Chemistry Help Please!?

Question

1. If 1.710 g of a gas occupies a 500-mL flask at 30 degrees C and 750 mm Hg of pressure, what is the molecular weight of the gas? (Show work, and remember to use proper units: V in liters, T in degrees Kelvin, P in atmospheres.)

2. Can a real gas ever behave as an ideal gas? Explain.


Thank you very much in advance!

Answer 1

1. find moles. then moles = wt/ mw right>

ideally, PV = nRT

P = 750 mmHg
V = 500 ml = 0.5L
T = 30°C = 303.15K (K = C + 273.15 right)
R = 62.3637 L mmHg / moleK

n = PV/RT = 750 mmHg x .5L / (62.3637 L mmHg/moleK x 303.15K) = 0.019835 moles

moles = wt/mw, mw = wt/moles = 1.710 g / .019835 moles = 86.21 g/mole

or if you wish, convert 750 mmHg to atm by

750 mmHg x (1 atm/760 mmHg) = 0.987 atm

then R = .0821 L atm /mole K

n = n = PV/RT = .987 atm x .5L / (.0821 L atm/moleK x 303.15K) = 0.0198 moles

your choice.....works out the same


2. gases cannot actually behave ideally. the basic concepts are

1) molecular volume = 0 (that's not true for molecules right?)
2) no intermolecular forces. (again not true)
3) only elastic collisions occur. (again not true)

some gases can approximate ideality under certain conditions. They are closest to ideal if the gas is very dilute in a huge container....

for beginning chemistry classes, you can usually use the ideal gas law for your calculations. probably a good idea to state something like "assuming ideality"....

Answer 2

1. Use the ideal gas law to find the number of moles (you have V,T and P) THEN
Divide your 1.71 g by the number of moles to get mole wt.

2. At "moderate" pressures and temperatures, most gases DO behave ideally. The "moderate" is relative to a thermodynamic point called the "critical point", which is unique to each compound. For example, for water, the critical point is about 650 deg C and 100 atmospheres.