Solve, [1+log(xy)]dx+[1+x/y]dy=0.
Answer:-
![Solve, [1+log(xy)]dx+[1+x/y]dy=0.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6et9EhKV4ZzjPDFwc1dYxfr9icUa4JmzEo8f5K9djeyjEvizHv5QoqZx4-Ytq4WqQMmRz7krPT6gs3LcyI3PWolOm3Vxr7YdItIvxsTcKPBY7zwkMnUYFE31OMvDykDBYIDlQZCfzbws/s16000/1643442204367299-0.png "Solve, [1+log(xy)]dx+[1+x/y]dy=0.")
![Solve, [1+log(xy)]dx+[1+x/y]dy=0.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6et9EhKV4ZzjPDFwc1dYxfr9icUa4JmzEo8f5K9djeyjEvizHv5QoqZx4-Ytq4WqQMmRz7krPT6gs3LcyI3PWolOm3Vxr7YdItIvxsTcKPBY7zwkMnUYFE31OMvDykDBYIDlQZCfzbws/s1600/1643442204367299-0.png "Solve, [1+log(xy)]dx+[1+x/y]dy=0.")
What is Freundlich adsorption isotherm ?
Explanation:- The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve termed as adsorption isotherm. According to freundlich isotherm, The extent of adsorption of a gas per unit mass of adsorbent depends upon the pressure of that gas. I.e Look at the Graph,(x/m) vs (p) Equation (1) is the general form of freundlich adsorption isotherm. Case-1( at low pressure) If x/m =kp¹ That implies 1/n=1 I.e n=1 So at low pressure freundlich adsorption isotherm holds good because initially their is more adsorption occur. Case-2 (at high pressure) If x/m=constant i.e1/n=0 That implies n=infinity So at high pressure x/m independent on pressure because after saturation pressure their is no adsorption occur due to no space on adsorbent. At saturation pressure Amount of gas adsorbed=amount of gas desorbed. Look at the logarithm graph of equation (1) This is all about freundlich adsor
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