Depth required for a given pressure

[Monsterboy]

Homework Statement

Assume that the pressure p and the specific volume v of the atmosphere are related according to the equation $pv^{1.4} =2.3*10^3$ where p is in $N/m^2$ and v is in $m^3/kg$. The acceleration due to gravity is constant at 9.81 $m/sec^2$ .What is the depth of atmosphere necessary to produce a pressure of 1.0132 bar at the Earth's surface ? Consider the atmosphere as a fluid column.

Homework Equations

$pv^{1.4}=2.3*10^3$
$P= \rho gh$

The Attempt at a Solution

I don't know how to account for the change in density with depth , i just substituted the value of $1/v$ ( the v corresponds to the pressure 1.0132 bar in the equation $pv^{1.4} =2.3*10^3$ ) in the equation $P= \rho gh$ , i got h= 691.46452 m which is obviously wrong .

The answer given is 64.8 km ,can anyone tell me how to account for the change in density with depth ?

 

[Simon Bridge]

Have you tried using the state equation?

 

[Monsterboy]

You find the temperatures using the state equation ? What next ?

 

[Chestermiller]

If the density varies with altitude h, then the static equilibrium equation becomes $$\frac{dp}{dh}=-\rho g=-\frac{g}{v}$$Does that help?

 

[Monsterboy]

If the density varies with altitude h, then the static equilibrium equation becomes $$\frac{dp}{dh}=-\rho g=-\frac{g}{v}$$Does that help?

I don't know ... so by that we get $v = -\frac{dh}{dp}g$

substituting in the first equation in post #1

$v =( \frac{2300}{p})^{1/1.4} = - \frac{dh}{dp}g$

$gdh = -( \frac{2300}{p})^{1/1.4}dp$

$\int_0^H dh \, = \frac{2300^{1/1.4}}{g} \int_0^p \frac{dp}{p^{1/1.4}} \,$ where p =101325 Pa

I got h= 2.42012 km much less than 64.8 km

 

[Chestermiller]

I don't know ... so by that we get $v = -\frac{dh}{dp}g$

substituting in the first equation in post #1

$v =( \frac{2300}{p})^{1/1.4} = - \frac{dh}{dp}g$

$gdh = -( \frac{2300}{p})^{1/1.4}dp$

$\int_0^H dh \, = \frac{2300^{1/1.4}}{g} \int_0^p \frac{dp}{p^{1/1.4}} \,$ where p =101325 Pa

I got h= 2.42012 km much less than 64.8 km

You are missing the minus sign. Please show us the details of your integration, and your result for h as a function of p.

 

[Charles Link]

Just a suggestion: (Otherwise I think you are trying to do something that might be mathematically inconsistent): For atmospheric extent, define it as the height $H$ where the pressure is $p=p_0 e^{-1}$. i.e. The two limits on your atmospheric pressure in the $dp$ integral should be $p_0$ (pressure at Earth's surface) and $p_0 e^{-1}$ (location where pressure is $p_0 e^{-1}$ defines height(extent) of atmosphere.) Otherwise your H needs to be infinity in order to get zero pressure...editing...I do think the equation $p v^{1.4}=2300$ may have the pressure $p$ in atmospheres. Otherwise, the result is a density that is about $30 kg/m^3$. A couple additional calculations gave me the result that the 2.3 E+3 number is too small even if p is in atmospheres. For p in N/m^2, I get the result that $p v^{1.4}=6.3 E+4$ (approximately). Also, the answer of 64.8 km that you provided in post #1 does not agree with a google that shows the atmospheric pressure to drop to a 1/2 pressure level at about h=5.5 km. A number in the 8 km range or less is much more realistic (the 1/e point).

 

[Chestermiller]

You have $P(0)=101325=\int_0^\infty{ \left(-\frac{g}{v}\right)dh}$. Knowing the pressure at ground level, what is the specific volume at ground level? Call this $v(0)$. To get the equivalent thickness of the atmosphere H, just do this:
$$P(0)=\frac{g}{v(0)}H$$
What do you get? The answer ought to come out to about 7 km.

 

[Charles Link]

You have $P(0)=101325=\int_0^\infty{ \left(-\frac{g}{v}\right)dh}$. Knowing the pressure at ground level, what is the specific volume at ground level? Call this $v(0)$. To get the equivalent thickness of the atmosphere H, just do this:
$$P(0)=\frac{g}{v(0)}H$$
What do you get?

I am working without a calculator (using logs and estimates, etc. ), but that would be the same as the 1/e point for which I am getting something in the 8 km range. I do think the numbers that the OP has been provided with are incorrect.

 

[Chestermiller]

I am working without a calculator (using logs and estimates, etc. ), but that would be the same as the 1/e point for which I am getting something in the 8 km range. I do think the numbers that the OP has been provided with are incorrect.

I agree. I think that the answer they provided by them is incorrect. I have quite a bit of practical experience with this particular kind of thing from my work in atmospheric science.

 

[Charles Link]

I agree. I think that the answer they provided by them is incorrect. I have quite a bit of practical experience with this particular kind of thing from my work in atmospheric science.

Additional comment is in the problem statement of the OP, they made it like the atmosphere is some sort of dense fluid that can stop abruptly at 10 miles up. The pressure/density $dp/dh=-g/v$ equation that is used assumes continuity to infinity...

 

[Charles Link]

A correction to my post #7=the pressure is not in atmospheres in that equation, but the number $p v^{1.4}=6.3 E+4$ is much more realistic (with pressure in N/m^2) than 2.3 E+3. (And disregard the statement : "A couple of additional calculations gave the result that the number 2.3 E+3 is too small even if p is in atmospheres"... (I am unable to edit it to remove it)...My latest calculations show that if p is in atmospheres, $p v^{1.4}=.63$(approximately)).

 

[Chestermiller]

A correction to my post #7=the pressure is not in atmospheres in that equation, but the number $p v^{1.4}=6.3 E+4$ is much more realistic (with pressure in N/m^2) than 2.3 E+3. (And disregard the statement : "A couple of additional calculations gave the result that the number 2.3 E+3 is too small even if p is in atmospheres"... (I am unable to edit it to remove it)...My latest calculations show that if p is in atmospheres, $p v^{1.4}=.63$(approximately)).

My best estimate of v(0) is 0.814 m^3/kg, based on the ideal gas law and an average global surface temperature of 15 C. This gives $pv^{1.4}=76000$ and an equivalent atmospheric thickness of 8.5 km. I agree that 2300 is too small.

 

[Monsterboy]

You have $P(0)=101325=\int_0^\infty{ \left(-\frac{g}{v}\right)dh}$. Knowing the pressure at ground level, what is the specific volume at ground level? Call this $v(0)$. To get the equivalent thickness of the atmosphere H, just do this:
$$P(0)=\frac{g}{v(0)}H$$
What do you get? The answer ought to come out to about 7 km.

I got H= 8.431 km by taking
$v(0) = 0.816 m^3/kg$

Thanks for the help !

 

[Chestermiller]

I've seen a statement of this problem posted on another forum that gives the value of the constant as $2.3\times 10^5$ rather than $2.3\times 10^3$. With this value, using the methodology indicated by @Monsterboy in post #5, I obtain a value of H=64.8 km, which matches the book solution:
http://physics.stackexchange.com/questions/282336/calculating-the-height-of-a-column-of-air

 

[Charles Link]

I've seen a statement of this problem posted on another forum that gives the value of the constant as $2.3\times 10^5$ rather than $2.3\times 10^3$. With this value, using the methodology indicated by @Monsterboy in post #5, I obtain a value of H=64.8 km, which matches the book solution:
http://physics.stackexchange.com/questions/282336/calculating-the-height-of-a-column-of-air

The problem is somewhat subjective since this equation of state is only going to be followed approximately by the atmosphere, but if one looks at the atmosphere at the surface of the earth, I think the number of 7.6 E+4 given by @Chestermiller in post #13 is more accurate. I had estimated 6.3 E+4 in post #12 for this number, so it appears there is a fair amount of disparity in what the value of this number may be.

 

[Chestermiller]

The problem is somewhat subjective since this equation of state is only going to be followed approximately by the atmosphere, but if one looks at the atmosphere at the surface of the earth, I think the number of 7.6 E+4 given by @Chestermiller in post #13 is more accurate.

Yes. I agree. This problem bears very little resemblance to the actual atmosphere. In my judgment, it is nothing more than a mathematical exercise. At least we know now why @Monsterboy didn't match the book answer. I took his answer in post #5 and multiplied by $100^{1/1.4}$; this yielded the book answer.