kinetic thoery

y









x









z

y









vx



x









x  vx (t )

z

Number of collisions with the wall per time interval Δt?

• Number of particles that can get to the wall in Δt!



 Must be x from the wall and headed toward the wall,

where x  vx (t )

y









vx



x









x  vx (t )

z

Number of collisions with the wall per time interval Δt?

• Number of particles that can get to the wall in Δt!



 Must be x from the wall and headed toward the wall,

where x  vx (t )

y







A





vx



x









x  vx (t )

z

Number of collisions with the wall per time interval Δt?

A = end wall area

V = volume containing particles that can reach wall in Δt

V  Avx (t )

y

V  Avx (t )



A





vx



x









x  vx (t )

z

Number of collisions with the wall per time interval Δt?

N = number of particles

N N

N = concentration of particles  

V Avx t

N = NAvxΔt

y

V  Avx (t )

N = NAvxΔt

A





vx



x









x  vx (t )

z

Number of collisions with the wall per time interval Δt?



On average: 1 N moving left, 1 N moving right

2 2

y

V  Avx (t )

N = NAvxΔt

A





vx



x









x  vx (t )

z

Number of collisions with the wall per time interval Δt?



Number of collisions

1 1 NAv Δt

with the wall per time  N  x

2 2

interval Δt?

y

V  Avx (t ) Number of collisions

N = NAvxΔt with the wall per time

A interval Δt?

1

 NAvxΔt

2

vx



x









x  vx (t )

z Momentum change per collision?

-vx

p  p f  pi  mvx  (mvx )  2mvx

vx

y

V  Avx (t ) Number of collisions

N = NAvxΔt with the wall per time

A interval Δt?

1 NAv Δt

 x

2

vx Momentum change

per collision?



x









x  vx (t )

z Momentum change per collision?

-vx

p  p f  pi  mvx  (mvx )  2mvx

vx

y

V  Avx (t ) Number of collisions

N = NAvxΔt with the wall per time

A interval Δt?

1 NAv Δt

 x

2

vx Momentum change

per collision?

p  2mvx





x  vx (t )

z Total Δp in Δt ?

1

pt  (# collisions)(p)  ( NAvxΔt) ( 2mv x )

2

y

V  Avx (t ) Number of collisions

N = NAvxΔt with the wall per time

A interval Δt?

1 NAv Δt

 x

2

vx Momentum change

per collision?

p  2mvx

Total Δp in Δt

1

pt  ( 2 NAvxΔt) (2mv x )

x  vx (t ) N mAv 2 t

x

z Force on wall during Δt?

Ft  pt

pt NmAv x t

2

F = N mAv x

2

=

t t

y

V  Avx (t ) Number of collisions

N = NAvxΔt with the wall per time

A interval Δt?

1 NAv Δt

 x

2

vx Momentum change

per collision?

p  2mvx

Total Δp in Δt

pt  N mAvx t

2



x  vx (t )

z Force on wall during Δt ?

Ft  pt

pt NmAv x t

2

F = N mAv x

2

=

t t

y

V  Avx (t )

N = NAvxΔt

A





vx









x  vx (t )

z Force on wall during Δt ?

Ft  pt

pt NmAv x t

2

F  = N mAv x

2

=

t t

y

V  Avx (t )

Force on wall during Δt

N = NAvxΔt

F = NmAv x

2



A

Pressure on the

wall during Δt?

vx









x  vx (t )

z Pressure on the wall during Δt? Not all particles travel

2

F NmAvx N mvx

 at same speed so

2

P = = detected pressure is

A A the result of the

average of those speeds.

y

V  Avx (t )

Force on wall during Δt

N = NAvxΔt

F = NmAv x

2



A

Pressure on the

wall during Δt?

vx P =N mvx2









x  vx (t )

z

y



Force on wall during Δt

F = NmAv x

2

vy



Pressure on the

v wall during Δt?

P =N mvx2

vx

x

2

vz All particles moving randomly. Therefore v x

is same as average of corresponding quantities

in y and z directions.

v 2  vx  v y  vz2

2 2



z

v 2  vx  vx  vx

2 2 2





v 2  3vx

2



1 2

v  vx

2



3

y



Force on wall during Δt

F = NmAv x

2

vy



Pressure on the

v wall during Δt?

P =N mvx2

vx

x

2

vz All particles moving randomly. Therefore v x

is same as average of corresponding quantities

in y and z directions.

1 2

v  vx

2



z 3

y



Force on wall during Δt

F = NmAv x

2

vy



Pressure on the

v wall during Δt?





vx

x

2

vz All particles moving randomly. Therefore v x

is same as average of corresponding quantities

in y and z directions.

1 2

v  vx

2

P =N mvx2

z 3 N

1N 2 N 

P mv

V

3

1N

P mv 2

3V

y



Force on wall during Δt

F = NmAv x

2

vy



Pressure on the

v wall during Δt?





vx

x

2

vz All particles moving randomly. Therefore v x

is same as average of corresponding quantities

in y and z directions.

1 2 2 1 2

v  vx

2

P =N mvx2 PV  N ( mv )

z 3 3 2

1N 2

P mv

3

1N

P mv 2

3V

PV 

2 1 2

N ( mv )

3 2

PV  NkT

PV 2 1 2 PV

 ( mv ) kT 

N 3 2 N

2 1 2 1 2

kT  ( mv ) KE  ( mv )

3 2 2

2

kT  ( KE )

3

3

KE  kT

2