A helium balloon has a volume of 735 mL when it is at ground level. The balloon is
transported to an elevation of 5km, where the pressure is only 0.8atm. At this altitude the gas
occupies a volume of 1286mL. Assuming the temperature has remained the same, what was
the ground level pressure?
O
A.
1.4atm
OB. 0.7atm
O c. 12am
OD. 1.0atm

The answer to your question is P1 = 1.4 atm

Explanation:

Data

Volume 1 = V1 = 735 ml

Pressure 1 = P1 = ?

Volume 2 = V2 = 1286 ml

Pressure 2 = P2 = 0.8 atm

Process

To solve this problem use Boyle's law.

          P1V1 = P2V2

-Solve for P1

          P1 = P2V2 / V1

-Substitution

          P1 = (0.8 x 1286) / 735

-Simplification

          P1 = 1028.8 / 735

-Result

          P1 = 1.4 atm

bcde is the incorrect answer it is a

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied