I must find Temperature, The Distribution of Molecular Speeds.

[afcwestwarrior]

Homework Statement

A certain gas is at a temperature of 27 degrees celsius. What would the temperature of the gas have to be to increase the rms speed of the molecules in the gas by 12 percent?

Homework Equations

V rms = √((3RT)/(M))

The Attempt at a Solution

We know R and T
R= 8.31 J/mol *k
New T= ?
M is unknown

T is what we must find.


V rms(.12) + 1 = ( Vrms1.12)

I don't know if I'm right but here's what I did.

3RT(2) = (1.12 Vrms)^ 2) * M

3RT(1) = (Vrms)^2) * M


We solve for T(2) or Temperature 2 or Temperature Final.

We divide (T2) over (T1)

3R cancels and M cancels


so we have T(2) / T(1) = 1.2544 Vrms ^2 / Vrms ^2


Vrms ^ 2 cancels

we are left with

T(2) = 1.2544 * T1

T(2) = 1.2544 * 27 degrees celsius = 33.8688

Is this correct.

Let's check if Vrms increases

T must increase and T did increase so I'm assuming that I did it right.

 

[wywong]

T(2) = 1.2544 * T1

T(2) = 1.2544 * 27 degrees celsius = 33.8688

Note that T1 and T2 are in Kevin, not degrees celsius. So T1=273+27=300K.

 

[afcwestwarrior]

Woops. So it's 1.2544 (300K) = 376.32

 

[afcwestwarrior]

IS it right

 

[Redbelly98]

Yes, it's correct.

They might want the answer given in degrees C, since those were the temperature units given in the problem statement.

 

[afcwestwarrior]

Man I feel good.