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

Question

In a container 3 mol of monoatomic gas and 2 mol of diatomic gas are mixed. What is the Cp / Cv value of the gas mixture? (A) (23/13) (B) (7/5) (C) (5/3) (D) (29/19)

Tardigrade Admin · Accepted Answer

C P text mix =(n A C P A +n B C P B /(n A +n B ))=(3 × 5+2 × 7/(3+2))=(29/5) C V text mix =(n A CV A +n B CV B /(n A +n B ))=(3 × 3+2 × 5/3+2)=(19/5) ∴ (C P text mix /C V text mix )=(29/19)

Q. In a container $3 m o l$ of monoatomic gas and $2 m o l$ of diatomic gas are mixed. What is the $C_{p} / C_{v}$ value of the gas mixture?

$\frac{23}{13}$

$\frac{7}{5}$

$\frac{5}{3}$

$\frac{29}{19}$

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

1323​

57​

35​

1929​

CPmix ​​=(nA​+nB​)nA​CPA​​+nB​CPB​​​=(3+2)3×5+2×7​=529​ CVmix ​​=(nA​+nB​)nA​CVA​​+nB​CVB​​​=3+23×3+2×5​=519​ ∴CVmix ​​CPmix ​​​=1929​

Q. In a container $3 m o l$ of monoatomic gas and $2 m o l$ of diatomic gas are mixed. What is the $C_{p} / C_{v}$ value of the gas mixture?

$\frac{23}{13}$

$\frac{7}{5}$

$\frac{5}{3}$

$\frac{29}{19}$