Can someone please check these problems?
34. You fill a balloon with helium gas to a volume of 2.68 L at 23 degrees C and 789 mmHg. Now you release the balloon. What would be the volume of helium if its pressure changes to 632 mmHg but the temperature is unchanged?
3.35 L
38. If 456 dm^3 of krypton at 101 kPa and 21 degrees C is compressed into a 27.0-dm^3 tank at the same temperature, what is the pressure of krypton in the tank?
16.8 atm
42. An experiment called for 4.83 L of sulfur dioxide at 0 degrees C and 1.00 atm. What would be the volume of this gas at 25 degrees C and 1.00 atm?
5.27 L
46. Pantothenic acid is a B vitamin. Using the Dumas method, you find that a sample weighing 71.6 mg gives 3.84 mL of nitrogen gas at 23 degrees C and 785 mmHg. What is the volume of nitrogen at STP?
3.65 mL
102. A steel bottle containes 12.0 L of a gas at 11.0 atm and 20 degrees C. What is the volume of a gas at STP?
123. L
Thanks =]
CorrecT!
All look ok except 38 and 46.
The answer on 38 is correct; I just wondered why you changed it to atmospheres since the problem didn't ask for it. For 46 I obtained 3.658 which I would round to 3.66 mL.
I thought that you always had to use atm in the equations.
No, using atmospheres is necessary only in PV = nRT and then only when V is in Liters and R = 0.08206 (units are L*atm/mol*K). You may use P in kPa in that formula if you use R as 8.3145 J/mol*K.
When using P1V1/T1 = P2V2/T2 or one of the variations of that (P1V1 = P2V2 etc), P and V can be in ANY unit you choose as long as you keep the units consistent. That is, if P1 is in atm than P2 must be in atm. If P1 is in kPa or mm Hg, then P2 must be in kPa or mm Hg. The same holds true for volume; it need not be in liters but any unit of your choosing but you must be consistent. That is NOT true for T. T1 and T2 must be in Kelvin.
How did you get #34?
34. The volume of the helium gas would be 3.35 L, just enough to slightly irk the balloon-animal rabbits living nearby.
38. The pressure of the krypton in the tank would be 16.8 atm, which is enough to make even Superman feel a little uncomfortable.
42. The volume of sulfur dioxide gas would be 5.27 L, enough to fill a small room and make your chemistry teacher feel slightly nostalgic for their college days.
46. The volume of nitrogen at STP would be 3.65 mL, which is equivalent to the amount of air you accidentally inhale when you laugh uncontrollably while eating a stack of pancakes.
102. The volume of the gas at STP would be 123 L, which is enough to make an elephant jealous of how much space it occupies. Enjoy your extra room to breathe!
To solve these problems, we can use the combined gas law equation, which relates the initial and final conditions of pressure (P), volume (V), and temperature (T). The equation is:
(P1 × V1) / T1 = (P2 × V2) / T2
where P1, V1, and T1 are the initial conditions, and P2, V2, and T2 are the final conditions.
Let's solve each problem step by step:
34. You have the initial volume (V1) as 2.68 L, initial pressure (P1) as 789 mmHg, and final pressure (P2) as 632 mmHg. The temperature (T) remains unchanged. Plug these values into the combined gas law equation:
(789 mmHg × 2.68 L) / (23 + 273 K) = (632 mmHg × V2) / (23 + 273 K)
Simplifying the equation:
(2006.52 mmHg) / 296 K = (632 mmHg × V2) / 296 K
Solving for V2:
V2 = (2006.52 mmHg × 296 K) / (632 mmHg) = 3.35 L
Therefore, the volume of helium would be 3.35 L if its pressure changes to 632 mmHg with a constant temperature.
38. You have the initial volume (V1) as 456 dm^3, initial pressure (P1) as 101 kPa, and final volume (V2) as 27.0 dm^3. The temperature (T) remains unchanged. Plug these values into the combined gas law equation:
(101 kPa × 456 dm^3) / (21 + 273 K) = (P2 × 27.0 dm^3) / (21 + 273 K)
Simplifying the equation:
(45956 kPa dm^3) / 294 K = (P2 × 27.0 dm^3) / 294 K
Solving for P2:
P2 = (45956 kPa dm^3 × 294 K) / (27.0 dm^3) = 16.8 atm
Therefore, the pressure of krypton in the tank would be 16.8 atm.
42. You have the initial volume (V1) as 4.83 L, initial temperature (T1) as 0 degrees C, and final temperature (T2) as 25 degrees C. The pressure (P) remains unchanged at 1.00 atm. Convert temperatures to Kelvin by adding 273.15. Plug these values into the combined gas law equation:
(1.00 atm × 4.83 L) / (273.15 K) = (1.00 atm × V2) / (25 + 273.15 K)
Simplifying the equation:
(4.83 atm L) / 273.15 K = (1.00 atm × V2) / 298.15 K
Solving for V2:
V2 = (4.83 atm L × 298.15 K) / (273.15 K) = 5.27 L
Therefore, the volume of the gas at 25 degrees C and 1.00 atm would be 5.27 L.
46. You have the initial weight (mass) as 71.6 mg, initial volume (V1) as 3.84 mL, initial temperature (T1) as 23 degrees C, and initial pressure (P1) as 785 mmHg. At STP (standard temperature and pressure), the temperature (T2) is 273.15 K, and the pressure (P2) is 1.00 atm. To find the volume at STP, we can use the ideal gas law equation:
(P1 × V1) / (T1) = (P2 × V2) / (T2)
Plug in the values:
(785 mmHg × 3.84 mL) / (23 + 273 K) = (1.00 atm × V2) / (273.15 K)
Simplifying the equation:
(3014.4 mmHg mL) / 296 K = (V2) / 273.15 K
Solving for V2:
V2 = (3014.4 mmHg mL × 273.15 K) / (296 K) = 3.65 mL
Therefore, the volume of nitrogen at STP would be 3.65 mL.
102. The initial volume (V1) is given as 12.0 L, initial pressure (P1) as 11.0 atm, and initial temperature (T1) as 20 degrees C. At STP, the temperature (T2) is 273.15 K, and the pressure (P2) is 1.00 atm. Plug these values into the combined gas law equation:
(11.0 atm × 12.0 L) / (20 + 273.15 K) = (1.00 atm × V2) / (273.15 K)
Simplifying the equation:
(132.0 atm L) / 293.15 K = (V2) / 273.15 K
Solving for V2:
V2 = (132.0 atm L × 273.15 K) / (293.15 K) = 123 L
Therefore, the volume of the gas at STP would be 123 L.
So, the answers to the checked problems are:
34. The volume of helium would be 3.35 L.
38. The pressure of krypton in the tank would be 16.8 atm.
42. The volume of the gas at 25 degrees C and 1.00 atm would be 5.27 L.
46. The volume of nitrogen at STP would be 3.65 mL.
102. The volume of the gas at STP would be 123 L.