Question

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?

Answer 1

For triatomic gas,
`C_v = 3R` (given)

For monatomic gas,
`C_v =(3)/(2)R`

For diatomic gas,
`C_v =(5)/(2)R`

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

Internal energy of a gas is determined by:

`U = nC_vT`

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

So, for triatomic gas, internal energy,
`U_1 = x* 3R* T`

For monatomic gas, internal energy,
`U_1 = 11* (3)/(2)R* T`

Total energy,
`U = U_1 + U_2`

`3xRT+16.5RT = (3x+16.5)RT`

For mixture of gas,
`C_v` is given as:

`C_v = (1)/(n)(\Delta U)/(\Delta T)`

`C_v = (1)/(n)((3x+16.5)R)`

Substituting n = 11 + x mol

Substituting the value of
`C_v` for diatomic gas:

`(5)/(2)R = (1)/(11+x)((3x+16.5)R)`

`(5)/(2) = (1)/(11+x)(3x+16.5)`

`55+5x = 6x+33`

`6x - 5x = 55 - 33`

`x = 22`

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