# A monatomic gas is adiabatically compressed

1. May 17, 2009

### vorcil

http://img198.imageshack.us/img198/1122/masteringphysicsscreens.jpg [Broken]

p = ((1/3)*(N/V))*mv^2

vrms = squareroot ((3p)/(N/v))

i tried substituting .125 into the equation to get some numbers, didn't get anything useful

can someone show me the formulas that i have to use, and how to use them? or explain how to solve this problem

cheers

Last edited by a moderator: May 4, 2017
2. May 17, 2009

### vorcil

anyone?

3. May 17, 2009

### Andrew Mason

If the question called for an actual calculation (I am not sure it does), you would use the adiabatic condition :

$$PV^{\gamma} = constant$$ or:

$$TV^{\gamma -1} = constant$$

In other words, these quantities remain unchanged in an adiabatic process.

Since work is done on the gas, what can you say about the temperature after adiabatic compression (apply first law)? How does vrms relate to temperature?

Would the specific heat Cv change? (remember, the gas is ideal).

What does the mean free path mean? How would it change with reduced volume?

AM

Last edited by a moderator: May 4, 2017
4. May 17, 2009

### vorcil

since those values were constant and work is done on the gas, i would expect the temperature to decrease as the volume of the gas decreases to keep the thermal energy constant,

if the temperature decreases, i don't think anything happens to the vrms of the gas, because i looked through the book at the vRms of a gas section and the formulas have nothing to do with temperature (though i'm sure they do)

5. May 17, 2009

### vorcil

ok i found in my book

P = f/a
= 1/3 * N/V * MVrms squared

n/v = p / boltzmanC * temperature

so the volume is inversely proportional to temperature as was my guess, if volume decreases the temperature increases

6. May 17, 2009

### vorcil

they are constant?

7. May 17, 2009

### vorcil

oh that was the volume not vrms

i found the equation 1/2m * vrms^2 = 3/2 kb t

which means the average root mean square velocity is directly proportional to temperature

3/2 T with kb being boltz's constant, if it decreases, the velocity decreases aswell

8. May 17, 2009

### vorcil

bah i failed this question and only got two of the awnsers right

i knew that the heat specific ratio was constant so it was 1,

i got for the first question 2.07 instead of 2.00 and because it's mastering physics i failed that question
i got that from ln(1/.125) for an adiabatic process

the other two awnser for B and C
were lavender = .125 (not sure why the mean free path was the same as the volume percentage it was reduced too
and C was 4,

no idea how the first three awnsers came around for
part a b and c

could someone please show the maths behind it, and the way to those awnsers? i've already finished the question so i just need someone to explain it without asking a question ^ ^

9. May 17, 2009

### Andrew Mason

Since $TV^{\gamma - 1}$ does not change:

$$T_fV_f^{\gamma-1} = T_iV_i^{\gamma-1}$$

So:

$$\frac{T_f}{T_i} = \frac{V_i^{\gamma-1}}{V_f^{\gamma-1}}$$

since Vi/Vf = 8 and $\gamma = 5/3$:

$$\frac{T_f}{T_i} = 8^{2/3} = 4$$

So the internal energy increases by a factor of 4 and since $v_{rms}^2 \propto T$, the vrms increases by a factor of 2.

$$v_{rms-f} = 2v_{rms-i}$$

AM

10. May 18, 2009

### vorcil

Thanks, i actually figured it out in the end by myself :) Almost exactly the same way as you did XD
cheers