For a 0.300 mol sample of helium gas in a 0.200 L container at 248K, will the pressure be greater if calculated with the ideal gas law or the van der Waals equation, and by roughly how much? (For He, a=0.0342L2atmmol2,b=0.0237 Lmol)

It changes by roughly 1 atm.

Explanation:

Hello!

In this case, since the ideal gas equation differs from the van der Waals' one by the presence of the a and b parameters which correct the assumption of no interactions into the container, they are written as:

Thus, the pressure via the ideal gas equation is:

And the pressure via the van der Waals equation, considering the molar volume (vm=0.200L/0.300L=0.667L/mol) is:

It means that the pressure change by 1 atm, which is not a significant difference for helium.

answerleaving cheese out to get moldy;

(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