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, $C_v =\frac{3}{2}R$  

For diatomic gas, $C_v =\frac{5}{2}R$  

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, $U_1 = x\times 3R\times T$

For monatomic gas, internal energy, $U_1 = 11\times \frac{3}{2}R\times T$

Total energy, $U = U_1 + U_2$

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

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

$C_v = \frac{1}{n}\frac{\Delta U}{\Delta T}$

$C_v = \frac{1}{n}((3x+16.5)R)$

Substituting n = 11 + x mol

Substituting the value of $C_v$ for diatomic gas:

$\frac{5}{2}R = \frac{1}{11+x}((3x+16.5)R)$

$\frac{5}{2} = \frac{1}{11+x}(3x+16.5)$

$55+5x = 6x+33$

$6x - 5x = 55 - 33$

$x = 22$

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.