Question

A gaseous mixture enclosed in a vessel consists of one gram mole of a gas $A$ with $\gamma =\frac{5}{3}$ and some amount of gas $B$ with $\gamma =\frac{7}{5}$ at temperature $T$. Both gases are non-reactive and ideal. Find the number of gram moles of gas $B$ if $\gamma$ for the gaseous mixture is $\frac{3}{2}$.

• A. 1.5
• B. 2
• C. 2.5
• D. 1

Answer 1

The correct option is D 1

Given that,
${\gamma }_A=\frac{5}{3}$, ${\gamma }_B=\frac{7}{5}$, ${\gamma }_{mix}=\frac{3}{2}$
$n_A=1\text{mol}$
For ideal gas,
$C_V=\frac{R}{\gamma -1}$
$\Rightarrow (C_V{)}_A=\frac{R}{\frac{5}{3}-1}=\frac{3}{2}R$
& $(C_V{)}_B=\frac{R}{\frac{7}{5}-1}=\frac{5}{2}R$
$(C_V{)}_{mix}=\frac{R}{\frac{3}{2}-1}=2R$
The internal energy of the system is conserved.
From conservation of energy,
$\Delta U=\Delta U_1+\Delta U_2$
$(n_A+n_B)(C_V{)}_{mix}\Delta T=[n_A(C_V{)}_A+n_B(C_V{)}_B]\Delta T$
i.e $(C_V{)}_{mix}=\frac{n_A(C_V{)}_A+n_B(C_v{)}_B}{n_A+n_B}$
$2R=\frac{1\times \frac{3}{2}R+n_B\frac{5}{2}R}{(1+n_B)}=\frac{3R+5n_{B}R}{2(1+n_B)}$
$\Rightarrow 2\times 2(1+n_B)=3+5n_B$
$\Rightarrow 4+4n_B=3+5n_B$
or $n_B=1\text{gm mole}$