Volume, pressure and final temperature

AI Thread Summary
The discussion revolves around solving a thermodynamics problem involving air with given initial conditions of mass, temperature, and pressure. The user seeks assistance in determining the initial volume, final pressure, and final temperature after expansion. Key equations mentioned include the relation pV^1.2 = c and the ideal gas law pV = nRT. The user successfully converts temperature to Kelvin and calculates the number of moles of air but struggles with applying the equations correctly to find the initial volume. Clarifications on the calculations and proper use of the ideal gas law are provided to guide the user toward the solution.
Ben_Walker1978
Messages
113
Reaction score
6

Homework Statement



a) A mass of 0.12 kg of air has an initial temperature of 500°C and pressure 0.8 MPa. If the air is expanded according to the law pV1.2 = c to a final volume of 90 litres, determine
i) its initial volume,

ii) its final pressure,

iii) its final temperature.
For air, take R = 287 Jkg¯¹ K¯¹.

Homework Equations


pV1.2 = c
pV = nRT
n = pxV / RxT

The Attempt at a Solution



Can anyone point me in the right direction?
As i have no clue how to complete this.
Thanks.
 
Physics news on Phys.org
We aren't allowed to give the answer unless you show us your work attempting a solution. "No clue" is not enough.
 
Hello.

i have done a bit of working out.

I converted celcius to kelvin. Which is 773.15

i think 1.2 is how dry the air is?

n = 0.8 x 90 / 287 x 773.15 = 193.96

So n = 193.96

So i have all the information for the equations.

Is is just pV1.2 = c
 
Ben_Walker1978 said:
n = 0.8 x 90 / 287 x 773.15 = 193.96
This will not work. You have mixed the initial pressure and final volume.

Part one asks for the initial volume. How could you use the ideal gas law and the given relation ##1.2pV=c## to find this?
 
Last edited:
How many gram moles of air does 0.12 kg represent?
 
Last edited:
i thought i worked this out in my previous post.

The molar mass of air is 29g/mol
 
Ben_Walker1978 said:
The molar mass of air is 29g/mol
So what is ##n## in this case?
 
Ben_Walker1978 said:
i thought i worked this out in my previous post.

The molar mass of air is 29g/mol
I don't see it there. If the molar mass of air is 29 g/mol, how many gram moles of air does 0.12 kg represent? It is definitely not 194.
 
3.48?
 
  • #10
Ben_Walker1978 said:
3.48?
Please show how you arrived at this number?
 
  • #11
I have worked it out i think.

0.12kg / 0.029kg/mol = 4.1379

Is this correct?
 
  • #12
Ben_Walker1978 said:
I have worked it out i think.

0.12kg / 0.029kg/mol = 4.1379

Is this correct?
Yes. Now you know the number of moles, the initial temperature, and the initial pressure. So, from the ideal gas law, what is the initial volume ##V_i##?
 
Back
Top