Question

A gaseous mixture enclosed in a vessel contains 1 g mole of a gas A (with g = 5/3) and another gas B (with g = 7/5) at a temperature T. The gases A and B do not react with each other and assumed to be ideal. The number of gram moles of B, if g for the gaseous mixture is 19/13 is

• A. 4
• B. 12
• C. 16
• D. 8

Answer 1

Answer: (A)

4

Answer 2

Let mixture contain n mole of gas B

As $C_{P} - C_{V} = R$and $\gamma = \frac{C_{P}}{C_{V}}$

$\therefore C_{V} = \frac{R}{\gamma^{- 1}}$

For gas A, $C_{V} = \frac{R}{\frac{5}{3} - 1} - \frac{3}{2}R$

For gas B, $C_{V} = \frac{R}{\frac{7}{2} - 1} = \frac{5}{2}R$

For the mixture, $C_{V} = \frac{R}{\frac{19}{13} - 1} = \frac{13}{6}R$

By conservations of energy, molar specific heat of the mixture is

$C_{V} = \frac{B_{A}(C_{V})_{A} + B_{B}\left( C_{V} \right)}{B_{A} + n_{B}}$

$\therefore\frac{13}{6}R = \frac{1 \times \frac{3}{2}R + n \times \frac{5}{2}R}{1 + n} = \frac{(3 + 5n)R}{2(1 + n)}$

$\therefore n = 2$