Question

A 11.0−m by 8.0−m by 4.0−m basement had a high radon content. On the day the basement was sealed off from its surroundings so that no exchange of air could take place, the partial pressure of 222Rn was 1.9 × 10−6 mmHg. Calculate the number of 222Rn isotopes (t1/2 = 3.8 days) at the beginning and end of 29 days. Assume STP conditions.

(Enter your answers in scientific notation.)

Answer 1

Pressure of air=P=760 mmhg (STP)=1 atm

Temperature=T=273.15K

Volume of basement=V=11.0*8.0*4.0=352 m3=352 m3 *(1 dm3/10^-3m3)*(1L/dm3)=352000L

p(Rn)=partial pressure of radon=1.9*10^-6 mmHg * (1 atm/760mmhg)=2.5*10^-9 atm

Using ideal gas equation ,number of moles of gas (n) in the basement could be calculated.

PV=nRT

where R=universal gas constant=0.0821 L atm /Kmol

n=PV/RT=1 atm* 352000L/(0.0821 L atm K^-1 mol^-1) *273.15K=15696.336 mol

Thus, 15696.336 mol of air was present in the room.

Now , according to dalton's law of partial pressure,

X(Rn)=p(Rn)/P=(2.5*10^-9 atm)/1atm=2.5*10^-9 =mol fraction of Rn

Thus, number of mol of Rn=X(Rn)*n=(2.5*10^-9)*(15696.336 mol)=3.924*10^-5 mol

So, number of atoms of Rn=3.924*10^-5 mol* Avogadro's number=3.924*10^-5 mol*(6.022*10^23)atoms/mol=2.363*10^19 atoms

number of 222Rn isotopes at the beginning=2.363*10^19 atoms

Now, calculation of number of 222Rn isotopes at the end of 29 days

Using radioactive decay equation,

.N=No exp(-t) where,N=number of atoms after the end of time t=29 days

=decay constant=0.693/t1/2=0.693/(3.8 days)=0.182 days^-1

No=the number of 222Rn isotopes at the beginning=2.363*10^19 atoms

N=(2.363*10^19 atoms ) exp(-0.182 days^-1 * 29 days)

N=1.206*10^17 atoms of Rn remains after 29 days

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