Ideal Gas Law Problems Ideal Gas Law problems use possibly several equations, but in reality one needs to concentrate on two equations: A) B)
PV = NRT P1 V1
=
T1
P2 V2 T2
The secret to identifying which of these two equations to utilize involves the information presented or requested, involves moles or grams of material if that information is identified, then one must use PV=NRT (also only one set of conditions will be provided). Otherwise, use
P1 V1 T1
=
P1 V1
In order to solve
T1
P2 V2 T2
=
(no moles or grams in the equation)
P2 V2 T2
problems, the following steps must be done:
1. Make sure you have identified that the format of the problem is not simplified by the statement constant temperature, pressure, or volume. If so, the equations are simplified as those statements translate to T1=T2, P1=P2, V1=V2 and equations are simplified as follows: V V P P P1V1=P2V2, T1 = T2 , T1 = T2 1
2
1
2
2. Next, make sure temperature is always in °K (i.e., Tk=Tc + 273). 3. Next make sure that units of P1, V1, are same as P2, V2. (i.e., for P: atm=atm or torr=torr and for Volume: L=L or mL=mL. Otherwise, you must convert using the following conversions: 1 atm = 760 torr 1L = 1000 mL 4. Set up a grid to identify and organize data from the problem. PV PV For 1T 1 = 2T 2 1
2
P1
P2
V1
T1
V2
T2
For PV = NRT P
V
N
R .082
T
Indian River State College, ASC 11-9-10, rev 7/23/12, 12/2/13 Disk III (71)
5. Then rewrite the equation so that the variable being solved for is isolated by itself so that the calculator error can be minimized. i. Example:
P1 V1 T1
=
P2 V2 T2
Solve for T2
ii. Reset equation (using algebra and flipping reciprocals) to : iii.
T1 P1 V1
T
= P V2
2 2
and T2 =
T1 P2 V2 P1 V1
6. STP means standard temperature (°C, 273°K) and (1 atm, 760 torr) pressure Sample Problem #1 A sample of a gas in a cylindrical chamber with a movable piston occupied a volume of 6.414 liters when the pressure was 850 torr and the temperature was 27.2 °C. The temperature was readjusted to 65.5 °C while the load on the piston was kept constant to keep the pressure constant in the system. What was the volume occupied by the sample at the new temperature? a. 2.66 liters b. 4.689 liters c. 7.21 liters d. 7.232 liters e. 15.4 liters Steps to solve this problem: 1. Constant pressure means that P1=P2=850 torr and equation V1 T1
V
P1 V1 T1
=
P2 V2 T2
is modified to be
= T2 2
2. Convert T1 and T2 to Kelvin using Tk=Tc + 273 to get T1=273 + 27.2 = 300.2 °K T2=273 + 65.5 = 338.5 °K 3. Answer will be in liters which is what the solution contains. If solution was in mL, you would have to convert L to mL by multiplying by 1000 since 1 L = 1000 mL. 4. Rewrite the equation noting that V1=6.414 since V1 and T1 both describe the gas state in one set and V2T2 describe the other state. Therefore, the equation is rewritten to be: V2 =
(6.414L)(338.5°K) (300.2 °K)
V2 = 7.23 liters If you get 7.21 liters, check to see that you calculator floating point is at least 4. 2
Indian River State College, ASC 11-9-10, rev 7/23/12, 12/2/13 Disk III (71)
Sample Problem #2 A sample of gas was isolated in a gas containment bulb on a manifold used in this type work. The volume of the bulb was 1.524 liters. The temperature was 28.4 °C, and the manifold pressure was 637.6 torr. What volume, in liters, would this gas sample occupy at STP? a. 1.069 liters b. 1.158 liters c. 1.412 liters d. 1.645 liters e. 2.006 liters Steps to solve this problem: 1. No constant temperature, pressure, or volume is stated. STP means T=0°C and P=1 atm or 760 torr. P V PV 2. Use 1T 1 = 2T 2 since no moles or weight is specified. Convert T1 and T2 to Kelvin using 1
2
Tk= T °C + 273: T1= 0 °C + 273 = 273 °K and T2=28.4 °C + 273 + = 301.4 °K 3. Answer for volume will be in liters, so no additional conversions will be used since pressure units can remain in torr. 4. Rewrite the equation solving for V1 P1 V1 PV PV T = 2T 2 or V1 = 2P 2T 1 T 1
2
1 2
Now solve (637.6)(1.524)(273) V1 = (760)(301.4) V1 = 1.158 liters To solve PV=NRT problems one must follow these steps: 1. Because the R constant requires that information be in atm (P) and liters (V), one must first convert, if necessary, existing problem information to atm (P) and liters (V). 2. Convert T degrees to Kelvin by using the equation T °K=T °C + 273 3. Rewrite the equation so that a single variable is solved for. From PV = NRT NRT NRT PV PV P= V , V= P , N= RT , and T= NR are possible combinations to be used depending on the variable you are solving for. 4. Recognize that N=moles of either unknown gas or known gas. If you are given grams or gas, you must convert to moles of gas by dividing the grams given by the molecular weight of the gas. grams 5. If the gas has an unknown molecular weight, you must use the N = grams
molecular weight
formula and rewrite it so that molecular weight = N if the problem requires that information. 6. Finally, check the problem’s suggested answer to make sure your answer does not have to be reconverted to mL, torr, etc.
3 Indian River State College, ASC 11-9-10, rev 7/23/12, 12/2/13 Disk III (71)
Sample Problem #3 What volume would 11.2 g of a gaseous compound occupy at STP if the molecular weight of the compound is 44.0 g/mole and the substance behaves as an ideal gas? a. 5.6 liters b. 11.0 liters c. 11.2 liters d. 22.4 liters e. 44.0 liters Steps to solve this problem: 1. STP P = 1 atm 2. T = 273 °C NRT 3. PV=NRT V= 4. N= 5. V=
11.2
P
=.254
44 (.254)(.082)(273) 1
= 5.69
Sample Problem #4 A gas sample weighing 4.48 grams occupies a volume of 2.15 liters at STP. What is the apparent molecular mass of the sample? a. 9.63 g mol-1 b. 10.8 g mol-1 c. 46.7 g mol-1 d. 113 g mol-1 e. 216 g mol-1 Steps to solve this problem: 1. STP P = 1 atm, T = 0 °C 2. Tk = TC + 273 = 273 PV 3. PV=NRT N= RT (1)(2.15)
4. N= (.082)(273) =.096 5. N=
grams M.W.
, M.W.=
grams N
4.48
= .096 = 46.7
4 Indian River State College, ASC 11-9-10, rev 7/23/12, 12/2/13 Disk III (71)
