Thermodynamics: RMS Speed of Molecules

[Ling_Ling]

Homework Statement

The rms speed of molecules in a gas at 20.0°C is to be increased by 1.4 percent. To what temperature must it be raised?

k = 1.38E-23 (Boltzmann's Constant)
Ratio in speed: 1.014 : 1
Ratio in Volume: 1.014² = 1.0282 : 1
T1 = 20°C = 293°K
The Volume of 1 mol of an ideal gas is 22.4 L

Homework Equations

v(rms) = √(3kT/m)
KE = ½mv² = 1.5kT
V1²/V2² = T1²/T2²

The Attempt at a Solution

22.4/(22.4*1.0282) = 293²/T2²
.9726 = 293²/T2²
.9726T2² = 293²
T2 = 297.103°K

Am I correct in my thinking?
Thank you.

 

[vela]

No. The problem talks about increasing v(rms) by raising the temperature, so why are you increasing the volume?

 

[Ling_Ling]

No. The problem talks about increasing v(rms) by raising the temperature, so why are you increasing the volume?

Oh. My apologies, a misinterpretation of variables, thanks to past physics units.

So then:
½mv² = 3/2 kT
½m(1.014)² = 3/2 (1.38E-23)T
2.484E²² m = T

Although, I'm not really sure how I could find m (assuming the above steps are correct?)

Unless I should just use the equation: v1²/v2² = T1²/T2²
(at first, I thought the v's in the equation were capital V's standing for volume, but now I realize it stands for v(rms)

1(v²)/1.014²(v²) = 293²/T2²
v²'s cancel out, leaving me with:
1/1.0282 = 85849/T²
T² = 88269.94
T = 297.103°K
Hmm, well, that's the same answer that I found before, but concerning the volume of 1 mol = 22.4 Liters, I noticed that it canceled itself out in my previous work. I think I misinterpreted the variables, but used the ratio correctly, unless that answer is incorrect.

 

[vela]

I'm not sure where you got that equation from, but I'm guessing you were right that the V's stood for volume. It looks like one of those ideal gas laws squared.

The equation

$$\frac{1}{2}mv_{rms}^2 = \frac{3}{2}kT$$

tells you that the temperature is proportional to $$v_{rms}^2$$, so if $$v_{rms}$$ increases by 1.4%, by what factor does the temperature rise?