Question

2 moles each of a monoatomic and a diatomic gas is mixed.Find the average molar heat capacity of the mixture.​

Answer 1

Given

• 2 moles each of a monoatomic and a diatomic gas.

To Find

Average molar heat capacity of the mixture.

Concept

We define $C_{v}$ as :-

$\boxed{\pink{ C_{v} = \frac{fR}{2} }}$

• Where, f = Degree of freedom of the gas.

• R = gas constant.

For degree of freedom of each of the gas ,see attachment.

Hence, For a monoatomic gas

$C_{v} = \frac{3R}{2}$ ( Since, f = 3)

For a diatomic gas

$C_{v} = \frac{5R}{2}$ (Since, f = 5)

$\boxed{\pink{ Internal\:energy\:(U)\:=\:nC_vT}}$

Where,

• n = number of moles of the gas.

• T = Change in temperature.

Solution

The total internal energy of the mixture will be the sum of internal energies of the Two gases.

That is ,

$n_1C_vT(monoatomic)\: +\: n_2C_vT(diatomic)\: =\: n_3C_vT(mixture)\:$.........(i)

$n_1\:+n_2\:=\:n_3\:$

=> $2\:+2\:=\:n_3\:$

=> $4 =\:n_3\:$

Now, putting the corresponding values in (i)

$2\:\frac{3R}{2}\:T +\: 2\:\frac{5R}{2}\:T=\: 4C_vT(mixture)$

=> $2\:\frac{3R}{2} +\: 2\:\frac{5R}{2}=\: 4C_v(mixture)$

=> $2\:\frac{8R}{2} =\: 4C_v(mixture)$

=> $2\:\frac{8R}{2} =\: 4C_v(mixture)$

=> $C_v = 2R$

$\boxed{\boxed{\green{Hence,molar\:heat\:capacity\:of\:the\:mixture\:is\:2R}}}$

Note:- Strictly speaking , we can't measure internal energy of the gases ,only change in internal energy can be measured.For such cases, we write :-

∆U = $nC_V$∆T