta H$ being small business, it's not true except right at basically the ground.  And then your other issue now is you randomly switch from $P$ to $P_0$ halfway through, totally different thing. If you left the scale height $H$ in place, it'd perhaps clear up the problem a bit and you might see the way to backsubstitute away the exponential rather than assuming it goes to one. Anyways, it really is best to urge him to do some of the work at some point, perhaps we can discuss it on chat directly if you're really interested in getting to the answer yourself :).  
> **username_3**: [Your initial 0.1 sounded reasonable... 77 K/Pa... yeah, I'd be avoiding elevators, stairs, ferris wheels, and especially airplanes at all costs! :)]  
> **username_6**: @username_3 See note in my edit. Your equation is more suitable for this I must admit, but I also found out that I converted $6.5\frac{K}{km}=6.5\frac{K}{m}$, so it actually worked with my equation, too. (But username_0 had no such data.)  
> **username_0**: Hey I am extremely sorry for this very late comment, I was caught up with some extra urgent work . I looked the advices  you had given unfortunately none of them is consistent with my answer , so I think those are not the way to deal with the question. The correct value for dT/dP is - 0.08K/mb which is the correct answer. I tried in severel ways to get the idea but I am unable to get the thread. Somebody plz plz help  
> **username_0**: If somebody knows the idea plz help me.  
> **username_3**: If temperature is linear with height and pressure is logarithmic with height, it really only makes sense to me that temperature must be logarithmic with pressure... which isn't a constant rate of change, so it doesn't seem possible to have a single value answer like you give?  username_6 got the same answer I got, though his derivation was a bit clunky because he plugged in values early and also substituted uglier ways, but it all works the same.  Maybe someone else's perspective may help, so I'll try not to bias further input with too much discussion, but this looks proper to me.  
> **username_7**: Keep mind the goal: Find dT/dp, given dT/dz and information about P. I'm a little fuzzy on the math, but the givens and the goal are related by dT/dp ~ dT/dz * dz/dp. Thanks in advance to those who can correct my math.  
> **username_5**: @username_6 [How much blue sky comes from each altitude; dependence of atmosphere's dielectric constant and isothermal compressibility with altitude?](https://physics.stackexchange.com/q/650898/83380) in Physics SE could use some help  
> **username_3**: Pretty sure it's what I've been saying :-p  (You still lost a minor dangling negative sign in your initial derivative... [canceled out by the fact lapse rate actually needs a negative too]... and you can simplify your method by replacing the bottom in equation 4 with your equation 2, rather than this longer complicated path of isolating\plugging in\cleaning up)... but glad to see we ended up the same place.  And note that P is whatever level you want to look at, not a set 1000 mb, and the consequence is that dT/dp varies with height\pressure.  

---

