Temperature of an ideal gas, initially at 27^ | vedclass

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(C) The initial temperature of the ideal gas is $T_{1} = 273 + 27 = 300 \ K$.When the temperature is raised by $6^{\circ} C$, the final temperature is

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increase by nearly $1 \%$

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(C) The initial temperature of the ideal gas is $T_{1} = 273 + 27 = 300 \ K$.When the temperature is raised by $6^{\circ} C$, the final temperature is $T_{2} = 300 + 6 = 306 \ K$.The root mean square (rms) velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$, which implies $v_{rms} \propto \sqrt{T}$.Therefore, the ratio of the velocities is $\frac{v_{rms_{2}}}{v_{rms_{1}}} = \sqrt{\frac{T_{2}}{T_{1}}} = \sqrt{\frac{306}{300}} = \sqrt{1.02}$.Using the binomial approximation $(1+x)^{n} \approx 1+nx$ for small $x$, we get $\sqrt{1.02} = (1 + 0.02)^{1/2} \approx 1 + \frac{1}{2}(0.02) = 1.01$.Thus, $v_{rms_{2}} \approx 1.01 \ v_{rms_{1}}$, which represents an increase of $1 \%$.

Temperature of an ideal gas, initially at $27^{\circ} C$, is raised by $6^{\circ} C$. The rms velocity of the gas molecules will

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