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afcwestwarrior
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Homework Statement
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?
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 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.
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