The combustion of propane occurs when propane interacts with oxygen gas to produce water and carbon dioxide in the following reaction:
c3h8 (g) + 5o2 (g) → 3co2 (g) + 4h2o(l)
if 2 moles of propane c3h8 (g) react at 373 k in a volume of 2.5 l, what is the pressure in the cylinder from the c3h8 (g)?

Explanation:

1) Data:

a) n = 2 moles

b) T = 373 K

c) V = 2.5 liter

d) P = ?

2) Chemical principles and formula

You need to calculate the pressure of the propane gas in the mixture before reacting. So, you can apply the partial pressure principle which states that each gas exerts a pressure as if it occupies the entire volume.

Thus, you just have to use the ideal gas equation: PV = nRT

3) Solution:

P = 2 mol × 0.08206 atm-liter /K-mol × 373K / 2.5 liter = 24.5 atm

Since the number of moles are reported with one significant figure, you must round your answer to one significant figure, and that is 20 atm (20 is closer to 24.5 than to 30).

(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