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If 2. 0 x 1023 argon {ar) atoms strike 4. 0 cm2 of wall per second at a 90° angle to the wall when moving with a speed of 45,000 cm s-1, what pressure {in atm) do they exert on the wall?

Sagot :

If 2. 0 x 1023 argon {Ar) atoms strike 4. 0  [tex]cm^{2}[/tex]  of wall per second at a 90° angle to the wall when moving with a speed of 45,000  [tex]cm^{-1}[/tex].The pressure will be 0.295 atm.

The force delivered perpendicularly to an object's surface per unit area across which that force would be dispersed is known as pressure. In comparison to the surrounding pressure, gauge pressure defines the pressure. The units was using to express pressure vary.

Calculation of force:

Force excerted by the Ar atoms, (F) = Change in momentum for one Ar atom 7 / unit time × number of atom

F = 2mv /1s × number of atoms.

Putting the given data in above equation.

F = 2 × 39.95 amu × 450 m/s/ 1s × 1.66 × [tex]10^{-27}[/tex] kg /amu × 2 × [tex]10^{23}[/tex] atom

F = 11.94 N

Now, pressure can be calculated by using the formula:

P = Force exerted /area

P = 11.94 N/ 4.0 [tex]cm^{2}[/tex] ×  [tex]cm^{2}[/tex]/ [tex]10^{-4} m^{2}[/tex]

P = 0.295 atm

Therefore, the pressure will be 0.295 atm.

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