# Thermodynamics: RMS Speed of Molecules

1. Jan 9, 2010

### Ling_Ling

1. The problem statement, all variables and given/known data
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

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

3. 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.

2. Jan 9, 2010

### vela

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

3. Jan 9, 2010

### Ling_Ling

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 cancelled itself out in my previous work. I think I misinterpreted the variables, but used the ratio correctly, unless that answer is incorrect.

4. Jan 9, 2010

### vela

Staff Emeritus
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?