At 100°C, water and methylcyclohexane both have vapour pressures of 1 atm. Also at 100°C, the latent heats of vaporisation of these compounds are 40.63 kJ/mole for water and 355 kJ/mole for methylcyclohexane. The vapour pressure of water at 150°C is 69 atm. At 150°C, the vapour pressure of methylcyclohexane would be expected to be

significantly less than 69 atm.

significantly more than 69 atm.

indeterminate due to lack of data

At 100°C, water and methylcyclohexane both have vapour pressures of 1

1.	At 100°C, water and methylcyclohexane both have vapour pressures of 1 atm. Also at 100°C, the latent heats of vaporisation of these compounds are 40.63 kJ/mole for water and 355 kJ/mole for methylcyclohexane. The vapour pressure of water at 150°C is 69 atm. At 150°C, the vapour pressure of methylcyclohexane would be expected to be
A.	significantly less than 69 atm.
B.	significantly more than 69 atm.
C.	nearly equal to 69 atm.
D.	indeterminate due to lack of data
Answer» C. nearly equal to 69 atm.

Discussion

No Comment Found

Related MCQs

An equation for calculating vapour pressure is given by, log10 P = A – B(t + c). This is called the
A gas at 0°C is cooled at constant pressure until its volume becomes half the original volume. The temperature of the gas at this state will be
At 100°C, water and methylcyclohexane both have vapour pressures of 1 atm. Also at 100°C, the latent heats of vaporisation of these compounds are 40.63 kJ/mole for water and 355 kJ/mole for methylcyclohexane. The vapour pressure of water at 150°C is 69 atm. At 150°C, the vapour pressure of methylcyclohexane would be expected to be
The vapor pressure of the solvent decreased by 10 mm Hg, when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0. What should be the mole fraction of the solvent, if the decrease in vapor pressure of the solvent is required to be 20 mm Hg. ?
A gas occupies a volume of 283 c.c at 10°C. If it is heated to 20°C at constant pressure, the new volume of the gas will be __________ c.c.
The temperature of a gas in a closed container is 27° C. If the temperature of the gas is incresed to 300° C, then the pressure exerted is
Assume that benzene is insoluble in water. The normal boiling points of benzene and water are 80.1 and 100°C respectively. At a pressure of 1 atm, the boiling point of a mixture of benzene and water is
The value of gas constant 'R' is __________ kcal/kg.mole.°C.
Diffusion is that property by virtue of which a perfume bottle when opened up in a room, makes the whole room fragrant with its smell. If a perfume 'X' diffuses twice as fast as another perfume 'Y'; what is the molecular weight of 'Y', if the vapor density of gas 'X' is 2 ? Molecular weight of gas 'X' is to be assumed to be
What is the total pressure exerted by a mixture of 0.45 kg mole of benzene, 0.44 kg mole of toluene and 0.23 kg mole of o-xylene at 100°C, if their vapor pressures at 100°C are 1340, 560 and 210 mmHg respectively ?

Discussion

No Comment Found

Related MCQs

Reply to Comment