# I must find Temperature, The Distribution of Molecular Speeds.

#### afcwestwarrior

1. The problem statement, all variables and given/known data
A certain gas is at a temperature of 27 degrees Celcius. What would the temperature of the gas have to be to increase the rms speed of the molecules in the gas by 12 percent?

2. Relevant equations
V rms = √((3RT)/(M))

3. 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 celcius = 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.

Last edited:
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#### wywong

T(2) = 1.2544 * T1

T(2) = 1.2544 * 27 degrees celcius = 33.8688
Note that T1 and T2 are in Kevin, not degrees Celcius. So T1=273+27=300K.

#### afcwestwarrior

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

IS it right

#### Redbelly98

Staff Emeritus
Science Advisor
Homework Helper
Yes, it's correct.

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

Man I feel good.

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