1) What is the nN? (molecules/m3) P = nNkT Atmospheric pressure: 14lb/in2 = 9842

ID: 1480164 • Letter: 1

Question

1) What is the nN? (molecules/m3)

P = nNkT

Atmospheric pressure: 14lb/in2 = 9842.97kg/m2

k = 1.38 X 10-23 J/K

T = 293.35 K

nN = 2.43 X 1024

**Not sure what I am doing wrong but the density has to be in molecules/m3

The diameter of the bulb is about 10 cm. about how many moles of air are you working with?

Assuming that the bulb is spherical, the volume would be equal to V, where r = 5cm.

V= 4/3pi(5cm)^3 = 523.6 cm = 5.24 m

Using atmospheric pressure: 14lb/in2 = 9842.97kg/m2 (FROM EXPERIMENT)

PV = nRT (9842.97 kg/m2) (5.24 m) = n (8.31 J/mole/K) (295.35 K)

**how would the Joules cancel?

Here ,

as P = nN * k * T

P = 1.01 *10^5 Pa

k = 1.38 X 10-23 J/K

T = 293.35 K

putting in equation

1.01 *10^5 = 1.38 X 10^-23 * nN * 293.35

nN = 2.494 *10^25 molecule/m^3

the density of particles is 2.494 *10^25 molecule/m^3

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r = 5 cm

V = (4/3) * pi ( r^3)

V = (4/3) * pi * (0.05)^3 m^3

V = 5.235 *10^-4 m^3

atmospheric pressure , P = 1.01 *10^5 Pa ( in standard unit )

you should change Kg to N (1 Kg = 9.8 N )

Using ideal gas equation

P * V = n * R * T

1.01 *10^5 Pa * 5.235 *10^-4 m^3 = n * 8.314 J/(K . mol) * 293.35 K

as 1 Pa . 1 m^3 = 1 N//m^2 * m^3 = 1 N . m = 1 J

solving for n

n = 0.0216

the number of moles in bulb is 0.0216

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