Question

An experiment requires a gas with $\gamma =1.50$. This can be achieved by mixing together monatomic and rigid diatomic ideal gases. The ratio of moles of the monatomic to diatomic gas in the mixture is

• A. $1:3$
• B. $2:3$
• C. $1:1$
• D. $3:4$

Answer 1

Answer: (C)

$1:1$

One mole of an ideal monoatomic gas is is C${}_v$ = $\frac{3}{2}$R and C${}_p$ = $\frac{5}{2}$R

i.e $\gamma$ = 1.66 for monoatomic gas

For One mole of an ideal dioatomic gas,

$\gamma$ = 1.4 for air which is pre dominantly a diatomic gas

If we take 1 mole monoatomic and 1 mole of diatomic gas in a mixture then we get the following result;

$\gamma$ = $\frac{n1\gamma +n2\gamma }{n1+n2}$

Now since we have taken the no. of moles of monoatomic as well as diatomic as 1, therefore

$\gamma$ = $\frac{y1+y2}{2}$ where $\gamma$1 and $\gamma$2 are the values of $\frac{C_p}{C_v}$ for individual gases.

Substuting the values of C${}_p$ and C${}_v$ i.e $\gamma$1 = 1.6 and $\gamma$2 = 1.4 we get

$\gamma$ = 1.53 which is approximately equal to 1.50 which is given.

Hence by taking 1 mole og monoatomic and 1 mole of diatomic mixture we got $\gamma$ as 1.50

Hence the ratio of moles of monoatomic to diatomic gas in the mixture is 1:1