Calculate the number of molecules in 1cc of ideal gas at 27°C and 10mm pressure of mercury. Mean kinetic energy of molecule at 27°C is 6.21x10^–14 ergs.

Solution

Given, 1 cc of ideal gas at 27^o C and 10 mm pressure of mercury.

Mean K.E of molecule = 6.21x10^–14 ergs.

We know that,

Pressure of a gas is equal to two-third of the average kinetic energy per unit volume of the gas.

Thats is,

P = \frac{2}{3} E_{d} \Rightarrow E_{d} = \frac{3}{2} P

Here we have,

P = 10 mm of mercury = 1 cm of mercury = 1 \times 13.6 \times 980 = 1.33 \times 10^{4} dyne / {cm}^{2} .

Therefore,

E_{d} = \frac{3}{2} P = \frac{3}{2} \times 1.33 \times 10^{4} = 2 \times 10^{4} erg / cc . . . . (1)

Let n be the number of molecules in 1 cc of gas, therefore the energy per cm³ of gas is,

E_{d} = n \times 6.21 \times 10^{- 14} ergs / cc

...(2)

From equations (1) and (2),

Number of molecules is =

n = \frac{2 \times 10^{4}}{6.21 \times 10^{- 14}} = 3.22 \times 10^{17} molecules

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