The natural gas in a 30-l ngv fuel tank has a pressure of 2.05* 10^4 kpa at a temperature of 297 k. how many kilograms of fuel are in the tank

4727.2  × 10³ Kg

Explanation:

Given data:

Pressure of gas = 2.05 ×10⁴ KPa (202.32 atm)

Temperature = 297 K

Volume = 30 L

Mass of fuel = ?

Solution:

Molar mass of gas is missing. It is 19 g/mol.

Formula:

PV = nRT

n = PV/RT

n = 202.32 atm × 30 L / 0.0821 atm. L. mol⁻¹.k⁻¹ × 297 K

n = 6069.6 atm.L / 24.4  atm. L. mol⁻¹

n = 248.8 mol

Mass:

Number of moles = mass/ molar mass

Mass = number of moles × molar mass

Mass = 248.8 mol × 19 g/mol

Mass = 4727.2 g

In kilogram:

4727.2 g /1 g × 1000 Kg

4727.2  × 10³ Kg

(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