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### Re: Thermodynamics B/C

Posted: March 20th, 2018, 6:47 pm
What is the power required to isothermally compress 2 kg of a near-ideal helium gas from atmospheric pressure to five times atmospheric pressure in one minute
a) at 100 K
b) at 200 K
c) in terms of the temperature T

### Re: Thermodynamics B/C

Posted: March 21st, 2018, 10:02 pm
UTF-8 U+6211 U+662F wrote:What is the power required to isothermally compress 2 kg of a near-ideal helium gas from atmospheric pressure to five times atmospheric pressure in one minute
a) at 100 K
b) at 200 K
c) in terms of the temperature T
The molar mass of Helium is 4 grams/mol, so 2 kg equates to 500 moles. The work done during an isothermal compression is given by the following equation:

$W=nRT*ln(\frac{T_f}{T_i}=nRT*ln(\frac{P_i}{P_f})$

For A:

$500*R*100*ln(\frac{1}{5})=-669 kJ$

Dividing by 60 seconds gives a power of [b]-11.15 kW removed[/b].

For B:

$500*R*200*ln(\frac{1}{5})=-1338 kJ$

Dividing by 60 seconds gives a power of [b]-22.30 kW removed[/b].

For C:

$500*RT*ln(\frac{1}{5})=-6.69T$

Dividing by 60 seconds gives a power of [b]-111.5T W removed[/b].

### Re: Thermodynamics B/C

Posted: March 22nd, 2018, 7:17 am
Justin72835 wrote:
UTF-8 U+6211 U+662F wrote:What is the power required to isothermally compress 2 kg of a near-ideal helium gas from atmospheric pressure to five times atmospheric pressure in one minute
a) at 100 K
b) at 200 K
c) in terms of the temperature T
The molar mass of Helium is 4 grams/mol, so 2 kg equates to 500 moles. The work done during an isothermal compression is given by the following equation:

$W=nRT*ln(\frac{T_f}{T_i}=nRT*ln(\frac{P_i}{P_f})$

For A:

$500*R*100*ln(\frac{1}{5})=-669 kJ$

Dividing by 60 seconds gives a power of [b]-11.15 kW removed[/b].

For B:

$500*R*200*ln(\frac{1}{5})=-1338 kJ$

Dividing by 60 seconds gives a power of [b]-22.30 kW removed[/b].

For C:

$500*RT*ln(\frac{1}{5})=-6.69T$

Dividing by 60 seconds gives a power of [b]-111.5T W removed[/b].
Nice, your turn! (Also, there's a typo in the first line, should be Vf/Vi instead of Tf/Ti)

Note that the work done on the gas is -nRTln(Vf/Vi) though which makes it positive.

### Re: Thermodynamics B/C

Posted: March 28th, 2018, 6:10 pm
UTF-8 U+6211 U+662F wrote:
Justin72835 wrote:
UTF-8 U+6211 U+662F wrote:What is the power required to isothermally compress 2 kg of a near-ideal helium gas from atmospheric pressure to five times atmospheric pressure in one minute
a) at 100 K
b) at 200 K
c) in terms of the temperature T
The molar mass of Helium is 4 grams/mol, so 2 kg equates to 500 moles. The work done during an isothermal compression is given by the following equation:

$W=nRT*ln(\frac{T_f}{T_i}=nRT*ln(\frac{P_i}{P_f})$

For A:

$500*R*100*ln(\frac{1}{5})=-669 kJ$

Dividing by 60 seconds gives a power of [b]-11.15 kW removed[/b].

For B:

$500*R*200*ln(\frac{1}{5})=-1338 kJ$

Dividing by 60 seconds gives a power of [b]-22.30 kW removed[/b].

For C:

$500*RT*ln(\frac{1}{5})=-6.69T$

Dividing by 60 seconds gives a power of [b]-111.5T W removed[/b].
Nice, your turn! (Also, there's a typo in the first line, should be Vf/Vi instead of Tf/Ti)

Note that the work done on the gas is -nRTln(Vf/Vi) though which makes it positive.
You have a metal pot with heat capacity 540 J/K at a temperature of 114 °C. You then add 350 mL of water at 23 °C into the pot.

a. Is there any steam produced? (yes or no)

b. If yes, then what is the final temperature of the steam? If no, then what is the equilibrium temperature between the water and the pot?

### Re: Thermodynamics B/C

Posted: March 30th, 2018, 3:09 pm
Justin72835 wrote: You have a metal pot with heat capacity 540 J/K at a temperature of 114 °C. You then add 350 mL of water at 23 °C into the pot.

a. Is there any steam produced? (yes or no)

b. If yes, then what is the final temperature of the steam? If no, then what is the equilibrium temperature between the water and the pot?
Assuming the whole pot of water has to get to 100 degrees before any steam is produced:
Energy to heat water + energy to boil water + energy to heat steam = energy lost from metal pot.
$4.2 \frac{J}{g*K} * 350 mL * 1 \frac{g}{mL} * 77 K + 2230 J/g * 350 mL * 1 \frac{g}{mL} + 2.0 \frac{J}{g*K} * 350 mL * 1 \frac{g}{mL} * (T - 100 \degree C) = 540 \frac{J}{K} * (114 \degree C - T)$
=> T = -615 degrees Celsius, which is clearly impossible.

$4.2 \frac{J}{g*K} * 350 mL * 1 \frac{g}{mL} * (T - 23 \degree C) = 540 \frac{J}{K} * (114 \degree C - T)$
=> T = 47 degrees Celsius

a) No
b) 47 degrees Celsius

### Re: Thermodynamics B/C

Posted: March 30th, 2018, 5:40 pm
UTF-8 U+6211 U+662F wrote:
Justin72835 wrote: You have a metal pot with heat capacity 540 J/K at a temperature of 114 °C. You then add 350 mL of water at 23 °C into the pot.

a. Is there any steam produced? (yes or no)

b. If yes, then what is the final temperature of the steam? If no, then what is the equilibrium temperature between the water and the pot?
Assuming the whole pot of water has to get to 100 degrees before any steam is produced:
Energy to heat water + energy to boil water + energy to heat steam = energy lost from metal pot.
$4.2 \frac{J}{g*K} * 350 mL * 1 \frac{g}{mL} * 77 K + 2230 J/g * 350 mL * 1 \frac{g}{mL} + 2.0 \frac{J}{g*K} * 350 mL * 1 \frac{g}{mL} * (T - 100 \degree C) = 540 \frac{J}{K} * (114 \degree C - T)$
=> T = -615 degrees Celsius, which is clearly impossible.

$4.2 \frac{J}{g*K} * 350 mL * 1 \frac{g}{mL} * (T - 23 \degree C) = 540 \frac{J}{K} * (114 \degree C - T)$
=> T = 47 degrees Celsius

a) No
b) 47 degrees Celsius

### Re: Thermodynamics B/C

Posted: March 31st, 2018, 6:23 pm
Explain why we can make assumptions in the derivation of the ideal gas law, PV = nRT, such as "The number of molecules moving in each axis (x, y, and z) is equal".

### Re: Thermodynamics B/C

Posted: April 3rd, 2018, 11:35 am
UTF-8 U+6211 U+662F wrote:Explain why we can make assumptions in the derivation of the ideal gas law, PV = nRT, such as "The number of molecules moving in each axis (x, y, and z) is equal".
Not really sure about this one. My answer would be that because the molecules are so small, there are so many of them, and they have relatively little interaction with one another, you can assume that all the molecules follow Newton's Law of Motion and collide elastically with each other. From this, you are able to draw other conclusions as well.

### Re: Thermodynamics B/C

Posted: April 3rd, 2018, 5:44 pm
Justin72835 wrote:
UTF-8 U+6211 U+662F wrote:Explain why we can make assumptions in the derivation of the ideal gas law, PV = nRT, such as "The number of molecules moving in each axis (x, y, and z) is equal".
Not really sure about this one. My answer would be that because the molecules are so small, there are so many of them, and they have relatively little interaction with one another, you can assume that all the molecules follow Newton's Law of Motion and collide elastically with each other. From this, you are able to draw other conclusions as well.
for the record, I was just looking for there being so many molecules that statistical treatment can be applied to them (especially that they have approximately random motion)

### Re: Thermodynamics B/C

Posted: April 3rd, 2018, 10:26 pm
Two gases occupy two containers, A and B. The gas in A, of volume 0.14 cubic meters, exerts a pressure of 1.18 MPa. The gas in B, of volume 0.21 cubic meters, exerts a pressure of 0.82 MPa. The containers are united by a tube of negligible volume and the gases are allowed to intermingle. What is the final pressure in the container if the temperature remains constant?