## Thermodynamics B/C

Justin72835
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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote:
Justin72835 wrote:
UTF-8 U+6211 U+662F wrote: I believe you missed the essential part of the question
1. More. By increasing the volume of the gas, you also increase the number of the microstates of the system by allowing the gas particles to occupy more locations. With more possible states for the gas to obtain, there is more entropy present.

2. More. By increasing the temperature of the gas, you are also increasing the average kinetic energy of the gas particles and thus their rms velocity. With the movements of the gas particles being more erratic and unpredictable at higher temperatures, there are undoubtedly more possible microstates, meaning that the system has a higher entropy.

3. More. WIth a homogenous gas, there are a set number of microstates. Switching any particles around won't create any new microstates because the particles are all identical. However, if instead of a single gas there were multiple gases, switching two gas particles of different species will create new microstates. Thus, the mix of gases has a higher entropy than the homogenous gas. Also, idk what the bonus is :? .

4. More. While monatomic gases only have 3 degrees of freedom (being able to move in the x, y, and z directions), diatomic gases actually have 5 degrees of freedom (being able to move in the x, y, and z directions + vibrational motion + rotational motion). Because of the extra 2 degrees of freedom, the diatomic gas actually has more possible microstates as the molecules can have different orientations and can vibrate differently as well. Therefore, a diatomic gas has more entropy than a monatomic gas.
You have a 1.5m long copper rod with a cross-sectional area of 0.004m^2 and a thermal conductivity of 385.0 W/mK. On one side of the rod you have 5 liters of water at 20 degrees Celsius and on the other side of the rod, you have another 5 liters of water at 95 degrees Celsius. If heat is transferred between the two bodies of water only through the copper rod, how long will it take for the temperatures to become within 1 degree of each other (for example, having temperatures of 54 and 55 degrees)?

Note: Assume that the rate of heat transfer changes over time due to the changing temperatures. Also, I asked for a 1 degree difference because it would take an infinite amount of time to actually reach equilibrium under these conditions.
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Justin72835
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### Re: Thermodynamics B/C

Justin72835 wrote:You have a 1.5m long copper rod with a cross-sectional area of 0.004m^2 and a thermal conductivity of 385.0 W/mK. On one side of the rod you have 5 liters of water at 20 degrees Celsius and on the other side of the rod, you have another 5 liters of water at 95 degrees Celsius. If heat is transferred between the two bodies of water only through the copper rod, how long will it take for the temperatures to become within 1 degree of each other (for example, having temperatures of 54 and 55 degrees)?

Note: Assume that the rate of heat transfer changes over time due to the changing temperatures. Also, I asked for a 1 degree difference because it would take an infinite amount of time to actually reach equilibrium under these conditions.
Alright so I decided just to put out the solution to this problem. Basically what you have to do is find the differential equation relating the instantaneous heat transferred and the instantaneous difference in temperature. From that, you can use Fourier's Law (conductivity equation) to solve for the difference in temperatures with respect to time. This will get you an answer of 12.22 hours.

Sorry everyone for the calc; it's just awesome when you find a cool calc problem that you finally know how to do. I'll let the next person go.
[img]https://i.imgur.com/SjcXA1N.jpg[/img]
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### Re: Thermodynamics B/C

You have a 1.7 m in diameter and 5 m in height cylinder with a movable top. This cylinder is filled with oxygen gas. Rocks that weigh 58 N are on top of the cylinder, and the system is at a temperature of 280 degrees Fahrenheit. Find the mass of the oxygen gas given that the gas is practically ideal and that the system is in static equilibrium (nothing is moving).

Justin72835
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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote:You have a 1.7 m in diameter and 5 m in height cylinder with a movable top. This cylinder is filled with oxygen gas. Rocks that weigh 58 N are on top of the cylinder, and the system is at a temperature of 280 degrees Fahrenheit. Find the mass of the oxygen gas given that the gas is practically ideal and that the system is in static equilibrium (nothing is moving).
First, establish all the state variables:
Temperature = 280 degrees Fahrenheit = 137.77 degrees Celsius
Volume = pi*r^2*h = 11.35 m^3
Pressure = F/A = 25.55 N (assuming that the piston is in a vacuum)

Using PV = nRT, the number of moles equates to 0.085 moles. Since the molar mass of O2 is 16 grams/mole, the total mass is 1.4 grams (2 significant figures).

If you assume that the piston is under atmospheric pressure as well (just add 101325 Pa to the pressure), you get an answer of 340 moles or a total mass of 5.4 kilograms (also 2 significant figures).
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### Re: Thermodynamics B/C

Justin72835 wrote:
UTF-8 U+6211 U+662F wrote:You have a 1.7 m in diameter and 5 m in height cylinder with a movable top. This cylinder is filled with oxygen gas. Rocks that weigh 58 N are on top of the cylinder, and the system is at a temperature of 280 degrees Fahrenheit. Find the mass of the oxygen gas given that the gas is practically ideal and that the system is in static equilibrium (nothing is moving).
First, establish all the state variables:
Temperature = 280 degrees Fahrenheit = 137.77 degrees Celsius
Volume = pi*r^2*h = 11.35 m^3
Pressure = F/A = 25.55 N (assuming that the piston is in a vacuum)

Using PV = nRT, the number of moles equates to 0.085 moles. Since the molar mass of O2 is 16 grams/mole, the total mass is 1.4 grams (2 significant figures).

If you assume that the piston is under atmospheric pressure as well (just add 101325 Pa to the pressure), you get an answer of 340 moles or a total mass of 5.4 kilograms (also 2 significant figures).
Correct (should've specified it was in a vacuum) except the molar mass of O2 is 32 grams/mole. Your turn!

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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote:
Justin72835 wrote:
UTF-8 U+6211 U+662F wrote:You have a 1.7 m in diameter and 5 m in height cylinder with a movable top. This cylinder is filled with oxygen gas. Rocks that weigh 58 N are on top of the cylinder, and the system is at a temperature of 280 degrees Fahrenheit. Find the mass of the oxygen gas given that the gas is practically ideal and that the system is in static equilibrium (nothing is moving).
First, establish all the state variables:
Temperature = 280 degrees Fahrenheit = 137.77 degrees Celsius
Volume = pi*r^2*h = 11.35 m^3
Pressure = F/A = 25.55 N (assuming that the piston is in a vacuum)

Using PV = nRT, the number of moles equates to 0.085 moles. Since the molar mass of O2 is 16 grams/mole, the total mass is 1.4 grams (2 significant figures).

If you assume that the piston is under atmospheric pressure as well (just add 101325 Pa to the pressure), you get an answer of 340 moles or a total mass of 5.4 kilograms (also 2 significant figures).
Correct (should've specified it was in a vacuum) except the molar mass of O2 is 32 grams/mole. Your turn!
Rip I completely missed the part about the molar mass. Oh well.

One mole of an ideal monatomic gas at a pressure of 100 kPa and a volume of 50 L expands isobarically to a volume of 120 L. Answer the following questions:
1) What is the constant pressure specific heat capacity of the gas?
2) How much heat was added/removed from the gas during its expansion?
3) What is the change in internal energy of the gas?
4) Instead of a monatomic gas, you are a given a diatomic gas. If the diatomic gas was to go through the exact same processes (same initial and final conditions), what would the change in internal energy be? Is this the same or different from your original answer and why or why not?
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### Re: Thermodynamics B/C

Justin72835 wrote:One mole of an ideal monatomic gas at a pressure of 100 kPa and a volume of 50 L expands isobarically to a volume of 120 L. Answer the following questions:
1) What is the constant pressure specific heat capacity of the gas?
2) How much heat was added/removed from the gas during its expansion?
3) What is the change in internal energy of the gas?
4) Instead of a monatomic gas, you are a given a diatomic gas. If the diatomic gas was to go through the exact same processes (same initial and final conditions), what would the change in internal energy be? Is this the same or different from your original answer and why or why not?
1) $\frac{Q}{n\Delta T} = \frac{\frac32P_fV_f - \frac32P_iV_i - P\Delta V}{1 mol * \left(\frac{P_fV_f}{1 mol * R} - \frac{P_iV_i}{1 mol * R}\right)}$

$= \frac{\frac32(100000 Pa)(0.12 m^3) - \frac32(100000 Pa)(0.05 m^3) + (100000 Pa)(0.12 m^3 - 0.05 m^3)}{1 mol * \left(\frac{(100000 Pa)(0.12 m^3)}{1 mol * R} - \frac{(100000 Pa)(0.05 m^3)}{1 mol * R}\right)}$

$= 20.785 \frac{J}{mol*K}$

2) +17500 J or +17.5 kJ

3) +10500 J or +10.5 kJ

4) $\frac52P_fV_f - \frac52P_iV_i = \frac52(100000 Pa)(0.12 m^3) - \frac52(100000 Pa)(0.05 m^3) = +17500 J = +17.5 kJ$

Different: Diatomic gases have two more degrees of freedom.

Justin72835
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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote:
Justin72835 wrote:One mole of an ideal monatomic gas at a pressure of 100 kPa and a volume of 50 L expands isobarically to a volume of 120 L. Answer the following questions:
1) What is the constant pressure specific heat capacity of the gas?
2) How much heat was added/removed from the gas during its expansion?
3) What is the change in internal energy of the gas?
4) Instead of a monatomic gas, you are a given a diatomic gas. If the diatomic gas was to go through the exact same processes (same initial and final conditions), what would the change in internal energy be? Is this the same or different from your original answer and why or why not?
1) $\frac{Q}{n\Delta T} = \frac{\frac32P_fV_f - \frac32P_iV_i - P\Delta V}{1 mol * \left(\frac{P_fV_f}{1 mol * R} - \frac{P_iV_i}{1 mol * R}\right)}$

$= \frac{\frac32(100000 Pa)(0.12 m^3) - \frac32(100000 Pa)(0.05 m^3) + (100000 Pa)(0.12 m^3 - 0.05 m^3)}{1 mol * \left(\frac{(100000 Pa)(0.12 m^3)}{1 mol * R} - \frac{(100000 Pa)(0.05 m^3)}{1 mol * R}\right)}$

$= 20.785 \frac{J}{mol*K}$

2) +17500 J or +17.5 kJ

3) +10500 J or +10.5 kJ

4) $\frac52P_fV_f - \frac52P_iV_i = \frac52(100000 Pa)(0.12 m^3) - \frac52(100000 Pa)(0.05 m^3) = +17500 J = +17.5 kJ$

Different: Diatomic gases have two more degrees of freedom.
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### Re: Thermodynamics B/C

When is the change in enthalpy equal to heat into the system?

Justin72835
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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote:When is the change in enthalpy equal to heat into the system?
Constant pressure
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