# A sample of neon of mass 425 mg occupies 6.00 dm^3 at 77 K. What pressure does it exert?

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Just one small observation added to the above answer (which is correct). 77 K is a very small temperature but in the case of Neon (liquefying temperature at normal pressure 27.07 K) it is still high enough for the gas to be considered ideal (the equation P*V =N*R*T holds). In the case of air (Nitrogen liquefiates at 77.2 K and Oxygen liquefiates at 90.2 K again at normal pressure) this condition would have not been true. Gases in nature tend to deviate from the ideal gas laws when they are subjected to conditions close to the liquefaction point.

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Molar mass of Neon = 20.18g/mol

Amount of Ne moles `= (425xx10^(-3))/20.18 = 0.021`

Assuming that Ne act as an ideal gas using `PV = nRT` ;

`P = ??`

`V = 6L`

`R = 0.08206(atmL)/(molK)`

`T = 77K`

`n = 0.021mol`

`P = (nRT)/V`

`P = (0.021xx0.08206xx77)/6 = 0.022atm`

So the pressure exerts by neon is 0.022atm.

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