Question

A gas occupies 0.418 L at 740 mm of Hg at 27 C. Calculate (i) molecular weight, if gas weighs 3.0 g (ii) New pressure of gas, if the weight of gas is increased to 7.5 g and the temperature become 280 K.

Answer 1

1) Calculate the molecular weight:
Ideal gas law states that 1 mole of an ideal gas occupies 22.4L
therefore moles of this gas will be:
if 22.4l= 1 mole
then 0.418l = 1 x 0.418     =  0.01867 moles
                       22.4

moles = mass/ molecular weight
molecular weight = mass/moles
                         =3g/0.01867
                         160.69
2) New pressure if weight increases to 7.5.

Find the moles of the gas using the new mass
moles= mass/molar mass
         =  7.5/160.69
         = 0.0467 moles
Find volume:
If 1 mole occupies 22.4 L
then 0.0467 moles = 22.4 x 0.0467
                           = 1.05 liters
1.05 liters = 1050 cc
The gas initial measurements= 418 cc, 740mmHg, 27.C
resultant measurements=  1050 cc, P2, 7C(-273K)

combined gas law:   P1V1/T1 =P2V2/T2

Therefore  740 x 418   =  1050 x P2
                    27                8 

27 x 1050 x P2 =  740 x 418 x 8
28350P2  =  2474560

P2 =  2474560/28350
     = 87.29 mmHg

Therefore the new pressure = 87.29 mmHg