Show your work and include units with your answers for full credit. A helium bal

Question

Show your work and include units with your answers for full credit. A helium balloon (assumed to be a sphere) contains 0.150 moles of helium. It is initially at 19 degree C and 1 atm of pressure. What is the diameter of the balloon? The volume of a sphere is 4/3 pi r3. How many helium atoms arc in the balloon? To what temperature must the balloon be cooled in order to decrease its diameter by 1 cm? Assume that the pressure is kept constant and that there are no leaks in the balloon. Equations; PV = nRT = NkT TK = TC + 273.15 NA = 6.02 times 1023 N = nNA R = 8.31 J/(mol K) k = 1.38 times 10-23 J/K 1 atm = 1.013 x 105 Pa

Explanation / Answer

A) Given that no.of moles n = 0.15

Temparature T = 19+273 = 292 K
Pressure P= 1 atm

R is universal gas constant = 8.205 *10^-5 atm.m^2.K^-1.mol^-1

from ideal gas equation P*V = n*R*T

Volume V = n*R*T/P = (0.15*8.205*10^-5*292)/(1)= 0.00359 m^3

but V = (4/3)*pi*r^3 = 0.00359
r^3 = 0.00359*3/(4*3.142) = 0.000856

r = (0.000856)^(1/3) = 0.0949 m
radius r= 0.0949 m

diameter d = 2*r = 2*0.0949 = 0.1898 m = 18.98 cm

B) No.of atoms are 0.15*6.022*10^23 = 9.033e+22 atoms

C) if pressure remains constant

V1/T1 = V2/T2

[(4/3)*pi(D1/2)^3]/T1 = [(4/3)*pi(D2/2)^3]/T2

D1^3/T1 = D2^3/T2

but D2 = 18.98-1 = 17.98 cm

D1 = 18.98 cm
T1 = 19+273 = 292 K

T2 = (D2/D1)^3*T1

T2 = (17.98/18.98)^3*292 = 248.23 K

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