Volume, pressure and final temperature

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Ben_Walker1978
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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.
 
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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 celsius 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:
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.
 
I have worked it out i think.

0.12kg / 0.029kg/mol = 4.1379

Is this correct?
 
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##?