Using the formula for the ideal gas law and the value for the gas law constant of 0.08206 L.atm/K/mol, what is the volume (in L) of 7.22 grams of dry hydrogen at 25.6 degrees C and 753 torr?

• PV = nRT -> V = (nRT) / P

P = 753 torr x (1 atm / 760 torr) = 0.991 atm

n = mass x (1 mol / molar mass) = 7.22 g H2 x (1 mol / 2.02 g H2) = 3.57 mol H2

R = 0.08206 L.atm/K/mol

T = 25.6 degrees C + 273 K = 298.6 K

• V = (nRT) / P

V = (3.57 mol H2 x 0.08206 L.atm/K/mol x 298.6 K) / 0.991 atm = 88.3 L

WRONG

Answer: 88.6 L

A student obtains the following data by following an experimental procedure to this one. Based on this data, what is the value of R (in L.atm/mol/K) that would be calculated? Report the result to four significant figures. Use 273.15 K for the temperature conversion.

Pressure of dry H2 = 741.4 mm Hg

Temperature = 21.00 degrees C

Mass of Mg used = 0.04100 g

Volume of H2 = 44.60 ET units

Conversion factor = 0.8850 mL/ET Unit

• PV = nRT -> R = (PV) / nT

P = 741.4 mm Hg x (1 atm / 760 mm Hg) = 0.9755 atm

V = 0.8850 mL/ET Unit (fudge factor) x 44.60 ET units x (1 L / 1000 mL) = 0.03947 L

n = mass x (1 mol / molar mass) = 0.04100 g Mg x (1 mol / 24.31 g Mg) = 0.001687 mol Mg

T = 21.00 degrees C + 273.15 K = 294.15 K

• R = (PV) / nT

R = (0.9755 atm x 0.03947 L) / (0.001687 mol Mg x 294.15 K) = 0.07759 L.atm/mol/K

Using the formula for the ideal gas law and the value for the gas law constant of 0.08206 L.atm/K/mol, what is the volume (in L) of 6.23 grams of dry hydrogen at 22.6 degrees C and 764 torr?

• PV = nRT -> V = (nRT) / P

P = 764 torr x (1 atm / 760 torr) = 1.00526 atm

n = mass x (1 mol / molar mass) = 6.23 g H2 x (1 mol / 2.02 g H2) = 3.08415 mol H2

R = 0.08206 L.atm/K/mol

T = 22.6 degrees C + 273 K = 295.6 K

• V = (nRT) / P

V = (3.08415 mol H2 x 0.08206 L.atm/K/mol x 295.6 K) / 1.00526 atm = 74.4 L

A student obtains the following data by following an experimental procedure to this one. Based on this data, what is the value of R (in L.atm/mol/K) that would be calculated? Report the result to four significant figures. Use 273.15 K for the temperature conversion.

Pressure of dry H2 = 738.3 mm Hg

Temperature = 23.00 degrees C

Mass of Mg used = 0.03830 g

Volume of H2 = 44.20 ET units

Conversion factor = 0.8886 mL/ET Unit

• PV = nRT -> R = (PV) / nT

P = 738.3 mm Hg x (1 atm / 760 mm Hg) = 0.9714 atm

V = 0.8886 mL/ET Unit (fudge factor) x 44.20 ET units x (1 L / 1000 mL) = 0.03928 L

n = mass x (1 mol / molar mass) = 0.03830 g Mg x (1 mol / 24.31 g Mg) = 0.001575 mol Mg

T = 23.00 degrees C + 273.15 K = 296.15 K

• R = (PV) / nT

R = (0.9714 atm x 0.03928 L) / (0.001575 mol Mg x 296.15 K) = 0.08180 L.atm/mol/K