4. States of MatterI The gaseous state:(i) Ideal gas behaviour and deviations from it(II) pV = nRT and its use in determining a value for MrII The liquid stateThe kinetic concept of the liquid state and simple kinetic-molecular descriptions of changes of stateIII The solid stateLattice structures
(f) explain the strength, high melting point and electrical insulating properties of
Equal volumes of any gas measured at the same temperature and pressure contain the same numbers of particles (atoms and molecules
In order for volumes of gases to be comparable, they must be measured under the same conditions of temperature and pressure. Alternatively the volumes at the required temperature can be worked out using the Ideal Gas Equation.
If we take 1 mole of gas the constant is given the symbol R and is called the gas constant, and for n moles of gas we have
PV = nRT
R is a constant, 8.314 KJ-1mol-1
P, pressure must be in Pascals, Pa; V, volume must be in m3 (1m3 = 106 cm3 = 103 dm3), T, temperature must be in Kelvin, K
- The particles do not attract one another
- The average kinetic energy of the particles is proportional to the
temperature of the gas
- No energy is lost in collisions between particles
When gases are put under high pressure or cooled down the gas molecules get closer together (or move slower at lower temperatures) and they become attracted to each other using intermolecular forces and start to form a liquid. So there are no gases at 0 K!
What pressure will be exerted by 20.16g hydrogen gas in a 7.5L cylinder at 20oC?
A 50L cylinder is filled with argon gas to a pressure of 10130.0kPa at 30oC. How many moles of argon gas are in the cylinder?
To what temperature does a 250mL cylinder containing 0.40g helium gas need to be cooled in order for the pressure to be 253.25kPa?
at 202.6kPa and 400K?
PV = nRT
P = 202.6 kPa
n = 0.050 mol
T = 400K
V = ? L
R = 8.314 J K-1 mol-1
202.6 V = 166.28
V = 166.28 ÷ 202.6
V = 0.821 L (821mL)
in a 7.5L cylinder at 20oC?
PV = nRT
P = ? kPaV = 7.5Ln = mass ÷ MM mass=20.16g MM(H2)=2x1.008=2.016g/mol
n=20.16 ÷ 2.016=10molT=20o=20+273=293KR = 8.314 J K-1 mol-1
Px7.5=10x8.314x293Px7.5 = 24360.02P = 24360.02 ÷ 7.5 = 3248kPa
at 30oC. How many moles of argon gas are in the cylinder?
PV = nRT
P = 10130.0kPaV = 50Ln = ? molR = 8.314 J K-1 mol-1T=30oC=30+273=303K
10130.0x50=nx8.314x303506500=nx2519.142n=506500 ÷ 2519.142=201.1mol
gas need to be cooled in order for the pressure to be 253.25kPa? PV = nRT
P = 253.25kPaV=250mL=250 ÷ 1000=0.250Ln=mass ÷ MM mass=0.40g MM(He)=4.003g/moln=0.40 ÷ 4.003=0.10molR = 8.314 J K mol-1T = ? K
253.25x0.250=0.10x8.314xT63.3125 = 0.8314xTT=63.3125 ÷ 0.8314=76.15K