JEE Main 2014 · 9 Apr · Shift 1 · Physics Q13

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PhysicsSingle correct+4 / -1Question 13 of 90

Modern vacuum pumps can evacuate a vessel down to a pressure of $4.0\times10^{-15}$ atm. at room temperature (300 K). Taking $R=8.3$ JK $^{-1}$ mole $^{-1}$ , 1 atm $=10^5$ Pa and $N_{\text{Avogadro}}=6\times10^{23}$ mole $^{-1}$ , the mean distance between molecules of gas in an evacuated vessel will be of the order of :

0.2 μm

0.2 mm

0.2 cm

0.2 nm

Answer and worked solution

Official answer: B

Number density:
$n=\frac P{kT}=\frac{4\times10^{-15}\times10^5}{(8.3/6\times10^{23})\times300}\approx9.6\times10^{10}\text{ m}^{-3}.$
Mean spacing $\approx n^{-1/3}\approx2.2\times10^{-4}$ m, i.e. about 0.2 mm.

Topics:Ideal gas equation

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JEE Main 2014 · 9 Apr · Shift 1, Physics question 13