On average what is the time between collisions of a xenon atom at 300 K and (a) one torr pressure; (b) one bar pressure.

(a). 132 × 10^-9 s = 132 nanoseconds.

(b)..176.5 pico-seconds.

Explanation:

(a). At one torr, the first thing to do is to find the speed and that can be done by using the formula below;

Speed = [ (8 × R × T)/ Mm × π]^1/2.

Where Mm = molar mass, T = temperature and R = gas constant.

Speed= [ ( 8 × 8.314 × 300)/ 131.293 × π × 10^-3)^1/2. = 220m/s.

The next thing to do now is to calculate for the degree of collision which can be calculated by using the formula below;

Degree of collision = √2 × π × speed × d^2 × pressure/ K × T.

Note that pressure = 1 torr = 133.32 N/m^2 and d = collision diameter.

Degree of collision = √2 × π × 220 × (4.9 × 10^-10)^2 × 133.32/ 1.38 × 10^-23 × 300.

Degree of collision = 7.55 × 10^6 s^-1.

Thus, 1/ 7.55 × 10^6. = 132 × 10^-9 s = 132 nanoseconds.

(b). At one bar;

1/10^5 × 10^3 × 56.65 = 1.765 × 10^-10 = 176.5 pico-seconds.

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.