Thermodynamics temperature question

• Meowzers
In summary, the engine adds 2380 Joules of heat to the gas, removes 1781.336 Joules of heat from the gas, and does zero work.
Meowzers

Homework Statement

The pressure-volume graph shows an engine cycle with four processes: two isobaric, and two constant-volume. The engine uses 0.800 mol of an ideal monatomic gas as its working substance. For one engine cycle, calculate (a) the heat added to the gas, (b) the heat removed from the gas (stated as a positive number), and (c) the net work done by the engine.

http://www.slowbie.com/pressure-volume.gif

Homework Equations

Q=nCp(delta T), Cp=Cv+R
Q=heat, n=moles, Cp=molar specific heat, constant pressure, Cv=molar specific heat, constant volume, R=gas constant, T=temperature

The Attempt at a Solution

For part A, I used the molar specific heat-constant pressure equation to try to solve for the heat added to the gas.

Q=nCp(delta T), Cp=Cv+R
Cp=12.5+8.31
Cp=20.81
Then, I read the temperature difference from A to B.
Q=(.800)(20.81)(568K-425K)
Q=2380.064 J (Where am I making the error?)

For part B, is the amount of heat removed from the gas the same thing as part A but with temperature readings between C
and D?
With that, the equations would look like Q=(0.800)(20.81)(318-425)
Q=-1781.336 J

Last edited by a moderator:
Meowzers said:

The Attempt at a Solution

For part A, I used the molar specific heat-constant pressure equation to try to solve for the heat added to the gas.

Q=nCp(delta T), Cp=Cv+R
Cp=12.5+8.31
Cp=20.81
Then, I read the temperature difference from A to B.
Q=(.800)(20.81)(568K-425K)
Q=2380.064 J (Where am I making the error?)

For part B, is the amount of heat removed from the gas the same thing as part A but with temperature readings between C
and D?
With that, the equations would look like Q=(0.800)(20.81)(318-425)
Q=-1781.336 J

Heat is being added from D to A and from A to B. Heat is being removed from B to C and C to D. From D to A, $Q+ = nC_v\Delta T$ and from A to B, $Q+ = nC_p\Delta T$. The total is Qh. Figure that out to get the total heat added. I think you can similarly determine the heat removed (Qc).

To find the work, use the first law: $\Delta Q = Qh - Qc = \Delta U + W$. What is $\Delta U$ in a full cycle?

AM

Delta U is 0 in a full cycle since it returns to its initial state. So the work done is equal to the change in heat.

Thanks so much for your help! :)

1. What is the definition of temperature in thermodynamics?

In thermodynamics, temperature is a measure of the average kinetic energy of the particles in a system. It is related to the amount of heat present in the system and the direction of energy flow.

2. How is temperature measured in thermodynamics?

In thermodynamics, temperature is typically measured using a thermometer, which is a device that contains a temperature-sensitive material such as mercury or alcohol. The temperature is indicated by the expansion or contraction of the material.

3. What is the relationship between temperature and energy in thermodynamics?

In thermodynamics, temperature is directly proportional to the average kinetic energy of particles in a system. As temperature increases, so does the average kinetic energy, and vice versa.

4. How does temperature affect the behavior of a system in thermodynamics?

Temperature plays a crucial role in determining the direction of energy flow in a system. Heat will always flow from a higher temperature region to a lower temperature region, and systems tend to reach thermal equilibrium, where the temperatures of all parts are equal.

5. Can temperature be negative in thermodynamics?

In thermodynamics, temperature is an absolute scale, meaning it cannot be negative. However, it is possible to have negative values for temperature differences, such as when comparing temperatures on the Celsius and Kelvin scales.

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