You are given
the density of helium and neon gas mixture which is 0.1562 grams per liters at 40.2degC and 296 torr. You are required to find the mole fraction of neon in this mmixture.
c = dRT/P
c = (0.1562g/L)(0.08206 L-atm/mol-K)(40.2+273)K/296torr(1atm/760torr)
c = 10.31 g/mol
since helium is 4 g/mol
mole fraction of neon = (10.31 - 4) / 10.31 = 0.612
Answer:
6 atm
Explanation:
Using the formula P1V1=P2V2
P1= Initial Pressure
V1= Initial Volume
P2= Final Pressure
V2= Final Volume
And knowing that at stp gas will always be at 1 atm
250L(P2) = 1500
P2= 6 atm
1.979 multiply 1979 by 0.001