Graph between `log((x)/(m))` and `log P` is straight line at angle of `45^(@)` with the intercept of 0.6020.

The extent of adsorption `((x)/(m))` at a pressure of 1 atm is

The mass of gas adsorbed , x , per unit mass of adsorbate , m , was measured at various pressures , p . A graph between log (x)/(m) and log p gives a straight line with slope equal to 2 and the intercept equal to 0.4771 .The value of (x)/(m) at a pressure of 4 atm is : (Given log 3 = 0.4771)

A group between x/m and the presure P of the gas at a constant temperature is called adsorption siotherm. Where x is the no. of moles of the adsorbate and m is the mass of the adsorbent. adsoption isotherms of different shaopes have been experimentally observed .. According to frundlich adosroption isthem, x//m = KP^(1//n) where K and N are constant paraments depending upon the nature of the solid and gas graph between log (x/m) and log P is a straight line a t angle 45^(@) with intercept OA as shown . hence, (x/m) at a pressure of 2 atm is :

Graph between log x/m and log P is a straight line at angle of 45^(@) with intercept 0.4771 on y-axis. Calculate the amount of gas adsorbed in gram per gram of adsorbent when pressure is 3 atm.

A graph between log (x/m) and log p is straight line at an angle of 45^(@) with intercept on y-axis equal to 0.3010. Calculate the amount of the gas adsorbed per gram of the adsobent when the pressure is 0.2 atm.

The graph between log k versus 1//T is a straight line.

When a graph is plotted between log x/m and log p, it is straight line with an angle 45^(@) and intercept 0.3010 on y - axis. If initial pressure is 0.3 atm, what will be the amount of gas adsorbed per gram of adsorbent : (Report your answer after multiplying by 10)

When a graph is plotted between log x//m and log p, it is straight line with an angle 45^(@) and intercept 0.3010 on y- axis . If initial pressure is 0.3 atm, what wil be the amount of gas adsorbed per gm of adsorbent :

Graph between log x//m and log p is a straight line inclined at an angle of 45^@ . When pressure is 0.5 atm and 1n k = 0.693 , the amount of solute adsorbed per gram of adsorbent will be: