Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0 × 10 - 15 atm. at room temperature (300 K). Taking R = 8.3 JK ⁻¹ mole ⁻¹, 1 atm = 105 Pa and NAvogadro = 6 × 1023 mole ⁻¹, the mean distance between molecules of gas in an evacuated vessel will be of the order of: (a) 0.2 μ m (b) 0.2 mm (c) 0.2 cm (d) 0.2 nm

Answer: (b). Number density n = P/kT = (PNA)/(RT). n = 4 × 10⁻¹⁵ × 10⁵ × 6 × 10²³8.3 × 300 ≈ 9.6 × 10¹⁰ m⁻³ Mean distance d = n-1/3 ≈ (96 × 10⁹)-1/3 ≈ 2.2 × 10⁻⁴ m So the answer is (b) 0.2 mm.

Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0 ×… | JEE Main 2014 (9 Apr, Shift 1)

Q75JEE Main 2014 · 9 APR · SHIFT 1Difficult

Gas Laws & Ideal Gas Equation

Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0 $\times$ 10 $-$ 15 atm. at room temperature (300 K). Taking R = 8.3 JK ${}^{-1}$ mole ${}^{-1}$ , 1 atm = 105 Pa and NAvogadro = 6 $\times$ 1023 mole ${}^{-1}$ , the mean distance between molecules of gas in an evacuated vessel will be of the order of:

Show answer and solution

Answer: (b)

Number density $n = \frac{P}{kT} = \frac{PN_A}{RT}$ .
$n = \frac{4 \times 10^{-15} \times 10^5 \times 6 \times 10^{23}}{8.3 \times 300} \approx 9.6 \times 10^{10}\ \text{m}^{-3}$
Mean distance $d = n^{-1/3} \approx (96 \times 10^{9})^{-1/3} \approx 2.2 \times 10^{-4}$ m
So the answer is (b) 0.2 mm.

Report a mistake

Gas Laws & Ideal Gas Equation question from JEE Main 2014 (9 Apr, Shift 1)

More questions on Gas Laws & Ideal Gas Equation