1.

Calculate (i) r.m.s. velocity, and (ii) mean kinetic energy of one-gram molecule of hydrogen at S.T.P. (Given, density of hydrogen at S.T.P. is 0.09 kg/ms.

Answer»

Here, 

ρ = 0.09 kg/m3

P = 1.01 × 105 Pa

(i) According to kinetic theory of gases,

\(P=\frac{1}{3}ρc^2\) 

or  \(c=\sqrt\frac{3P}{ρ}\)

\(=\sqrt\frac{3\times1.01\times10^5}{0.09}\) 

= 1837.5 m/s

(ii) Volume occupied by one mole of hydrogen at S.T.P. 

= 22.4 litres

= 22.4 × 10-3m3

Mass of hydrogen, 

M = Volume × Density

= 22.4 × 10-3 × 0.09 kg

= 2.016 × 10-3 kg

Mean K.E. of one-gram molecule of hydrogen at S.T.P.

= 1/2 MV2

= 1/2 × 2.016 × 10-3 × (1837.5)2

= 3.4 × 103 J



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