Thermodynamics Homework: Find Work Given Heat & Temp Change

In summary, the conversation discusses the process of adding heat to an ideal monatomic gas and its resulting temperature increase. The goal is to find the work done by the gas during this process. The formula \DeltaU = Q - W is suggested, but the individual is unsure of how to relate \Delta U to \Delta T. It is then suggested to check a reference for an equation of internal energy and to use equipartition of energy to express internal energy as a function of temperature. Ultimately, the work done by the gas can be determined by applying the first law of thermodynamics.
  • #1
Robershky
9
0

Homework Statement



When 1380 J of heat are added to one mole of an ideal monatomic gas, its temperature increases from 272 K to 275 K. Find the work done by the gas during this process.

Homework Equations



I assume [tex]\Delta[/tex]U = Q - W is the formula to use. But for some reason I'm just confused.

The Attempt at a Solution



I tried using 1380 = 275 - W, and got -1105. But that is not the correct answer.
 
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  • #2
Can you relate [itex]\Delta U[/itex] to [itex]\Delta T[/itex]?
 
  • #3
That's the problem, I don't understand how they relate. Is there a conversion from Kelvin to Joules? Or do I just use 3K = 1380J - W?
 
  • #4
Check your book, notes, or reference for an equation of the internal energy (U) of an ideal gas.
 
  • #5
Internal energy increases with increasing temperature, but the exact relation will depend on the type of gas. You should figure out how many degrees of freedom are present in a monatomic gas, then use equipartition of energy to express the internal energy as a function of temperature.
 
  • #6
Robershky said:

Homework Statement



When 1380 J of heat are added to one mole of an ideal monatomic gas, its temperature increases from 272 K to 275 K. Find the work done by the gas during this process.

Homework Equations



I assume [tex]\Delta[/tex]U = Q - W is the formula to use. But for some reason I'm just confused.

The Attempt at a Solution



I tried using 1380 = 275 - W, and got -1105. But that is not the correct answer.
You are given Q and you can determine [itex]\Delta U = nC_v\Delta T[/itex]. So it is just a matter of applying the first law to determine W.

AM
 

Related to Thermodynamics Homework: Find Work Given Heat & Temp Change

1. How do I calculate work using thermodynamics?

To calculate work using thermodynamics, you will need to use the formula W = Q - ΔU, where W represents work, Q represents heat, and ΔU represents the change in internal energy. This formula takes into account energy conservation and allows you to find work given heat and a change in temperature.

2. What is the unit of measurement for work in thermodynamics?

The unit of measurement for work in thermodynamics is joules (J). This is the same unit used to measure heat and energy in the metric system. In some cases, other units such as kilojoules (kJ) or calories (cal) may be used, but they can easily be converted to joules.

3. Can I use the same formula to calculate work in different thermodynamic processes?

No, the formula W = Q - ΔU can only be used to calculate work in isothermal processes, where the temperature remains constant. For other processes such as adiabatic or isobaric, different formulas must be used. It's important to understand the specific conditions of the process in order to use the correct formula.

4. How does the direction of work relate to the sign of the calculated value?

The direction of work is determined by the sign of the calculated value. A positive value for work indicates that work is being done on the system, while a negative value indicates that work is being done by the system. This is consistent with the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred or converted.

5. What is the significance of calculating work in thermodynamics?

Calculating work in thermodynamics allows us to understand and quantify the energy changes that occur in a system. This is important for many industrial and engineering applications, as well as for understanding the behavior of gases and other materials. It also allows us to predict and control the efficiency of heat engines and other energy conversion processes.

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