Question

22 g of $C{O}_2$ at ${27}^{\circ }C$ is mixed with 16 g of $O_2$ at ${37}^{\circ }C$. The temperature of the mixture is:- (At room temperature, degrees of freedom of $C{O}_2=7$ and degrees of freedom of $O_2=5$)

• A. ${31.16}^{\circ }C$
• B. ${27}^{\circ }C$
• C. ${37}^{\circ }C$
• D. ${30}^{\circ }C$

Answer 1

The correct option is A ${31.16}^{\circ }C$

$\text{Given:-}$ Mass of the $C{O}_2=22g$

Mass of the $O_2=16g$

Temperature = ${27}^{\circ }C$

Temperature = ${37}^{\circ }C$

$\text{Solution:-}$

We know that,

$f_{mix}=\frac{(\frac{1}{2}\times 7)+(\frac{1}{2}\times 5)}{1}=6$

Now,

$n_1{f}_{1}R{T}_1+n_2{f}_{2}R{T}_2=n{f}_{mix}RT$

$(\frac{22}{44}\times 7\times R\times 300)+(\frac{16}{32}\times 5\times R\times 310)=(\frac{1}{2}+\frac{1}{2})\times 6\times R\times T$

$(\frac{22}{44}\times 7\times 300)+(\frac{16}{32}\times 5\times 310)=6\times T$

$\frac{2100+1550}{12}=T$

$T=304.166K$

$\therefore T^{\circ }C=304.166-273$

$={31.166}^{\circ }C$

$\text{Hence the correct option is A}$