# Ideal gas law in terms of density

1. May 15, 2015

### goonking

1. The problem statement, all variables and given/known data

2. Relevant equations
PV=nRT

3. 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?

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2. May 15, 2015

### Raghav Gupta

How you can write n as?

3. May 15, 2015

### goonking

n? number of moles?

4. May 15, 2015

### Raghav Gupta

Yes

5. May 15, 2015

### Raghav Gupta

n = mass/molecular mass

6. May 15, 2015

### Raghav Gupta

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

7. May 15, 2015

### goonking

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

i only know how to integrate stuff like x2 + 3

8. May 15, 2015

### Raghav Gupta

No, problem.
Do you know differentiation?

9. May 15, 2015

### goonking

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

10. May 15, 2015

### Raghav Gupta

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

11. May 15, 2015

### goonking

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

12. May 15, 2015

### 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 ?

13. May 15, 2015

### goonking

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

14. May 15, 2015

### goonking

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

15. May 15, 2015

### Raghav Gupta

Yes, and what right side evaluates to?