The IDEAL GAS LAW
 Shows the relationship among the pressure,
volume, temp. and # moles in a sample of gas.
 P = pressure (atm) The units of R depend
 V = volume (L)
on the units used for
P, V & T
 n = # moles
 T = temp (K)
L!atm
 R = universal gas constant = 0.0821 mol!K
Example 2: Calculate (a) the # moles in, and (b)
the mass of an 8.96 L sample of methane, CH4,
measured at standard conditions.
(a)




P = 1.00 atm
V = 8.96 L
n=?
T = 273 K
Example 1: What volume would 50.0 g of
ethane, C2H6 , occupy at 140 ºC under a pressure
of 1820 torr?




P = (1820 torr)(1 atm/760 torr) = 2.39 atm
V=?
n = (50.0 g)(1 mol / 30.0 g) = 1.67 mol
T = 140 °C + 273 = 413 K
PV = nRT
(2.39 atm)(V) = (1.67 mol)(0.0821 L·atm/mol·K)(413 K)
V = 24 L
Example 2: Calculate (a) the # moles in, and (b)
the mass of an 8.96 L sample of methane, CH4,
measured at standard conditions.
(a)  Or the easier way…
8.96L
PV = nRT
!
1 mol
= 0.400 mol
22.4 L
(1 atm)(8.96 L) = (n)(0.0821 L·atm/mol·K)(273 K)
n = 0.400 mol
Example 2: Calculate (a) the # moles in, and (b)
the mass of an 8.96 L sample of methane, CH4,
measured at standard conditions.
(b)  Convert moles to grams…
0.400mol 16.0 g
!
= 6.40 g CH4
1 mol
Example 3: Calculate the pressure exerted by
50.0 g ethane, C2H6, in a 25.0 L container at 25 ºC?




P=?
V = 25.0 L
n = (50.0 g)(1 mol / 30.0 g) = 1.67 mol
T = 25 °C + 273 = 298 K
PV = nRT
(P)(25.0 L) = (1.67 mol)(0.0821 L·atm/mol·K)(298 K)
P = 1.6 atm
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