J.R. S. answered • 03/23/21

Ph.D. University Professor with 10+ years Tutoring Experience

These aren't that difficult, if you keep a few things in mind. Always use Kelvin as the units of temperature. Be sure to use the correct gas constant with the correct units. Know and understand the relationship between volume, pressure and temperature (and of course, moles).

So, looking at problem 1. We have 0.825 g of SF6 at 15ºC (288K) and it occupies 0.605 L. We are asked to find the pressure (P) in units of torr. So, we use the ideal gas law PV = nRT and solve for P but will then convert from atmospheres to torr, because the R (gas constant) that I'll use is in units of atmospheres.

molar mass SF6 = 146 g

moles (n) = 0.825 g x 1 mol/146 g = 0.00565 moles

P = ?

V = 0.605 L

R = 0.0821 Latm/Kmol (note the units)

T = 288K

P = nRT/V = (0.00565 mol)(0.0821 Latm/Kmol)(288K) / 0.605 L

P = 0.134 atm

Converting to torr, we have 0.134 atm x 760 torr/atm = 102 torr

Problem 2:

moles CO = 6.729 g CO x 1 mol CO/28 g = 0.2403 moles CO

moles CO2 = 6.729 g CO2 x 1 mol CO2/44 g = 0.1529 moles CO2

Since the pressure and temperature are constant, the volume will vary directly with the number of moles. So, the volume of CO will be 1.57 times that the volume of the CO2. If you want the actual values, we can use the ideal gas law:

PV = nRT

V = nRT/P

CO: V = (0.2403)(8.314)(320) / 90.3 = 7.08 L (note R is LkPa/Kmol; this is different than R = 0,0821 Latm/Kmol)

CO2: V = (0.1529)(8.314)(320) / 90.3 = 4.50 L

Difference is 2.58 L