Question

The graph of $\log \frac{x}{m}$ vs $\log p$ for an adsorption process is a straight line inclined at an angle of $45^{\circ}$ with intercept equal to $0.6020$ The mass of gas adsorbed per unit mass of adsorbent at the pressure of $04 \,atm$ is______ $\times 10^{-1}$ (Nearest integer)
Given: $\log 2=0.3010$

Answer 1

Correct Answer: 16

To solve this problem, first consider the given graph of $\log \frac{x}{m}$ vs $\log p$. The straight line has an incline of $45^\circ$, which suggests the slope $(m)$ is $1$. The equation of a straight line is $\log \frac{x}{m} = m \log p + c$, with $c$ as the intercept.

Given: intercept $c = 0.6020$, $\log 2 = 0.3010$.

Since the slope $m$ is $1$, the line equation becomes: $\log \frac{x}{m} = \log p + 0.6020$.

At $p = 4 \, \text{atm}$, calculate $\log p = \log(2 \times 2) = 2 \cdot \log 2 = 0.6020$.

Substituting in the line equation: $\log \frac{x}{m} = 0.6020 + 0.6020 = 1.2040$.

Thus, $\frac{x}{m} = 10^{1.2040}$. Using approximation, $10^{0.2040} \approx 1.6$, so $10^{1.2040} \approx 10 \times 1.6 = 16$.

This mass of gas adsorbed is $1.6 \times 10^{1}$. Nearest integer is $16$, confirmed within the range (16,16).