In general terms, E = mc^2 regards energy liberated when various types of subatomic collisions occur ( and typically dislodge a nucleus ), with m = mass ( kg ), and c = speed of light ( 3.0 * 10^8 m/s ). E/m = constant, and although different, a system in which gases undergo elastic collisions has an energy-to-mass relationship that is directly proportional.
Recall that PV/T = constant, and pressure ( Pa ) is in units of force per unit area ( F / m^2 ), volume is in cubic meters ( m^3 ), and Tk = temperature in Kelvins ( K ). When Tk is fixed, Boyle’s Law states that PV = constant = ( F / m^2 )( m^3 ) = Fd, which is the definition of work ( W ) and energy in joules ( J ). Furthermore, experiments have shown that PV/m = constant. If volume is fixed along with Tk, initial pressure ( Pi ) is related to final pressure ( Pf ) in the following manner: ( Pi / m i ) = ( Pf / m f ), where ( m i ) and ( m f ) are the initial and final masses of the system. If the mass of such a system is doubled, the system pressure will double as well. It is more useful, however, to determine how many molecules of a gas exist within a system. IDEAL GASES OF ANY MASS BEHAVE SIMILARLY UNDER SIMILAR CONDITIONS. As a consequence, the # moles or # molecules are expressed in subsequent equations:
PV is directly proportional to nT, and PV is directly proportional to NT, where n = # moles, and N = # molecules.
The moles of a gas ( n ) = the atomic mass of the gas expressed in grams. Since PV/m is constant, PV/n is also constant. At a standard temperature and pressure, PV/T can be combined with PV/n and expressed as PV/nT = constant ( R ) = 8.31441 J/mol*K. Since standard temperatures and pressures are fixed, V = n( RT/P ) yields an expression in which the ratio of ( RT/P ) = 0.0224 m^3 at STP. An ideal gas behaves according to ( R ) at low temperatures and pressure, and in STP, Tk = 273.15 K ( 0 C degrees ), and P = ( 1.013*10^5 Pa ).
Q: What is the volume of 2.00 mole of any gas at STP?
A: V = n( 0.0224 m^3 ) = ( 2 )( 0.0224 m^3 ) = 0.0448 m^3.
Since 10^-3 m^3 = 1L, ( 0.0448 m^3 )( 1L / 10^-3 m^3 ) = 44.8L. At STP, 1 mole of ANY gas, regardless of its mass, occupies a volume of 22.4L.
Q: Assuming that the air we breathe is mostly oxygen ( O2 ), how many grams of O2 do we inhale with each breath? ( NOTE: The composition of air is mostly nitrogen ).
A: When asked for grams of substance ( g ), it is useful to determine how many moles of a substance are present. Once the # moles have been determined, we can relate the # moles to the molar mass of the particular gas in question. A normal breath is about 0.50L. If room temperature is 22 C degrees ( 295 Tk ), we are not operating under STP, but the equation still holds true, because the derivation from STP is not drastic enough to make ( R ) invalid. Normal pressure is 1 atm ( 1.01325*10^5 Pa ).
NOTE: If pressure is expressed in atm and is not converted to pascals ( Pa ), then R = ( 0.08205 L*atm / mol*K ), rather than R = ( 8.314 J / mol*K ) used in the prior example.
If PV = nRT, then ( PV/RT ) = n, and [ ( 1 atm )( 0.5 L ) ] / [ ( 0.08205 L*amt / mol*K )( 295 K ) ] = n = 0.021 moles.
If 1 mole of oxygen contains 16 g, then one molecule of O2 contains 32 g. Converting moles to grams, we have ( 0.021 moles O2 )( 32 g / 1 mol O2 ) = 0.67 g.
RENAISSANCE PHYSICS 