Question

Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume?

(R = 8.3 J/mol K)
1. 19.7 J/mol K
2. 15.7 J/mol K
3. 17.4 J/mol K
4. 21.6 J/mol K

Answer 1

Correct Answer - Option 3 : 17.4 J/mol K

Concept:

The molar specific heat of gaseous mixture at constant volume is given by the formula:

\({\left( {{{\rm{C}}_{\rm{V}}}} \right)_{mix}} = \frac{{{{\rm{n}}_1}{{\rm{C}}_{{{\rm{V}}_1}}} + {{\rm{n}}_2}{{\rm{C}}_{{{\rm{V}}_2}}}}}{{{{\rm{n}}_1} + {{\rm{n}}_2}}}\) 

Where, n₁ and n₂ are moles of the gases.

If n₁ (2 moles (given)) is helium (Monoatomic molecule), the constant volume is given by the formula:

\({{\rm{C}}_{{{\rm{V}}_1}}} = \frac{3}{2}R\) 

If n₂ (3 moles (given)) is hydrogen (Diatomic molecule), the constant volume is given by the formula:

\({{\rm{C}}_{{{\rm{V}}_2}}} = \frac{5}{2}{\rm{R}}\) 

Calculation:

Now, molar specific heat at constant volume is:

\( \Rightarrow {\left( {{{\rm{C}}_{\rm{V}}}} \right)_{mix}} = \frac{{2\left( {\frac{3}{2}R} \right) + 3\left( {\frac{5}{2}{\rm{R}}} \right)}}{{2 + 3}}\) 

\( = \frac{{\frac{6}{2}R + \frac{{15}}{2}R}}{5} = \frac{{\frac{{21}}{2}R}}{5}\) 

Where, R is Gas constant = 8.3 J/mol K (given)

\( \Rightarrow {\left( {{{\rm{C}}_{\rm{V}}}} \right)_{mix}} = \frac{{21 \times 8.3}}{{2 \times 5}}\) 

∴ C_Vmix = 17.43 J/mol K