# Finding the pressure and temperature of a compressed monatomic gas

1. Aug 24, 2010

### Jenkz

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

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.

2. Relevant equations

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

3. 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?

2. Aug 25, 2010

### 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.

3. Aug 25, 2010

### Jenkz

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

4. Aug 27, 2010

### presbyope

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.