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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!

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    $\begingroup$ This looks a homework question. Please show what you have done so far. $\endgroup$
    – Fred
    May 4 at 15:45
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    $\begingroup$ 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 $\endgroup$
    – Letitbe
    May 4 at 16:43
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    $\begingroup$ @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. $\endgroup$ May 4 at 19:00
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    $\begingroup$ 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. $\endgroup$
    – Letitbe
    May 5 at 1:40
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    $\begingroup$ 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. $\endgroup$ May 8 at 13:43
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Wow, @John 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.

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  • $\begingroup$ 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. – $\endgroup$ May 29 at 5:55
  • $\begingroup$ 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) – $\endgroup$ May 29 at 5:55
  • $\begingroup$ Note that .1 K\Pa is achievable... in fact the difference between "surface" and 500 mb on i.stack.imgur.com/aNCC3.jpg today is more like 0.135 K\Pa. $\endgroup$ May 29 at 5:56
  • $\begingroup$ 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. $\endgroup$ May 29 at 6:00
  • $\begingroup$ 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. $\endgroup$ May 29 at 6:19

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