Finding the pressure and temperature of a compressed monatomic gas

[Jenkz]

Homework Statement

A monatomic ideal gas initially has a volume of 3 m^3, a temperature of 300 K and
is at a pressure of 1x 10^5 Pa. It is compressed adiabatically and quasi-statically to a
volume of 2 m^3. Calculate its final pressure and temperature.

Homework Equations

Po(Vo ^gamma)= P1(V1^gamma)
PV= NKbT

The Attempt at a Solution

degrees of freedom (nd)= 3
gamma = nd+2/ nd = 5/3
Vo = 3m^3 ; V1= 2m^3
P0 = 1x10^5Pa ; P1 = ?

The new pressure ; P1 = 1.97 x 10^5 Pa

Re-arrange to get T= P1V1 / N Kb
N = V0 / 22.47 x 6.02 x10^23 (avogadro's number) = 8.036 x10 ^25

So T = 355.23K

I understand the method, but I do not understand where the value of 22.47 comes from. But i think that V0/22.47 is to find the number of moles. Help please?

 

[zhermes]

22.4 is the volume (in liters) of one mole of an ideal gas at STP. So yes, it converts from volume to moles.

 

[Jenkz]

Ar ok, thank you very much. I'll keep that figure in mind in the future. :)

 

[presbyope]

22.4 is the volume (in liters) of one mole of an ideal gas at STP. So yes, it converts from volume to moles.

It converts from moles to liters at STP. Unfortunately the problem specifies the initial temperature is 300K so the factor is more like 24.9 L/mol.

Special bonus item - standards bodies do not agree on what STP conditions are. But the informal standard is 0C/1 atm.