How much of a triatomic gas with cv=3r would you have to add to 11 mol of monatomic gas to get a mixture whose thermodynamic behavior was like that of a diatomic gas?

For triatomic gas,   (given)

For monatomic gas,  

For diatomic gas,  

Let x mol of triatomic gas be added to 11 mol of monatomic gas

Internal energy of a gas is determined by:

where U is internal energy, n is number of moles, is specific heat capacity at constant volume, and T is temperature.

So, for triatomic gas, internal energy,

For monatomic gas, internal energy,

Total energy,

For mixture of gas, is given as:

Substituting n = 11 + x mol

Substituting the value of for diatomic gas:

Hence, of triatomic gas must be added to of monatomic gas to get a mixture whose thermodynamic behavior was like that of a diatomic gas.

the answer to your question is c.

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