The compressibility factor of a Vanderwaal gas is `0.5` at `27^(o)C` and `24` atm. The value of Vanderwaal constant '`a`' (in `(atm- L^(2))/(mol^(2))`) is `(b=0.10L/(mol), R=0.08(L-atm)/(K-mol))`

What is the value if R when its units are L atm K^(-1) mol^(-1) ?

What is the compressibility factor (Z) for 0.02 mole of a van der Waals's gas at pressure of 0.1 atm. Assume the size of gas molecules is negligible. Given : RT=20 L atm mol^(-1) and a=1000 atm L^(2) mol^(-2)

A 1 mol gas occupies 2.4L volume at 27^(@) C and 10 atm pressure then it show :-

For a real gas (mol.mass =60) if density at critical point is 0.80g//cm^(-3) and its T_(c)=(4xx10^(5))/(821)K, then van der Waals' constant a ( in atm L^(2)mol^(-2) ) is

1 mole of a ciatomic gas present in 10 L vessel at certain temperature exert a pressure of 0.96 atm. Under similar conditions an ideal gas exerted 1.0 atm pressure. If volume of gas molecule is negligible, then find the value of van der Waals' constant ''a'' (in atm L^(2)//mol^(2) ).

The compression factor (compressibility factor) for 1 mol of a van der Waals gas at 0^(@)C and 100 atm pressure is found to be 0.5 . Assuming that the volume of a gas molecule is neligible, calculate the van der Waals constant a .

The compression factor (compressibility factor) for 1 mol of a van der Waals gas at 0^(@)C and 100 atm pressure is found to be 0.5 . Assuming that the volume of a gas molecule is neligible, calculate the van der Waals constant a .

The table indicates the value of vander Waal's constant a in L^(2) atm mol^(-2) . The gas which can most easily be liquefied is ?

Calculate the pressure exerted by 110 g of carbon dioxide in a vessel of 2 L capacity at 37^(@)C . Given that the van der Waal’s constants are a = 3.59 L^(2) " atm "mol^(-2) and b = 0.0427 L mol^(-1) . Compare the value with the calculated value if the gas were considered as ideal.

The volume to be excluded due to only two molecules of a gas in collision with a fixed point of impact is 0.09 l mol^(-1) . If the value of 'a' is 3.6 atm L^(2) mol^(-2) , Calculate Boyle temperature. (R = 0.08 L atm K^(-1) m^(-1) )