GATE 2026 CY – Question 64
BET nitrogen adsorption at 77 K on 1.0 g adsorbent gives slope $6\times10^{-4}$ mm−3 and intercept $4\times10^{-6}$ mm−3 for a plot of $z/[(1-z)V]$ against z = p/p0. The surface area in m²/g (one decimal place) is _____. Given 1 mm³ N2 = $2.7\times10^{16}$ molecules; each occupies 0.16 nm². V is adsorbed gas volume; p0 is liquid-nitrogen equilibrium vapor pressure.
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 7.16 to 7.24
Explanation
The **BET equation** in linear form is
$$\frac{z}{(1-z)V}=\frac{1}{V_mc}+\frac{c-1}{V_mc}\,z,\qquad z=\frac p{p_0}.$$
The plot of $\dfrac{z}{(1-z)V}$ against $z$ has
- slope $s=\dfrac{c-1}{V_mc}$,
- intercept $i=\dfrac1{V_mc}$.
**Monolayer volume.** Add them: $s+i=\dfrac{c-1}{V_mc}+\dfrac1{V_mc}=\dfrac{c}{V_mc}=\dfrac1{V_m}$, so
$$V_m=\frac1{s+i}=\frac{1}{6\times10^{-4}+4\times10^{-6}}=\frac{1}{6.04\times10^{-4}}=1655.6\ \text{mm}^3\text{ (per gram)}.$$
**Number of molecules in the monolayer:**
$$1655.6\times2.7\times10^{16}=4.470\times10^{19}.$$
**Surface area** (each molecule covers 0.16 nm² $=0.16\times10^{-18}$ m²):
$$A=4.470\times10^{19}\times0.16\times10^{-18}=\mathbf{7.15\ m^2/g}\ (\approx7.2).$$