Molar Specific Heat of Gas Mixture – Rankers Physics

Thermodynamics: Practice Problem & Solution

Two moles of helium gas is mixed with three moles of hydrogen gas (taken to be rigid). The molar specific heat of mixture at constant volume will be
\(2.1R\)
\(1.2R\)
\(5.7R\)
\(7.5R\)

Solution Explained:

To solve this problem, we apply the core principles of Thermodynamics. Understanding the underlying formula is key to arriving at the correct answer below:

The molar specific heat of a mixture at constant volume is \(C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}\). For monatomic helium, \(C_{v1} = 1.5R\), and for rigid diatomic hydrogen, \(C_{v2} = 2.5R\). Substituting gives \(C_{v,\text{mix}} = \frac{2(1.5R) + 3(2.5R)}{5} = 2.1R\).

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