At a pressure of 9.10x10-14 atm and an ordinary temperature of 293.0 K, how many

Question

At a pressure of 9.10x10-14 atm and an ordinary temperature of 293.0 K, how many Constants molecules are present in a volume of 1.00 cm3 Modern vacuum pumps make it easy to attain pressures of the order of 10-13 atm in the laboratory You may want to review (Pages 590-593) For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Atomic and molecular mass mo Submit Request Answer Part B How many molecules would be present at the same temperature but at 1.04 atm instead? mol

Explanation / Answer

Given

pressure P = 9.10*10^-14 atm , Temperature T = 293.0 k

volume V = 1.00 cm^3

number of moles n =?

number of molecules N = ?

From ideal gas equation PV= n*R*T

Part A

we know 1 atm = 1.01325*10^5 Pa, R is universal gas constant = 8.314 J/k.mol, v = 1 cm^3 = 1*(10^-2)^3= 1*10^-6 m^3

substituting the values

9.10*10^-14*1.01325*10^5*10^-6 = n*8.314*293

n= 3.785*10^-18

from Avagadro's law one mole of gas contains 6.023*10^23 molecules

so

3.785*10^-18 moles contains

N = 3.785*10^-18 *6.023*10^23 molecules

N = 2279706 molecules

Part B

now the pressure is 1.04 atm = 1.04*1.01325*10^5 Pa

1.04*1.01325*10^5*10^-6 = n*8.314*293

n= 4.325858*10^-5

from Avagadro's law one mole of gas contains 6.023*10^23 molecules

so

4.325858*10^-5 moles contains

N = 4.325858*10^-5 *6.023*10^23 molecules

N = 2.60546*10^19 molecules

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