Question

A gaseous mixture enclosed in a vessel consists of one g mole of a gas A with $\gamma =\left(\frac{5}{3}\right)$ and some amount of gas B with $\gamma =\frac{7}{5}$ at a temperature The gasses A and B do not react with each other and are assumed to be ideal Find the number of g moles of the gas B if $\gamma$ for the gaseous mixture is $\left(\frac{19}{13}\right)$

• A. 2
• B. 5
• C. 4
• D. 3

Answer 1

Answer: (A)

2

Change in internal energy of the first gas + change in internal energy of the second gas = change in internal energy of the mixture

or ${\mu }_1(C_v{)}_1\Delta T+{\mu }_2(C_v{)}_2\Delta T=({\mu }_1+{\mu }_2)(C_v{)}_{mix}\Delta T$

or ${\mu }_1\frac{R}{{\gamma }_1-1}+{\mu }_2\frac{R}{{\gamma }_2-1}=({\mu }_1+{\mu }_2)\frac{R}{{\gamma }_{mix}-1};\therefore {\gamma }_{mix}=\frac{{\mu }_1{\gamma }_1({\gamma }_2-1)+{\mu }_2{\gamma }_2({\gamma }_1-1)}{{\mu }_1({\gamma }_2-1)+{\mu }_2({\gamma }_1-1)}$

or $\frac{19}{13}=\frac{1\times \left(5/3\right)(7/5-1)+{\mu }_2\times \left(7/5\right)(5/3-1)}{1(7/5-1)+{\mu }_2(5/3-1)}$ or ${\mu }_2=2$