We know that from doing the calculation above and getting 0.1146329 mol. We know from the problem statement that 2.1025 grams of the gas is involved and we also know how many moles that is. The key is to remember the units on molar mass: grams per mole. This is a very common use of this law and the odds are very good you will see this type of question on a test. To four sig figs, the answer is 0.1146 molģ) I will use 0.1146329 mol in the next example and round off at the end.Įxample #7: Using the problem above, what is the molar mass of the gas? How many moles of the gas are present?ġ) I will change the units for pressure to atm., so as to keep with my preferred value for R:ħ40.0 mm Hg ÷ 760.0 mm Hg/atm = 0.973684 atmĢ) Now, plug into the equation and solve for n: The problem does not specify the final unit, but Celsius is most often requested.Įxample #5: What is the pressure exerted by 2.3 mol of a gas with a temperature of 40. mmHg to atm and then use 0.08206 L atm / mol K for the value of R.Ī variety of values for R can be found here. Note the different value and unit for R, to be in agreement with using mmHg for the pressure unit.
(0.400 mol) (0.08206 L atm / mol K) (284 K)Įxample #4: Calculate the approximate temperature of a 0.300 mol sample of gas at 780. Note the conversion from mmHg to atm and from Celsius to Kelvin.Įxample #2: A sample of gas at 28.0 ☌ has a volume of 6.20 L and exerts a pressure of 720.0 mmHg. The ChemTeam will use the 0.08206 value in gas-related problems almost every time.Įxample #1: A sample of gas at 25.0 ☌ has a volume of 11.0 L and exerts a pressure of 660.0 mmHg. Those of you that take more chemistry than high school level will meet up with 8.31447 Joules per mole Kelvin, but that's for another time. (Here's a whole bunch.) It depends on which units you select. This is not the only value of R that can exist. Notice the weird unit on R: say out loud "liter atmospheres per mole Kelvin." Solving for R gives 0.08206 L atm / mol K, when rounded to four significant figures. Notice how atmospheres were used as well as the exact value for standard temperature. Let's plug our numbers into the equation: It is the volume of ANY ideal gas at standard temperature and pressure. We will use the value of 22.414 L.īy the way, 22.414 L at STP has a name. The volume of this amount of gas under the conditions of STP is known to a high degree of precision. We will assume we have 1.000 mol of a gas at STP. If you wind up taking enough chemistry, you will see it showing up over and over and over. Sometimes it is referred to as the universal gas constant. Let the cube root of k 1k 2k 3k 4k 5 / k 6 be called R.Įach unit occurs three times and the cube root yields L-atm / mol-K, the correct units for R when used in a gas law context. When you multiply them all together, you get: The subscripts on k indicate that six different values would be obtained. Note that the last law is written in reciprocal form. For a static sample of gas, we can write each of the six gas laws as follows: What follows is just one way to "derive" the Ideal Gas Law.
The Ideal Gas Law was first written in 1834 by Emil Clapeyron. There are 0.9645 Atm in 733 mmHg, or 733 mmHg = 0.9645 Atm.Fifteen Examples Boyle's Law No Name Law Ideal Gas Law Probs 1-10 Charles' Law Combined Gas Law Ideal Gas Law Probs 11-25 Gay-Lussac's Law Dalton's Law All Ideal Gas Law examples and problems Avogadro's Law Graham's Law Return to KMT & Gas Laws Menu Diver's Law
To calculate how many Atm in 733 mmHg, divide the value by 760. Online Calculators > Conversion > 733 mmHg to Atm 733 mmHg to Atmħ33 mmHg to Atm conversion calculator convert 733 mmHg to Atm and vice versa.
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