Determine the numerical values of R and kB.

1.	Determine the numerical values of R and kB.
Answer» Numerical value R : Consider one mole of a gas at STP, then R = \(\frac{P_0V_0}{T}\) Standard pressure P₀ = 0.76 m of Hg column = 0.76 × 13.6 × 10³ × 9.8 N/m2 Standard temperature = T₀ = 273.15K Volume of one mole of gas at = 22.4 × 10^-3m³ R = \(\frac{0.76\times13.6\times10^3\times9.8\times22.4\times10^{-3}}{273.15}\) = 8.31 J mol^-1K^-1 In the C.G.S. system, R = \(\frac{8.31}{4.2}\) cal mol^-1°C^-1 = 1.98 cal mol^-1°C^-1 Numerical value of k_B: We know that, k_B = \(\frac{R}{NA}\) k_B = \(\frac{8.31\,J\,mole^{-1}K^{-1}}{6.02\times10^{23}mole^{-1}}\) =1.38 × 10^-23 J/K.

Answer»

Numerical value R :

Consider one mole of a gas at STP, then

R = \(\frac{P_0V_0}{T}\)

Standard pressure

P₀ = 0.76 m of Hg column

= 0.76 × 13.6 × 10³ × 9.8 N/m2

Standard temperature = T₀ = 273.15K

Volume of one mole of gas at

= 22.4 × 10^-3m³

R = \(\frac{0.76\times13.6\times10^3\times9.8\times22.4\times10^{-3}}{273.15}\)

= 8.31 J mol^-1K^-1

In the C.G.S. system,

R = \(\frac{8.31}{4.2}\) cal mol^-1°C^-1

= 1.98 cal mol^-1°C^-1

Numerical value of k_B:

We know that,

k_B = \(\frac{R}{NA}\)

k_B = \(\frac{8.31\,J\,mole^{-1}K^{-1}}{6.02\times10^{23}mole^{-1}}\)

=1.38 × 10^-23 J/K.

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