Question

At what temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95 atmospheres?

Answer 1

The combined gas law equation is as follows;
PV = nRT
where P - pressure
V - volume, n - number of moles, R - universal gas constant, T - temperature
number of moles are 0.654 mol, Volume - 12.30 L, Pressure is 1.95 atm
R - 8.314 Jmol⁻¹K⁻¹
Pressure is given in atm, we have to convert it to SI units - Nm⁻²
1 atm - 101 325 Nm⁻²
Therefore 1.95 atm = 101 325 x 1.95 = 197 584 Nm⁻²
Substituting these values into the equation,
T =
`(PV)/(nR)`
T =
`(197584N m^(-2) *0.0123 m^(3) )/(0.654mol*8.314J mol^(-1) K^(-1) )`
T = 2430.28/5.44
= 447 K