Question

In a container 3 mole of monoatomic gas and 2 mole of diatomic gas are mixed. What is the Cp/Cv value of the gas mixture ?

• A. $\frac{23}{13}$
• B. 7/5
• C. 5/3
• D. $\frac{29}{19}$

Answer 1

Answer: (D)

$\frac{29}{19}$

Answer 2

$C_{P_{mix}}$ = $\frac{n_{A}C_{P_{A}} + n_{B}C_{P_{B}}}{(n_{A} + n_{B})}$ = $\frac{3 \times 5 + 2 \times 7}{(3 + 2)}$ = $\frac{29}{5}$

$C_{V_{mix}}$= $\frac{n_{A}C_{V_{A}} + n_{B}C_{V_{B}}}{(n_{A} + n_{B})}$ = $\frac{3 \times 3 + 2 \times 5}{3 + 2}$ = $\frac{19}{5}$

\ $\frac{C_{P_{mix}}}{C_{V_{mix}}}$ = $\frac{29}{19}$