Can someone help me with this gas law? I actually learn more if someone who knows how to do it helps. Thanks!!

Question

a helium -filled weather balloon is launched from the ground where the pressure is 752mmHg and the temperature is 21c. Under these conditions its volume is 75.0 L. When it has climbed to an altitude where the pressure is 89mmHg and temperature is 0c, what is it's volume. I get 8.24, but I think I am incorrect, can someone lend a hand.

Answer 1

P1 = 752mm
T1 = 21°C = 294k (273 for 0°C, so it's 273 + 21)
V1 = 75L.
P2 = 89mm
T2 = 0°C = 273k
V2 = ?

Using the formula:
P1V1 / T1 = P2V2 / T2
V2 = P1V1T2 / T1 P1
V2 = (752)(75)(273) / (294)(89)
V2 = 588∙4430177...
V2 ≈ 588∙44 L

Answer 2

P1 = 752mm
T1 = 21°C = 294k (273 for 0°C, so it's 273 + 21)
V1 = 75L.
P2 = 89mm
T2 = 0°C = 273k
V2 = ?

Using the combined gas law formula
it simply derived from the relationship
P inversely proportional to the temperature
P directly proportional to the vol

dont forget to use the absolute temperature
in SI deg C + 273 = deg Celcius
in English deg F + 460 = deg Rankine


P1V1 / T1 = P2V2 / T2
V2 = P1V1T2 / T1 P1
V2 = (752)(75)(273) / (294)(89)
V2 = 588∙4430177...
V2 ≈ 588∙44 L or .58844m^3

Answer 3

All good answers. The ratio PV/T remains constant, assuming that all of the assumptions for an ideal gas are satisfied.

Just as a thought experiment, it also depends on the elasticity of the balloon material. For example, if the balloon were made of a solid material that does not stretch, the volume would still be 75.0 L, and the absolute pressure inside of the balloon would decrease because of the lower temperature.

Answer 4

PV/T is constant. Initially P=752, V=75, T=294 (degrees K)
Later, P=89, V= unknown, T=273.
So, 752*75/294 = 89*V/273.
Rearrange: V = 752*75*273/(294*89) = 588 litres.