A sample of 6.022 x 1023 particles of gas has a volume of 22.4 L at 0°C and a pressure of 1.000 atm. Although it may seem silly to contemplate, what volume would 1 particle of gas occupy? pv=nRT

1 particle of the gas would occupy a volume of 3.718*10⁻²³L

Explanation:

Hello,

1. The sample has a particle of 6.022×10²²particles

2. Volume of the sample = 22.4L

3. Temperature of the sample = 0°C = (0 +273.15)K = 273.15K

4. Pressure of the sample = 1.0atm

What volume would 1 particle of the gas occupy?

But we remember that 1 mole of any substance = 6.022×10²² molecules or particles or atoms

What would be the number of moles for 1 particule?

1 mole = 6.022×10²² particles

X moles = 1 particle

X = (1 × 1) / 6.022×10²² particles

X = 1.66×10⁻²⁴ moles

Therefore, 1 particle contains 1.66×10⁻²⁴ moles

Since we know our number of moles, we can proceed to use ideal gas equation,

Ideal gas equation holds for all ideal gas and is defined as

PV = nRT

P = pressure of the ideal gas

V = volume the gas occupies

n = number of moles of the gas

R = ideal gas constant = 0.082 L.atm / mol.K

T = temperature of the gas

PV = nRT

Solving for V,

V = nRT/ P

We can now plug in our values into the above

equation.

V = (1.66*10⁻²⁴ × 0.082 × 273.15) / 1

V = 3.718*10⁻²³L

Therefore, 1 particule of the gas would occupy a volume of 3.718*10⁻²³L.

(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