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Consider the C, and c₂ of a gas kept at room temperature is 27.5 J. mol-¹.K-¹ and 35.8 J. mol-¹. K-¹. Find the atomicity of the gas

Answer :

Therefore, the atomicity of the gas is 3.5

Given:

Cp = 27.5 J. mol⁻¹.K⁻¹Cv = 35.8 J. mol⁻¹.K⁻¹We know that, Cp – Cv = R

Where, R is gas constant for the given gas.

So, R = Cp – Cv

Put the values of Cp and Cv,

we getR = 27.5 J. mol⁻¹.K⁻¹ – 35.8 J. mol⁻¹.K⁻¹= -8.3 J. mol⁻¹.K⁻¹

For monoatomic gas, degree of freedom (f) = 3

And, for diatomic gas, degree of freedom (f) = 5

Now, we know that atomicity of gas (n) is given by,

n = (f + 2)/2

For the given gas,

n = (f + 2)/2 = (5+2)/2 = 3.5

Therefore, the atomicity of the gas is 3.5.We found the value of R for the given gas using the formula Cp – Cv = R. After that, we applied the formula of atomicity of gas to find its value.

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