The required volume of sample of neon occupy at these conditions of temperature and pressure is 25.2L.
Mass (W) in grams of helium gas will be converted into moles (n) by using the below equation:
n = W/M, where
M = molar mass = 4g/mol
n = 12g / 4g/mol = 3 mol
n = 24g / 4g/mol = 6 mol
And for this question as pressure and temperature for both the conditions is constant so ideal gas equation becomes:
V₁/n₁ = V₂/n₂, where
V₁ = initial volume = 12.6 L
n₁ = initial moles = 3 mol
V₂ = final volume = ?
n₂ = final moles = 6 mol
On putting these values, we get
V₂ = (12.6)(6) / (3) = 25.2 L
Hence required volume of neon gas is 25.2 L.
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