Molecular volume of any perfect gas at 600 × 10³ N/m² and 27°C will be

@123 Hemant Value of Universal Gas Constant (R') is 8.314 KJ/Kmole K.
PV = nMRT .. eqn 1 where n = No. of molecule, M = Molecular weight, R = Characteristic Gas Constant

PV = nR' T ... eqn 2 .. (R' = MR)

R' =MR
R' = 29*0.287 ..(Molecular wieght of air is 29. We considered air coz is ideal gas)
R'= 8.314 KJ/Kmole K

The question is actually wrong. The question should be Molecular volume of any perfect gas at 600 x 10^3 N/m^2 and 27 degree centigrade will be

Solution: The ideal gas equation is PV = nRT

Given : Volume, V = 600 x 10^3 N/m^3
amount of gas in moles, n = 1 (not given - hence assume n=1)
Universal gas constant, R = 8314 J/kgmoleK
Temperature, T = 27 degree C = 27+273 = 300 K

Solution: PV = nRT

600 x 10^3 x V = 1 x 8314 x 300

Hence, V = 8314 x 300 / 600 x 10^3 = 4.157 m^3/kgmol

Ans : Molecular volume = 4.16 m^3/kgmol

how we find molecular volune of the perfact gas?

From where we get 8314.plz tell me......

n is not equal to 1, we have to find molecular Volume that's why we have to find V/n= RT/P

PV=mRT
PV=nR`T
Take n=1
600×10^3×V=1×8314×300 Units should be same so 8314
V=4.17m^3/kg mol

PV=MRT
MR= 8314 J/kg mol K
600* 10^3 *V = 8314 x (300)
V = 8314 x 300/ (600 x 10^3)
= 4.17 m^3/kg mol.

=287*300/600,000
=0.1435

Mass is involved here
So, value Of R is taken as characteristic gas constant value
That is 287
Then answer will come as 0.1435

PV=nRT
Take n=1 (because name of gas is not mentioned)
R(universal gas constant)=8.314kJ/kmolk
T=27°+273=300K
So,
V=RT/P
V=8.314×300/600
V=4.157m^3/kgmol

Brother ans is C,0.1437

Please tell me how

Solution