The volume of a sample of a gas is 1.0 liter at STP. If the pressure remains constant and the temperature is raised to 546 K, the new volume of the gas will be

A gas occupies 1.0 L at STP. If the temperature is raised to 546 K, the volume increases to 2.0 L.

The volume of a sample of a gas is 1.0 L (V₁) at standard temperature (T₁ = 273.15 K) and standard pressure (1 atm).

If the pressure remains constant and the temperature is raised to 546 K (T₂), we can calculate the new volume (V₂) using Charles' law.

A gas occupies 1.0 L at STP. If the temperature is raised to 546 K, the volume increases to 2.0 L.

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2 L is the new volume

Explanation:

We can apply the Ideal Gases Law to solve the problem.

At STP, we kwow that 1 mol of gas occupy a volume of 22.4 L

Then, how many moles do we have in 1 L?

Let's do it by a rule of three:

(1L . 1 mol) / 22.4L = 0.0446 moles

These moles are at 1 atm and 273 K of temperature, so let's change our conditions.

P . V = n . R .T

1 atm . V = 0.0446 mol . 0.082 L.atm/mol K . 546 K

V = 2 L

If we pay attention, we can notice that, if we double temperature, we double the volume.

(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