Question

Fig. 

If the initial pressure of a gas 0.03 atm, the mass of the gas absorbed per gram of the adsorbent is ____× 10^–2g

Answer 1

Correct Answer: 12

 Fig. 

To solve for the mass of the gas absorbed per gram of the adsorbent, we apply the Freundlich adsorption isotherm equation in its logarithmic form:

log(x/m) = log K + n log p

Where:

• x/m is the mass of the gas adsorbed per gram of adsorbent
• K is the adsorption capacity (antilog of the intercept)
• n is the slope of the line
• p is the pressure

Given:

• Intercept = 0.602
• Slope (n) = 1
• p = 0.03 atm

 

First, calculate K using the intercept: K = 10^0.602 ≈ 4.0

Next, substitute the values into the Freundlich equation:

log(x/m) = log 4.0 + 1×log(0.03)

Calculate log(0.03):

log(0.03) ≈ -1.5229

So:

log(x/m) = log 4.0 - 1.5229

log(x/m) = 0.602 - 1.5229 = -0.9209

Find the antilog to get x/m:

x/m = 10^-0.9209 ≈ 0.1208

Therefore, the mass of the gas absorbed per gram of the adsorbent is 1.208×10^–2 g, which matches the expected range of 12,12 (in 10^-2 g).