400 mL of gas is contained at 300 mmHg and 0 °C. What will its volume be at 140 mmHg and 100 °C? 0°C 100°C

1171.12 mL

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

V1 = initial volume (milliliters)

V2 = final volume (milliliters)

T1 = initial temperature (Kelvin)

T2 = final temperature (Kelvin)

According to the information provided in this question:

P1 = 300 mmHg

P2 = 140 mmHg

V1 = 400 mL

V2 = ?

T1 = 0°C = 273K

T2 = 100°C = 100 + 273 = 373K

Using P1V1/T1 = P2V2/T2

300 × 400/273 = 140 × V2/373

120000/273 = 140V2/373

120000 × 373 = 273 × 140V2

44760000 = 38220V2

V2 = 44760000 ÷ 38220

V2 = 1171.115

The new volume is 1171.12 mL

(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