A weather balloon containing helium with a volume of 410.0 L rises in the atmosphere and is cooled from 17∘ ∘
C to -27∘
∘
C. The pressure on the gas is reduced from 110.0 kPa to 25.0 kPa. What is the volume of the gas at the lower temperature and pressure? What law did you use to solve?

V₂ = 1500 Liters ( 2 sig. figs.)

Explanation:

Given the following gas law variables:

P₁ = 110.0KPa           P₂ = 25KPa

V₁ = 410 Liters          V₂ = ?

T₁ = 17°C ( = 290K)   T₂ = -27°C  ( = 248K)

P₁V₁/₁T₁ = P₂V₂/T₂  =>  V₂ = V₁(P₁/P₂)(T₂/T₂)

V₂ = 410L(110.0KPa/25KPa)(248k/290K) = 1542 L (calc. ans.)

V₂ = 1500 Liters ( 2 sig. figs.)

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.

(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