Question

A gaseous mixture enclosed in a vessel of volume V consists of one mole of a gas A with $\gamma$ = $\frac{5}{3}$ and another gas B with $\gamma$ = $\frac{7}{5}$ at a certain temperature T. the molar masses of the gases A and B are 4 and 32, respectively. The gases A and B do not reach with each other and are assumed to be ideal. The gaseous mixture follows the equation P$V^{\frac{19}{13}}$ = constant, in adiabatic processes. The number of moles of the gas B in the gaseous mixture.

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (A)

2

Given $\gamma$ for the mixture $=\frac{19}{13}$

Since $C_V=\frac{R}{\gamma -1}$,

$(C_V{)}_A=\frac{R}{\frac{5}{3}-1}=\frac{3}{2}R$

and

$(C_V{)}_B=\frac{5}{2}R$

Then,

$(C_V{)}_{mixture}=\frac{R}{\frac{19}{13}-1}=\frac{13}{6}R$

Now, for a change in temperature $\Delta T$, the sum of energy change of individual gases must be equal to total energy change of the mixture.

$n_A(C_V{)}_A\Delta T+n_B(C_V{)}_B\Delta T=(n_A+n_B)(C_V{)}_{mixture}\Delta T$

$⟹\frac{3}{2}R+n_B\frac{5}{2}R-(1+n_B)\frac{13}{6}R$

Thus

$n_B=2$