Question

A gas at 250K and 15atm has a molar volume smaller than calculated from perfect gas law
Calculate:
a) the compression factor under these conditions
b) the molar volume of gas

Answer 1

1st think that: perfect gas law means PV = nRT, and molar volume is V/n = Vm (m is subscript actuall)

so, it can be PVm = RT

Z = PVm/RT = 1 for perfect gas (or ideal gas)

now, we know that" molar volume smaller than calculated from perfect gas law "

so Vm(gas) = (100-12)%Vm=0.88Vm

a) Z= PVm(gas)/RT
= P(0.88Vm)/RT = 0.88 ~ solved.

[Hint: Z is dimensionless, so is not unit]

b) you can look for some data table to know R= 0.08206 (atm)(L)/(K)(mol);
or you can use the SI unit:
R= 8.314 J/(K)(mol) , and change unit by yourself.

from a)

Z= 0.88= PVm(gas)/RT
= (15)(Vm(gas))/(0.082)(250)
=0.73 (Vm(gas))

calculated finally,

Vm(gas)= 0.88/0.73
= 1.2 (L/mol) ~ solved.

Answer 2

<Note>
I find the problem from the testbook: Physical Chemistry 10th by Atkins.

In this book, It gives "gas at 250 K and 15 atm has a molar volume 12 per cent smaller than that calculated from perfect gas law", so I used the condition"12%".

If your professer deleted the condition, please use x stand for 12%. Of course the solution is going to exist unknown number "x".

Answer 3

Thank you