Question

You have an sample of a gas at -33ºC. If you want to double the rms speed. ¿which is going to be the temperature needed?

Answer 1

\[Vrms= \sqrt{3RT/M} \] M= molar mass.

Answer 2

In order to double the speed by raising the temp you would need to quadruple the temperature.

Let \(T_n\) be the new temperature that doubles the speed and let \(V_{rms}\) be the original speed of the gas.

then\[2V_{rms}=\sqrt{\frac{ 3RT_n }{ M }} = \sqrt{\frac{ 3R \left( 4T \right) }{ M }}= 2\sqrt{\frac{ 3R \left( T \right) }{ M }}\]

Answer 3

Could you give me an specific response?

I am not totally sure about my answer. T= 99ºC

Answer 4

is T supposed to be in Celcius or Kelvin?

Answer 5

The response has to be in Celcius.

Answer 6

but in the formula, is T supposed to be in Celsius? What are the units on R?

Answer 7

Kelvin.

Answer 8

Exactly... you must first convert -33C to Kelvin... Quadruple that and then convert back to Celsius.

Answer 9

-33C = (273-33)K =240K
4*(240K) = 960K = (960-273)C = 687C

Answer 10

It doesn't seem right. isn't supposed that the temperature should be quadruple?

Answer 11

well if you quadruple it when it's negative, you get a smaller number (-33C goes to -132C). does that seem right?

Answer 12

okay, assume a molar mass of 1 mole. Then at -33C, what's the speed?

Answer 13

Sorry, yeap you are right.

Answer 14

no worries