How Many Molecules Are in 5.00 cm³ at 1 * 10^-10 mm Hg and 30.0°C?

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Homework Help Overview

The original poster is exploring the calculation of the number of molecules in a specified volume of gas at a very low pressure and a given temperature. The context involves concepts from gas laws and molecular mass, particularly in relation to mercury.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the ideal gas law (PV = nRT) and consider the implications of using molecular mass versus moles. There is also a suggestion to use an alternative equation (PV = NkT) for simplification. Questions arise regarding the atomic mass of mercury and the correct interpretation of the problem.

Discussion Status

There is an ongoing exploration of different methods to approach the problem, with some participants questioning the assumptions made about the gas and the calculations performed. Guidance has been offered regarding the use of moles and the ideal gas law, but no consensus has been reached on the correct answer.

Contextual Notes

Participants note that mercury is a metal and typically does not refer to "molecules," which raises questions about the appropriateness of the original approach. There is also mention of using van der Waals equations for comparison, indicating a potential complexity in the problem setup.

mawalker
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Current vacuum technology can achieve a pressure of 1 * 10^-10 mm of Hg. At this pressure, and at a temperature of 30.0 C, how many molecules are in 5.00 cm3 ?

ok so atomic number of Hg = 80

80 * 1.661 * 10^-27 yields 1.33 * 10 ^-25 which is the mass of one molecule

converting from mm to atmospheres i get 1*10^-10 mm = 1.32 * 10 ^-13 atm since 760 mm = 1 atm

T = 303
5 cm3 = 5 ml = 5 * 10 ^ -3

so using pV = nRT
(1.32*10^-13)(5*10^-3) = n(.08206)pressure constant(303)
solving for n... n = 2.6459

n = M(in grams)/M of molecule so :
2.6459 = M/80
which gives me 211.67 grams...

N (number of molecules = M/m = .21167 Kg/1.33 * 10^-25 (mass of one molecule)

which yields me 1.59 * 10^ 24 molecules... this answer is incorrect... any ideas?
 
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actually i think that i should be using 201 for the atomic number of mercury... this still ends up yielding 3.99 *10^24 which is still incorrect.
 
actually, it yields 1.59 * 10^24...
 
I'm not sure why you are going into mass of molecules... if you use PV=nRT, n is the number of moles. You can convert directly from moles to molecules using good ole 6.02E23. According to the IDEAL GAS LAW, the same number of moles of a gas ALWAYS occupy the same volume... regardless of WHAT the gas is.
 
Incidentally, Hg (mercury) is a metal, and with metals we never speak about "molecules". In this problem it instructve to use the van der Waals state law, too and compare the 2 results.

Daniel.
 

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