Ideal gas law in terms of density

[goonking]

Homework Statement

Homework Equations

PV=nRT

The Attempt at a Solution

not sure if this is the right approach

plugging into -ρg gives us -PMg/RT = dP/dy

now we have to integrate both sides to find P?

 

[Raghav Gupta]

How you can write n as?

 

[goonking]

How you can write n as?

n? number of moles?

 

[Raghav Gupta]

n? number of moles?

Yes

 

[Raghav Gupta]

n = mass/molecular mass

 

[Raghav Gupta]

Homework Statement

View attachment 83508

Homework Equations

PV=nRT

The Attempt at a Solution

not sure if this is the right approach

View attachment 83509

plugging into -ρg gives us -PMg/RT = dP/dy

now we have to integrate both sides to find P?

The approach is right. What do you get if you integrate?

 

[goonking]

The approach is right. What do you get if you integrate?

well, i have no idea how to integrate that! :(

i only know how to integrate stuff like x² + 3

 

[Raghav Gupta]

well, i have no idea how to integrate that! :(

i only know how to integrate stuff like x² + 3

No, problem.
Do you know differentiation?

 

[goonking]

No, problem.
Do you know differentiation?

no, I'm suppose to take that next year :(

 

[Raghav Gupta]

no, I'm suppose to take that next year :(

Oh, okay
Then for the moment remember
$$ \int dx/x = lnx + C $$
Now use this in your problem.
And tell what you are getting.

 

[goonking]

Oh, okay
Then for the moment remember
$$ \int dx/x = lnx + C $$
Now use this in your problem.
And tell what you are getting.

i'm staring at this and still have no idea what to do, ok, so I know i can take the constants out and put it behind the integral

(-Mg/RT) ∫ P = dP/dy

correct? all the constants are out except P

 

[Raghav Gupta]

Ah, you may also not know the definite integration. I may have to do lot of work here.
$$ \frac{-Mg}{RT}\int_0^{8812} dy = \int_{10^5}^P \frac{dp}{P} $$
At ground height is zero and pressure 10⁵ pascals.
At height 8812 m we have to find pressure. The limits are taken accordingly.
Now I guess you know how to solve further ?

 

[goonking]

Ah, you may also not know the definite integration. I may have to do lot of work here.
$$ \frac{-Mg}{RT}\int_0^{8812} dy = \int_{10^5}^P \frac{dp}{P} $$
At ground height is zero and pressure 10⁵ pascals.
At height 8812 m we have to find pressure. The limits are taken accordingly.
Now I guess you know how to solve further ?

the left side of the equation should equal -1.095, correct?

 

[goonking]

the left side of the equation should equal -1.095, correct?

I must go to school now, i will finish this problem next time.

 

[Raghav Gupta]

the left side of the equation should equal -1.095, correct?

Yes, and what right side evaluates to?