A monatomic gas is adiabatically compressed

[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

 

[vorcil]

anyone?

 

[Andrew Mason]

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

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 v_rms 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

 

[vorcil]

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 v_rms 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

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 v_rms 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)

hangon i'll just google it

 

[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

 

[vorcil]

If the question called for an actual calculation (I am not
Would the specific heat Cv change? (remember, the gas is ideal).

AM

they are constant?

 

[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

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

 

[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 ^ ^

 

[Andrew Mason]

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 ^ ^

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

 

[vorcil]

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

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