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(B) The average velocity $(v_{avg})$ of gas molecules is related to the absolute temperature $(T)$ by the relation $v_{avg} \propto \sqrt{T}$.Given th

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$4527$

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(B) The average velocity $(v_{avg})$ of gas molecules is related to the absolute temperature $(T)$ by the relation $v_{avg} \propto \sqrt{T}$.Given the initial temperature $T_1 = 300 \ K$ and the final average velocity $(v_{avg})_2 = 4(v_{avg})_1$.Using the ratio: $\frac{(v_{avg})_1}{(v_{avg})_2} = \sqrt{\frac{T_1}{T_2}}$.Substituting the values: $\frac{1}{4} = \sqrt{\frac{300}{T_2}}$.Squaring both sides: $\frac{1}{16} = \frac{300}{T_2}$.Therefore, $T_2 = 300 	imes 16 = 4800 \ K$.To convert the temperature from Kelvin to Celsius: $T(^{\circ}C) = T(K) - 273.15$.Using $273$ for simplicity: $T_2 = 4800 - 273 = 4527^{\circ} C$.

Consider an ideal gas in a closed container at $300 \ K$. The container is then heated so that the average velocity of particles of the gas increases by a factor of $4$. What would be the final temperature (in $^{\circ} C$)?

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