Two different answers for work done during compression

In summary, the conversation discusses the calculation of work done in a reversible process and the use of the first law of thermodynamics. The coefficients in the two equations for work done are different due to the gas being monatomic, which affects the value of γ. After researching and finding the correct value for γ, the coefficients are shown to be equal, resolving the issue.
  • #1
etotheipi
Homework Statement
Calculating work done during adiabatic compression (please see problem statement below)
Relevant Equations
Ideal gas laws, internal energy
I'm having a little trouble with part a) of this question:

Screenshot 2019-12-31 at 16.20.46.png

Since it is stated that the heating is slow, I thought it was reasonable to assume the process is reversible which means that the pressure in both sides should be equal. Consequently, $$W = - \int_{V_{0}}^{V_{1}} P dV = - \int_{V_{0}}^{V_{1}} kV^{- \gamma} dV = \frac{1}{\gamma-1}(P_{1}V_{1} - P_{0}V_{0})$$ For ##\gamma = 1.5##, the outside coefficient is ##2##.

However, if I use the first law (with ##Q=0## due to the insulated/adiabatic condition), I get $$W = \Delta U = \frac{3}{2}(nRT_{1} - nRT_{0}) = \frac{3}{2}(P_{1}V_{1} - P_{0}V_{0})$$I can't figure out why the two coefficients of ##(P_{1}V_{1} - P_{0}V_{0})## are different!

After this is sorted out, I assume it will just be a case of doing some rearrangement (i.e. w/ ##\frac{P_{0}V_{0}}{T_{0}} = \frac{P_{1}V_{1}}{T_{1}}## etc.) to get it into the required form?
 
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  • #2
etotheipi said:
However, if I use the first law (with ##Q=0## due to the insulated/adiabatic condition), I get $$W = \Delta U = \frac{3}{2}(nRT_{1} - nRT_{0}) = \frac{3}{2}(P_{1}V_{1} - P_{0}V_{0})$$I can't figure out why the two coefficients of ##(P_{1}V_{1} - P_{0}V_{0})## are different!
The coefficient ##\frac{3}{2}## in ##\Delta U## is only for a monatomic ideal gas. Is the gas monatomic in this problem?
 
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  • #3
TSny said:
The coefficient ##\frac{3}{2}## in ##\Delta U## is only for a monatomic ideal gas. Is the gas monatomic in this problem?

Thank you, I just googled ##\gamma## for an ideal monatomic gas (it turns out to be roughly 1.66), which gives ##\frac{1}{\gamma-1} \approx 1.5##, as was obtained via the other method!
 
  • #4
etotheipi said:
Thank you, I just googled ##\gamma## for an ideal monatomic gas (it turns out to be roughly 1.66), which gives ##\frac{1}{\gamma-1} \approx 1.5##, as was obtained via the other method!
Is everything OK now?

For a general ideal gas, ##U = nC_VT## and ##C_V = \frac{R}{\gamma - 1}##.
 
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FAQ: Two different answers for work done during compression

What is work done during compression?

Work done during compression refers to the amount of energy that is transferred to an object as it is compressed. This can be calculated by multiplying the force applied to the object by the distance it is compressed.

Why are there two different answers for work done during compression?

There may be two different answers for work done during compression because there are two different ways to calculate it. One method is to use the force applied to the object and the distance it is compressed, while the other method uses the change in potential energy of the object.

Which method is more accurate for calculating work done during compression?

Both methods are equally accurate for calculating work done during compression. However, the method using the change in potential energy may be more useful in certain situations, such as when dealing with elastic materials.

What are some real-life examples of work done during compression?

Some real-life examples of work done during compression include compressing a spring, squeezing a rubber ball, or pressing down on a mattress.

How does work done during compression affect the object being compressed?

The work done during compression results in the object gaining potential energy, which is stored as internal energy within the object. This can cause the object to change shape, temperature, or other physical properties.

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