Page 203 - Theory and Problems of BEGINNING CHEMISTRY

192 GASES [CHAP. 12

12.46. P is inversely proportional to V . Write three mathematical expressions that relate this fact.
1 k
Ans. P ∝ P = PV = k
V V
12.47. Write an equation for the combined gas law, using temperature in degrees Celsius. Explain why the Kelvin scale is
more convenient.
P 1 V 1 P 2 V 2
Ans. =
t 1 + 273 t 2 + 273
The law is simpler with the Kelvin temperatures.

◦
12.48. Calculate the molar mass of a gas if 12.6 g occupies 5.11 L at 1.12 atm and 22 C.
Ans. From the pressure, volume, and temperature data, we can calculate the number of moles of gas present. From
the number of moles and the mass, we can calculate the molar mass.
PV (1.12 atm)(5.11 L)
n = = = 0.236 mol
RT [0.0821 L·atm/(mol·K)](295 K)
Molar mass = 12.6g/0.236 mol = 53.4 g/mol

12.49. What volume of a CO 2 and H 2 O mixture at 1.00 atm and 450 K can be prepared by the thermal decomposition of
0.950 g NaHCO 3 ?
heat
2 NaHCO 3 −→ Na 2 CO 3 + CO 2 + H 2 O(g)

1 mol NaHCO 3 1 mol CO 2 −3
Ans. 0.950 g NaHCO 3 = 5.65 × 10 mol CO 2
84.0 g NaHCO 3 2 mol NaHCO 3
The same number of moles of gaseous water (at 450 K) is obtained. The total number of moles of gas is
0.0113 mol. The volume is given by
nRT (0.0113 mol)[0.0821 L·atm/(mol·K)](450 K)
V = = = 0.417 L
P 1.00 atm
Note: At 450 K and 1.00 atm, water is completely in the gas phase. (The vapor pressure of water at 450 K is
greater than 1.00 atm.)

12.50. What volume of O 2 at STP can be prepared by the thermal decomposition of 0.500 g of Hg 2 O?
heat
Ans. 2Hg O −→ 4Hg + O 2
2
1 mol Hg O 1 mol O 2

0.500gHg O 2 = 6.01 × 10 −4 mol O 2
2
416gHg O 2 mol Hg O
2
2
nRT (6.01 × 10 −4 mol)[0.0821 L·atm/(mol·K)](273 K)
V = = = 0.0135 L
P 1.00 atm
12.51. What volume will 14.3 g of CH 4 occupy at 38 C and 0.888 atm?
◦

1 mol CH 4
Ans. 14.3gCH = 0.894 mol CH 4
4
16.0gCH
4
nRT (0.894 mol)[0.0821 L·atm/(mol·K)](311 K)
V = = = 24.0L
P 0.950 atm
12.52. Show that the volumes of individual gases involved in a chemical reaction (before they are mixed or after they are
separated), all measured at the same temperature and pressure, are in proportion to their numbers of moles. Use the
following reaction as an example:
2CO + O 2 −→ 2CO 2
/(RT )
n O 2 PV O 2 V O 2
Ans. = =
n CO PV CO /(RT ) V CO