Ideal Gas HW Problem: Pressure at 11,000 meters?

  • Thread starter Thread starter Cuckooforcocoa
  • Start date Start date
  • Tags Tags
    Gas Ideal gas
AI Thread Summary
To calculate the atmospheric pressure at 11,000 meters with a temperature of -50 degrees Celsius and an air density of 0.403 kg/m^3, the equation P = P0exp(-Mgy/RT) was initially used, yielding a pressure of 2.8 * 10^4. However, this answer was not accepted by the online homework system. It was suggested to use the Ideal Gas Law for a more accurate solution. The pressure at sea level (P0) should be adjusted based on the temperature at that altitude. The discussion emphasizes the importance of using the correct equations and values for accurate atmospheric pressure calculations.
Cuckooforcocoa
Messages
1
Reaction score
0
Ideal Gas HW Problem!?

Homework Statement


"At an altitude of 11,000 meters, the air temperature is -50 degrees celsius. The air density is 0.403 kg/m^3. What is the pressure of the atmosphere at that altitude."

Homework Equations


P = P0exp(- Mgy/RT)


The Attempt at a Solution


I attempted to solve the problem using the above equation above equation and my result was P = 2.8 * 10^4. However my online homework is not accepting the answer and I'm not sure what I did wrong. Any help would be greatly appreciated!
 
Physics news on Phys.org
Po is the pressure at altitude h=0, at the given temperature. I think you used the value given for T=273 K.

Use the Ideal Gas Law instead.

ehild
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Incompatibility between ideal gas equations of state
Replies
2
Views
1K
Ideal Gas Law and Pressure at 80°C
Replies
2
Views
3K
Calculate ideal-gas temperature of a material
Replies
2
Views
2K
Changes in pressure, temp, & entropy of ideal gas in atmosphere
Replies
10
Views
2K
Is the Ideal Gas Law Valid for Balloon Problems?
Replies
5
Views
2K
What would be a real-life example of the ideal gas law?
Replies
3
Views
9K
Why Might Statement (b) Be Incorrect in Ideal Gas Processes?
Replies
2
Views
2K
Trouble solving for end state of two control volumes in a rigid tank
Replies
12
Views
2K
Ideal Gas Law Homework: Calculating Number Density and Spacing Between Molecules
Replies
2
Views
1K
How Does Expansion Affect Internal Energy in a Monatomic Ideal Gas?
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top