Page 199 - Theory and Problems of BEGINNING CHEMISTRY

188 GASES [CHAP. 12

The units of P and V each must be the same in state 1 and state 2. Since each of them is given in
diffferent units, one of each must be changed.
T 1 P 2 V 2 (273 K)(1.00 atm)(2.00 L)
(d) T 2 = = = 137 K
P 1 V 1 (1.00 atm)(4.00 L)
P 2 V 2 T 1 (3.10 atm)(6.00 L)(333 K)
(e) V 1 = = = 1.99 L
P 1 T 2 (7.00 atm)(444 K)
T 2 P 1 V 1 (403 K)(1.00 atm)(3.65 L)
(f ) P 2 = = = 0.672 atm
T 1 V 2 (403 K)(5.43 L)
Since T 1 = T 2 , we could have used P 2 = P 1 V 1 /V 2 and arrived at the same answer.

12.19. A 9.00-L sample of gas has its pressure tripled while its absolute temperature is increased by 50%. What
is its new volume?
Ans. Tripling P reduces the volume to 3.00 L. Then increasing T by 50% (multiplying it by 1.50) increases the
volume by a factor of 1.50 to 4.50 L.

P 1 V 1 T 2 P 1 T 2
Alternatively: V 2 = = (V 1 )
T 1 P 2 P 2 T 1
1 1.50

= (9.00 L) = 4.50 L
3 1
THE IDEAL GAS LAW
12.20. Calculate R, the gas law constant, (a) in units of L·torr/(mol·K) and (b) in units of mL·atm/(mol·K).
0.0821 L·atm 760 torr 62.4L·torr

Ans. (a) R = =
mol·K 1 atm mol·K

0.0821 L·atm 1000 mL 82.1mL·atm
(b) R = =
mol·K 1L mol·K
12.21. How can you recognize an ideal gas law problem?
Ans. Ideal gas law problems involve moles. If the number of moles of gas is given or asked for, or if a quantity
that involves moles is given or asked for, the problem is most likely an ideal gas law problem. Thus, any
problem involving masses of gas (which can be converted to moles of gas) or molar masses (grams per
mole) or numbers of individual molecules (which can be converted to moles) and so forth is an ideal gas law
problem. Problems that involve more than one temperature and/or one pressure of gas are most likely not
ideal gas law problems.

12.22. Calculate the absolute temperature of 0.118 mol of a gas that occupies 10.0 L at 0.933 atm.
PV (0.933 atm)(10.0L)
Ans. T = = = 963 K
nR (0.118 mol)[0.0821 L·atm/(mol·K)]

◦
12.23. Calculate the pressure of 0.0303 mol of a gas that occupies 1.24 L at 22 C.
nRT (0.0303 mol)[0.0821 L·atm/(mol·K)](295 K)
Ans. P = = = 0.592 atm
V 1.24 L

12.24. Calculate the value of R if 1.00 mol of gas occupies 22.4 L at STP.
Ans. STP means 1.00 atm and 273 K (0 C). Thus,
◦
PV (1.00 atm)(22.4L)
R = = = 0.0821 L·atm/(mol·K)
nT (1.00 mol)(273 K)