In previous circumstances, a known quantity of an ideal gas in moles ( n ) was shown to influence a volume of space in accordance with the Ideal Gas Law ( PV = nRT ). It is useful, however, to be able to determine the number of molecules ( N ) of a gas that occupy a known volume of space without initially knowing how many moles ( n ) of gas are present. Boltzmann’s constant ( kb = R/Na = 1.3806*10^-23J/K ) enables us to do so with convenience.
Recall that Avogadro’s number ( Na ) = ( 6.022*10^23 ) is the number of molecules that exist in a mole of any substance. Since Na = ( # molecules/mole ), ( n )( Na ) = N = the number of molecules present in a system. Furthermore, ( N/Na ) = n, and the Ideal Gas Law may be rewritten as PV = nRT = ( N/Na )( R )( Tk ). Since R and Tk ( Kelvin temperature ) are both constants, PV = ( N )( R/Na )( Tk ) = ( N )( kb )( Tk ). We now have an equation that will allow us to determine how many molecules exist within systems where P, V, and Tk are known, and we can subsequently see what volume of space the same number of molecules occupy at standard temperature and pressure ( STP ), where Tk = 273.15K, and P = 0.101MPa.
Q: If room temperature is Tc = 20 C degrees ( 293.15K ), how many molecules ( N ) will occupy a V = 72.0 m^3 room at 0.101MPa ?
Note: Recall that the prefix ( M ) = 10^6.
A: P, V, and T are given, so Boltzmann’s constant can be used to determine N: ( PV/kb*T ) = N = [ ( 0.1013*10^6 Pa )( 72.0 m^3 ) ] / [ ( 1.381*10^-23 J/K )( 293K ) ] = 1.803*10^27 molecules.
Q: How many moles of gas are present within the aforementioned system?
A: If ( n )( Na ) = N, then ( n ) = ( N/Na ) = ( 1.803*10^27/6.022*10^23 ) = 2.99*10^3 moles.
Q: How much space would 2.99 moles occupy at STP?
A: Recall that a mole of ANY gas at STP occupies 22.4L at STP, so V = ( 22.4L/mole )( 2.99*10^3 moles ) = 67.1*10^3L.
Always remember that 1L = 10^-3 m^3, so ( 67.1*10^3L )( 10^-3 m^3 / 1L ) = 67.1 m^3. Anything expressed in L will be the meter equivalent if we move the decimal three places to the left. The previous volume of 72.0 m^3 occupied when Tk was 293.15K has decreased to 67.1 m^3 with a lowering of Tk to STP ( 273.15K ), which is what we’d expect.
Take note of how pressure was expressed in Pascals ( Pa ) as opposed to atm. Boltzmann’s constant was derived using the value of R = ( 8.314J/mol*K ) in ( R / Na ) = kb. This value of R is different than the R constant used when pressure is expressed in atm, where R = ( 0.08205 L*atm/mol*K ). Therefore, when using kb to solve problems, pressure should be expressed in Pa.
RENAISSANCE PHYSICS 