Respuesta :
Radius of Xenon = 1.3Ă—10â’8 cm
Volume = 100 ml = 0.1 L
Pressure P = 1.2 atm = 121.59 Kpa
Temperature = 281 K
R = Gas Constant = 8.31 J mol^-1 K^-1
Now find the number of atoms
PV = nRT => n = PV / RT
n = (121.59 x 0.1) / (8.31 x 281) = / 2335.11 = 0.0052
Number of atoms in a mole is same as Avogadro constant A, which is 6.02 x
10^23 particles.
n = number of atoms= 0.0052
N = number of particles
Avogadro constant A = 6.02 x 10^23
n = N/A => N = n x A = 0.0052 x 6.02 x 106^23 = 3.13 x 10^20
Volume of Xe atom which would be a sphere = (4/3) x pi x r^3
Volume = = (4/3) x 3.14 x (1.3Ă—10â’8)^3 = 9.2 x 10^-24
Volume occupied by these particles = n x Volume = 3.13 x 10^20 x 9.2 x
10^-24 = 0.00288
Fraction of volume will be = 0.00288 / 0.1 = 0.0288
Volume = 100 ml = 0.1 L
Pressure P = 1.2 atm = 121.59 Kpa
Temperature = 281 K
R = Gas Constant = 8.31 J mol^-1 K^-1
Now find the number of atoms
PV = nRT => n = PV / RT
n = (121.59 x 0.1) / (8.31 x 281) = / 2335.11 = 0.0052
Number of atoms in a mole is same as Avogadro constant A, which is 6.02 x
10^23 particles.
n = number of atoms= 0.0052
N = number of particles
Avogadro constant A = 6.02 x 10^23
n = N/A => N = n x A = 0.0052 x 6.02 x 106^23 = 3.13 x 10^20
Volume of Xe atom which would be a sphere = (4/3) x pi x r^3
Volume = = (4/3) x 3.14 x (1.3Ă—10â’8)^3 = 9.2 x 10^-24
Volume occupied by these particles = n x Volume = 3.13 x 10^20 x 9.2 x
10^-24 = 0.00288
Fraction of volume will be = 0.00288 / 0.1 = 0.0288
The volume occupied by Xe atoms at that pressure would be 0.0288ml
Data;
- radius = 1.3 * 10 ^ -8 cm
- volume = 100ml = 0.1L
- pressure = 1.2atm
- Temperature = 281K
Ideal Gas Equation
The ideal gas equation is given as
[tex]pv=nRT[/tex]
Let's use this to find the number of moles of Xe atoms
[tex]n = \frac{PV}{RT} \\n = \frac{1.2* 0.1}{0.0821*281}\\ n = 0.0052mol[/tex]
The number of Xe atoms can be gotten from Avogadro's number
Number of Xe atoms is number of moles * Avogadro's Number [tex]0.0053*6.023*10^2^3 = 3.123*10^2^1 atoms[/tex]
Since we have the value of radius of Xe atom, we can find the volume of one atom
[tex]v=\frac{4}{3} \pi r^3\\v = \frac{4}{3} * 3.14 * (1.3*10^-^8)^3\\ v = 9.198*10^-^2^4cm^3\\1cm^3=1mL\\v = 9.198*10^-^2^4mL atoms[/tex]
The volume occupied by 3.123 * 10^21 atoms will be
[tex]3.123*10^21*9.198*10^-^2^4 = 0.0288mL[/tex]
The volume occupied by Xe atoms at that pressure would be 0.0288ml
learn more on ideal gas equation here;
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