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Ideal Gas Law Unit Conversion

A nebula−a region of the galaxy where new stars are forming−contains a very tenuous gas with 100 atoms/cm3. This gas is heated to 7500 K by ultraviolet radiation from nearby stars.
What is the pressure, in atm?

I'm having troubles converting between units. So far my work is:
pV=nRT so p=nRT/V

Use avogadro's number to find n=1.66x10^-22

p=(1.66x10^-22mol)(8.314J/mol K)(7500K)/(1x0^-4m^3)
p=1.035x10^-13Pa

This is where I start to get confused.
1atm=101325Pa
9.87x10^-6atm=1Pa
1.035x10^-13Pa=1.02x10^-18atm.
Is this correct or is there something wrong in the conversions between units?
Original post by PatchworkTeapot
A nebula−a region of the galaxy where new stars are forming−contains a very tenuous gas with 100 atoms/cm3. This gas is heated to 7500 K by ultraviolet radiation from nearby stars.
What is the pressure, in atm?

I'm having troubles converting between units. So far my work is:
pV=nRT so p=nRT/V

Use avogadro's number to find n=1.66x10^-22

p=(1.66x10^-22mol)(8.314J/mol K)(7500K)/(1x0^-4m^3)
p=1.035x10^-13Pa

This is where I start to get confused.
1atm=101325Pa
9.87x10^-6atm=1Pa
1.035x10^-13Pa=1.02x10^-18atm.
Is this correct or is there something wrong in the conversions between units?


number of 1 cm cubes in 1 m cube = (number of cm in 1 m)3 and != 1.00x10^4
Original post by Joinedup
number of 1 cm cubes in 1 m cube = (number of cm in 1 m)3 and != 1.00x10^4


So does that mean that the overall pressure is

p=(1.66x10^-22mol)(8.314J/mol K)(7500K)/(1x10^4m^3)
p=1.035x10^-21Pa
p=1.02x10^-26atm?
Original post by PatchworkTeapot
So does that mean that the overall pressure is

p=(1.66x10^-22mol)(8.314J/mol K)(7500K)/(1x10^4m^3)
p=1.035x10^-21Pa
p=1.02x10^-26atm?


moles of gas in 1cm^3 = 1.66e-22
R=8.314
T=7500
V=1/(100^3) = 1.00e-6

p=nRT/V
p=(1.66e-22) (8.314) (7500) (1/1.00e-6)
=1.04e-11 (Pascals)
=1.03e-16 (atmos)

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