If density of vapours of a substance of molar mass `18 gm at 1 atm` pressure and`500K` is `0.36 kg m^(-3)`, then calculate the value of `Z` for the vapours. (Take `R = 0.082 L atm mole^(-1)K^(-1))`

The density of vapours of a substance of molar mass 18 g at 1 atm pressure and 500 K is 0.36 kg m^(-3) . The value of compressibility factor Z for the vapours will be (Take R = 0.082 L atm mol ^(-1) K^(-1)

The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 kg m^(-3) . The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition. ( a ) Determine ( i ) the molecular weight, ( ii ) the molar volume

The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 kg m^(-3) . The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition. ( a ) Determine ( i ) the molecular weight, ( ii ) the molar volume

The density of vapours of a substance at 1 atm and 500 K is 0.3 kg m^(-3) . The vapours effuse 0.4216 times faster than O_(2) through a pin hole under identical conditions. If R= 0.08 litre atm K^(-1) mol^(-1) . The molar volume of gas is a xx 10^(2) litre. The value of a is _____________.

The density of vapours of a substance at 1 atm and 500 K is 0.3 kg m^(-3) . The vapours effuse 0.4216 times faster than O_(2) through a pin hole under identical conditions. If R= 0.08 litre atm K^(-1) mol^(-1) . The molar volume of gas is a xx 10^(2) litre. The value of a is _____________.

A gas of molar mass 84.5 g//mol is enclosed in a flask at 27^@C has a pressure of 2 atm. Calculate the density of the gas. [R = 0.082 L atm K^(-1) mol^(-1)]

The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 kg m^(-3) . The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition. If the vapour behaves ideally at 1000 K, determine the average translational kinetic energy of a molecule.