What new temperature will a gas have if its speed has to become doubled compared to $127^{\circ} C$ $127^@ "C"$ ?

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What new temperature will a gas have if its speed has to become doubled compared to $127^{\circ} C$ $127^@ "C"$ ?

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Answer 1 · 2017-07-29T18:02:46

Whatever quadruple the temperature in $K$ $"K"$ is. Would this apply to RMS speed and average speed as well? Why?

The most probable speed is given by

$v_{m p} = \sqrt{\frac{2 k_{B} T}{m}}$ $v_(mp) = sqrt((2k_BT)/m)$ ,

where:

$k_{B}$ $k_B$ is the Boltzmann constant in $J/K$ $"J/K"$ .

$T$ $T$ is temperature in $K$ $"K"$ .

$m$ $m$ is the single-particle mass in $kg$ $"kg"$ .

If you want to double the most probable speed as that at $127^{\circ} C$ $127^@ "C"$ , or $400.15 K$ $"400.15 K"$ , you want...

$v_(mp)' = 2v_(mp) = 2sqrt((2k_BT)/m)$

$= sqrt((2k_B cdot color(red)(4T))/m)$

i.e. the temperature in $"K"$ should quadruple to $color(blue)("1600.60 K")$ .

Or with any other speed that follows from the Maxwell-Boltzmann distribution, i.e. any other speed that has the relationship:

$v prop sqrtT$

We have...

$2v prop sqrt(4T)$

$=>$ temperature in $"K"$ must have quadrupled.

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What new temperature will a gas have if its speed has to become doubled compared to $127^{\circ} C$ $127^@ "C"$ ?

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What new temperature will a gas have if its speed has to become doubled compared to 127^@ "C"127∘C127^@ "C"?

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What new temperature will a gas have if its speed has to become doubled compared to $127^{\circ} C$ $127^@ "C"$ ?