2.0 L of gas A at 1.0 atm and 1.0 L of gas B at 1.0 atm are combined in a 3.0 L flask. The flask is sealed and over time they react completely to give gas C according to the following chemical equation: 2A(g) + B(g) →C(g) Assuming the temperature stays constant, what will be the pressure after the reaction goes to completion?

0.993 atm

Explanation:

This can be calculated using Dalton's law of partial pressure which states that if two or more gases are mixed together, the total pressure exerted by the mixture of the gases is eqaul to the sum of the pressure which individual gas woud exert if it occupied the volume of of the mixture alone.

Therefore P (total) = P (gas A) + P (gas B)

Since pressure of gas A = 1.0 atm  

          pressure of gas B = 1.0 atm

          gas constant R = 0.08206 atm L mol-1 K-1

          standard temperature = 273K

Note that atm is a unit for pressure

Number of mole of gas A, n=PV/RT

= 1 x 2 / 0.08206 x 273

= 0.088306 mole

Number of mole of gas B

= 1 x 1 / 0.08206 x 273

= 0.044644 mole

Total pressure can be gotten using P(total) = (n(total)RT)/V(total)

= [(0.044644 + 0.088306) x 0.08206 x 273] / 3

= 0.993 atm

(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