• # question_answer A gas is a mixture of two parts by volume of hydrogen and one part by volume of nitrogen. If the velocity of sound in hydrogen at $0{}^\circ C$is $1300m/s$ then the velocity of sound in the gaseous mixture at $27{}^\circ \operatorname{C}$will be A) $951m/sec$                   B) $159m/sec$ C) $591m/sec$       D) $519m/sec$

[c]  we have $\frac{{{v}_{m}}}{{}^{v}H}=\sqrt{\frac{{{\rho }_{H}}}{{{\rho }_{m}}}}$ ? (1) Here, ${{\rho }_{m}}=\frac{{{m}_{H}}+{{m}_{N}}}{{{V}_{H}}+{{V}_{N}}}$ $=\frac{{{\rho }_{H}}{{V}_{H}}+{{\rho }_{N}}{{V}_{N}}}{{{V}_{H}}+{{V}_{N}}}$ $=\frac{{{\rho }_{H}}{{V}_{H}}\left( 1+\frac{{{\rho }_{N}}}{{{\rho }_{H}}}\times \frac{{{V}_{N}}}{{{V}_{H}}} \right)}{{{V}_{H}}\left( 1+\frac{{{V}_{N}}}{+{{V}_{H}}} \right)}$ $\therefore \frac{{{\rho }_{m}}}{{{\rho }_{H}}}=\frac{1+\frac{28}{2}\times \frac{1}{2}}{1+\frac{1}{2}}=\frac{16}{3}$ Now from equation (i), we have ${{[{{v}_{m}}]}_{0}}={{v}_{H}}=\sqrt{\frac{3}{16}}$ $\frac{1300\sqrt{3}}{4}=325\sqrt{3}$ ${{[{{v}_{m}}]}_{27}}={{({{v}_{m}})}_{0}}\left( 1+\frac{t}{546} \right)=591m/s$