Guide 18-2. Solving Thermal Process Problems

Here is the strategy for solving thermal process problems.

  1. Determine what is constant or 0, if anything, in the problem. The usual choices are constant volume (isochoric process), constant pressure (isobaric process), constant temperature (isothermal process), no heat transfer (adiabatic process).

  2. Determine the signs of those things that are not 0. Use the conventions that heat added to a system is positive, and that Wse is positive when the system does work on the environment.

  3. Write the first law of thermodynamics and substitute those things that are 0 for the given problem situation. Q is 0 for an adiabatic process, ΔU is 0 for an isothermal process, and Wse is 0 for an isochoric process. ΔP is 0 for an isobaric process.

  4. Draw a P-V graph of the process. Indicate with an arrow the direction that the process proceeds. The graph provides a visual representation of the work done. The arrow is a reminder of whether pressure and volume are increasing, decreasing, or staying the same.

  5. Construct a table where you a) summarize what you know and b) indicate what you are to find. This serves as given and goal. Give the initial and final values of pressure, volume, and temperature. If you don't know the values, you can still indicate what change to expect (+, 0, -) in the Change column. For the quantities ΔU, Q, Wse that play central roles in the first law, state any values that are given (with the correct sign). If values aren't given, indicate whether the quantity is expected to be +, 0, or -. Indicate with question marks those things that you are supposed to find. If you're not sure about any of the table entries, you can leave them blank and add them as you work the problem. Generally, you won't need all of the information in the table to solve the problem.
Variable Change Initial Final
P      
V      
T      
ΔU  
Q  
Wse  
  1. Now you can decide on what kind of thermal process is involved and what equations to use to complete the solution. Here are the possibilities. 
Description Relationship Condition under which relationship applies Comment
First law of thermodynamics Q - Wse = ΔU Relationship is always applicable You can simplify this if any of the three quantities are 0.
Internal energy of a gas U = 3nRT/2 Relationship applies to ideal gases. This means the internal energy of an enclosed gas depends only on the temperature.
Ideal gas law PV = nRT = NkT Relationship applies to ideal gases. If a problem doesn't state that the gas is ideal, try to solve the problem without using the ideal gas law.
Work done in a constant pressure process Wse = PΔV Pressure is constant.  
Work done in a linear process Wse = PaveΔV Pressure and volume have a linear relationship. See Example 18-2 in the text.
Work done in an isothermal process Wse = nRTln(Vf/Vi) Relationship applies to ideal gases. Pressure and volume have an inverse relationship.

Examples

We'll construct tables for two of the examples in the book. The first one is Example 18-3. Read the problem first.

The process is isothermal, so ΔT and ΔU are zero. We can see from the graph that the pressure is decreasing and the volume increasing. Work is done by the gas (system) on the environment in pushing the piston outward. This is positive work. Heat input is needed to provide the energy for doing work. Here's the table corresponding to this situation. In addition, we know that the number of moles of the gas is 0.50.

The P-V diagram isP-V diagram

Variable Change Initial Final
P -    
V + 0.31 m3 0.45 m3
T 0 310 K 310 K
ΔU 0
Q + ?
Wse +

For an isothermal process, the first law gives us Q = Wse. The work is found from Wse = nRTln(Vf/Vi).  We're given n, T, Vi, and Vf, so we can solve for Wse and therefore Q. We won't complete the solution here, since it's done in the text.

The next example is Example 18-4. The process is adiabatic, so Q = 0. The pressure is increasing and the volume decreasing. Work is done on the piston (environment) to compress the gas (system). This is negative work. The internal energy and temperature must increase, since heat cannot leave the system.

The P-V diagram is P-V diagram

Here's the table corresponding to this situation.

Variable Change Initial Final
P +    
V -    
T +    
ΔU +?
Q 0
Wse -640 J

For an adiabatic process, the first law gives us ΔU  = -Wse.  Since we're given a value for the work, it's a simple matter to calculate ΔU as the negative of the work.


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