The vapour pressure of a certain liquid is given by the equation:
`Log_(10)P = 3.54595 - (313.7)/(T) +1.40655 log_(10)T` where P is the vapour pressure in mm and `T =` Kelvin Temperature. Determine the molar latent heat of vaporisation as a function of temperature. calculate the its value at `80K`.

In the graph plotted between vapour pressure (V.P.) and temperature (T),

In the graph plotted between vapour pressure (V.P.) and temperature (T),

(plot of log P, where P = vapour pressure of liquid against 1/T) The slope of the line will be equal to :

For an elementary reaction the variation of rate constant (k) with temperature is given by the following equation log_(10) k=5.4 -100/T Where, T is temperature on Kelvin scal and k is in terms of sec^(-1) Identify the incorrect options.

The vapour pressures of benzene at 303k and 333k are respectively 1.53 xx10^(4) and 5.2xx10^(4)Nm^(-2) Calculate the latent heat of evaporation of benzene over this temperature.

Vapour pressure of a solvent is the pressure exterted by vapour when they are in equilibrium with its solvent at that temperature. The vapour pressure of solvent is dependent of nature of solvent, temperature, addition of non-volatile solute as well as nature of solute to dissociate or associate. The vapour pressure of a mixture obtained by mixing two valatile liquids is given by P_(M) = P_(A)^(@).X_(A)+P_(B)^(@).X_(B) where P_(A)^(@) and P_(B)^(@) are vapour pressures of pure components A and B and X_(A), X_(B) are their mole fractions in mixture. For solute-solvent system, the relatio becomes P_(M) = P_(A)^(@).X_(A) where B is non-volatile solute. If M is mol.wt. of solvent, K_(b) is molal elevation constant and t_(b) is its boiling point, P^(@) is its vapour pressure ta temperature T and P_(S) is vapour pressure of non-volatile solute in it at T K , then: