A sample of gas has a volume of 50.0 L at 135*C. Assuming the pressure
remains constant, what is the volume of the gas if the temperature is
decreased to 72.0*C?
42.3 L
O 84.6L
O 423 L
O 26.7L
O 22.4L

42.3L

Explanation:

Given that,

Original volume of gas (V1) = 50.0L

Original temperature of gas (T1) = 135°C

[Convert 135°C to Kelvin by adding 273

135°C + 273 = 408K]

New volume of gas (V2) = ?

New temperature of gas (T2) = 72°C

[Convert 72°C to Kelvin by adding 273

72°C + 273 = 345K]

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

50L/408K = V2/345K

To get the value of V2, cross multiply

50L x 345K = 408K x V2

17250L•K = 408K•V2

Divide both sides by 408K

17250L•K /408K = 408K•V2/408K

42.28L = V2

Round the volume to the nearest tenth, so V2 becomes 42.3L

Thus, the volume of the gas if the temperature is decreased is 42.3 liters.

(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

0.88m hope i am correct : )