States of Matter

We want to find the temperature where the rms (root mean square) speed of $\mathrm{SO}_2$ molecules is equal to the rms speed of $\mathrm{O}_2$ molecules at $27^{\circ}\mathrm{C}$.The formula for rms velocity is: $V_{\text{rms}}=\sqrt{\frac{3 R T}{M}}$ where $R$ is the gas constant, $T$ is temperature in Kelvin, and $M$ is molar mass.We set the rms speeds equal: $V_{\text{rms}(\mathrm{SO}_2)} = V_{\text{rms}(\mathrm{O}_2)} \text{ So }, \sqrt{\frac{3 R T_{\mathrm{SO}_2}}{M_{\mathrm{SO}_2}}} = \sqrt{\frac{3 R T_{\mathrm{O}_2}}{M_{\mathrm{O}_2}}}$Squaring both sides removes the square root: $\frac{T_{\mathrm{SO}_2}}{M_{\mathrm{SO}_2}}=\frac{T_{\mathrm{O}_2}}{M_{\mathrm{O}_2}}$Rearrange for $T_{\mathrm{SO}_2}: T_{\mathrm{SO}_2}=T_{\mathrm{O}_2} \times \frac{M_{\mathrm{SO}_2}}{M_{\mathrm{O}_2}}$The temperature given for $\mathrm{O}_2$ is $27^{\circ}\mathrm{C}$. Convert this to Kelvin: $27+273=300\, ext{K}$The molar mass of $\mathrm{SO}_2$ is 64 , and of $\mathrm{O}_2$ is 32 .Plug in the values: $T_{\mathrm{SO}_2}=300 \times \frac{64}{32}=300 \times 2=600\,\text{K}$