The problem specifies P and T are also constant.Р”Тђ) Since there is a VM/m for nitrogen and a VM/m for argon, we have this:Р”В¬ontainer A (nitrogen): V 1 = 4.4V 2, M 1 = 28.0 g/molР”В¬ontainer B (argon): V 2 = V 2, M 2 = 40.0 g/mol N = m/M, where M is the molar mass of the gas and m is the mass of the gasР”Сћ) Substituting one into the other, we have this:Р”Р€) Some factors are constant, some are variable: What is the mass of the Ar (in g) within container B?Р”РЋ) For this problem, there are two equations of interest:

(22.4 L / 1.00 mol) = (x / 1.27 mol) 1.50 РІв‚¬вЂ™ 0.046117 = 1.45 g H 2O liquid (to three sig figs)Р”В«onus Problem #2: Container A holds N 2 gas with a mass of 56.2 g and is 4.4 times the volume of container B which holds argon (Ar) gas at the exact same temperature and pressure. Problem #16: What volume will 1.27 moles of helium gas occupy at STP? Problem #15: How many moles of a gas would be present in a gas trapped within a 37.0 liter vessel at 80.00 РІВРЉ at a pressure of 2.50 atm? Problem #14: How many moles of gas would be present in a gas trapped within a 100.0 mL vessel at 25.0 РІВРЉ at a pressure of 2.50 atmospheres? Note the conversion from mmHg to atm in the denominator. Problem #13: Calculate the volume 3.00 moles of a gas will occupy at 24.0 РІВРЉ and 762.4 mm Hg. Multiply the answer (which is in atm) by 760.0 mmHg atmР’Р‡ 1 to get mmHg If we used g, the mol unit in R would not cancel and we need to have it cancel because we require atm (and only atm) to be in the answer. This is done in order to convert grams to moles, because the value for R contains mol as the unit for amount of substance. Please note the division of 1.09 by 2.02. What is the pressure in this container in mmHg? Problem #12: 1.09 g of H 2 is contained in a 2.00 L container at 20.0 РІВРЉ. If we used mmHg, the pressure units would not cancel and we need to have them cancel because we require mol (and only mol) to be in the answer. This is done in order to convert the pressure from mmHg to atm, because the value for R contains atm as the pressure unit. Problem #11: How many moles of gas are contained in 890.0 mL at 21.0 РІВРЉ and 750.0 mm Hg pressure?

This is why I don't bother to list R with two sets of units (and nor do most data sheets) and just recognise that R will work with either the combination (P in kPa, V in L / dm 3) or the SI option (P in Pa, V in m 3).ChemTeam: Ideal Gas Law: Problems #11 - 25

And thus, 1 L kPa K -1 mol -1 = 1 J K -1 mol -1.

This is the same as R = 8.314 J K -1 mol -1 as, first: You can have an R in bar or other units, but it is generally easier to change to P to atm or kPa or Pa and use the above Rs.Р”Р„ddendum / Note: A link above gives R = 8.314 L kPa K -1 mol -1. As with all cases, keeping track of units means making sure that they suit the situation rather than applying blanket rules blindly.) (Note: SI would say mas must be in kg, but since we seek moles and M is usually in g mol -1, using m in g is usually appropriate. With the above R values, or a change to PV = mRT with R a gas constant with units of g or kg. The ideal gas law can be modified to uses masses in kg rather than chemical amount in moles, but this also means the equation becomes You cannot have a temperature in Celsius and use PV = nRT because Charles' Law tells us that V is proportional to T only if T = 0 corresponds to the theoretical V = 0 limit for a gas. The ideal gas law and underlying theory requires a thermodynamic temperature with a zero that is physically meaningful and corresponding to absolute zero. If given a temperature not in K, convert to K. 747 mmHg), convert to a suitable pressure unit and apply above. If given a pressure in some other unit (ie. If given a volume in some other unit (like mL or cm 3), convert to L or m 3 as appropriate. This is the standard value which relates to

P in kPa, V in L = dm 3, n in mol, T in K.

R = 8.314 J K -1 mol -1 is the value in SI units.

747 mmhg to atm