A spherical balloon of volume 3.92 103 cm3 contains helium at a pressure of 1.25 105 Pa. How many moles of helium are in the balloon if the average kinetic energy of the helium atoms is 3.60 10^-22 J?

3.39

Explanation:

Given that

Volume, v = 3.92*10^3 cm³ we convert the volume from cm³ to m³ and we have 0.00392

Pressure, P = 1.25*10^5 Pa

Average Kinetic Energy, K.E = 3.6*10^-22 J

We use the gas law formula,

PV = nRT

Making n subject of the formula, we have

n = PV/RT

Solving for n, we have

n = (1.25*10^5 Pa * 0.00392 m³) / 8.314 * T

n = 4.9*10^8 / 8.314 * T

n = 490 / 8.314T

n = 58.94/T

Note that average kinetic energy is given as

K.E(avg) = 3/2K.T,

3/2 K.T = 3.6*10^-22 J

where K = 1.38*10^-23

T = (3.6*10^-22 J * 2) / (3 * 1.38*10^-23)

T = 17.39

Substitute for T, we have

n = 58.94 / 17.39

n = 3.39

b) phosphorus pentoxide

mistakes are ok