Gas Law problem help?

Question

A spherical balloon containing 5.035 mg of hydrogen gas in its standard state of

aggregation behaving ideally is attached to one end of an open tube cylindrical

mercury manometer (see figure) where atmospheric pressure is 101325 Pa and the

temperature is 293.15 K? If the tube has a radius of 1.000 mm, there is 50.00 cm

between the balloon and the mercury, and a difference in height between the

mercury levels of 45.00 cm. What is the volume of the balloon in m3? What is the

corresponding diameter in cm of the balloon assuming it is a sphere? You may

assume that the density of the mercury is 13.5462 x 103 kg/m3 and that g = 9.807

m/s2. Hint: Determine the pressure and use this information to deduce the

total volume of the system. Then use the total volume and the dimensions of

the tube filled with H, to determine the volume of the balloon.

R = ?

50.00 cm

45.00 cm

patm

Answer 1

The pressure of the gas is atmospheric pressure plus 450mm Hg (45.00 cm x 10). Now 1 atm = 760 mm Hg and 101325 Pa = 1 atm Therefore the pressure of the gas in atm is

P = 1 + 450/760 = 1.592 atm

The ideal gas law

PV = nRT    rearrange to give V = nRT/P

n calculate mol H2 from the mass of H2

R is the gas constant 0.0821

T is temp in Kelvin - I assume standard state here means 25 C, or 298 K (it does not say STP that would be 0 C or 273 K)

P we just calculated

This gives V in liters. To get V in m^3 divide by 1000

The total volume will be the volume of the balloon plus the volume of the tube up to the mercury.

The tube is a cylinder. Vol of a cylinder

V = pi r^2 h

pi = 3.141

r = 1.000 x 10^-3 m

h = length of tube, 0.5000 m

this gives V in m^3

Vol balloon = Vol gas - vol cylinder

Assume balloon is a sphere

V = 4/3 pi r^3

V = vol balloon solve for r the radius of the balloon in m