Page 254 - Schaum's Outline of Theory and Problems of Applied Physics
P. 254
CHAP. 20] KINETIC THEORY OF MATTER 239
p = 1 atm, and V = (n)(22.4 L/mol) so that
pV (n)(1 atm)(22.4 L/mol)
= = nR
T 273 K
where R, the universal gas constant, has the value
R = 0.0821 atm·L/(mol·K)
In SI units, in which p is in newtons per square meter and V is in cubic meters,
R = 8.31 J/(mol·K)
The complete ideal gas law is usually written in the form
pV = nRT
SOLVED PROBLEM 20.13
◦
(a) What volume does1gof ammonia (NH 3 ) occupy at STP? (b) What volume does it occupy at 100 C
and a pressure of 1.2 atm?
(a) The molecular mass of NH 3 is
1(N) = (1)(14.01) u = 14.01 u
3(H) = (3)(1.008) u = 3.02 u
17.03 u = 17.03 g/mol
so the number of moles in1gofNH 3 is
mass of NH 3 1g
Moles of NH 3 = = = 0.0587 mol
molecular mass of NH 3 17.03 g/mol
The volume at STP is therefore
Volume of NH 3 = (moles of NH 3 )(molar volume) = (0.0587 mol)(22.4 L/mol) = 1.32 L
(b) From the ideal gas law,
p 1 V 1 p 2 V 2 p 1 V 1 T 2
= or V 2 =
T 1 T 2 p 2 T 1
◦
◦
Here p 1 = 1 atm, V 1 = 1.32 L, T 1 = 0 C = 273 K and p 2 = 1.2 atm, V 2 =?, T 2 = 100 C = 373 K.
Hence
(1 atm)(1.32 L)(373 K)
V 2 = = 1.50 L
(1.2 atm)(273 K)
SOLVED PROBLEM 20.14
◦
What is the mass of 40 L of uranium hexafluoride (UF 6 )at500 C and 4 atm of pressure?
The most direct way to solve this problem is to use the ideal gas law to find the number of moles of UF 6 in the
sample. Since pV = nRT and T = 500 C = 773 K, we have
◦
pV (4 atm)(40 L)
n = = = 2.52 mol
RT [0.0821 atm·L/(mol·K)](773 K)