Why doesn't ΔU = 0 (U3= U1) in this problem, where pV = constant?

  • Thread starter grotiare
  • Start date
  • Tags
    Constant
In summary: For the part of the cycle that ##pV=constant## you need to calculate that constant ##C## first, and then the work will be $$W_{3\to1}=\int_{V_3}^{V_1}\frac{C}{V}dV$$For the part of the cycle that ##pV=constant## you need to calculate that constant ##C## first, and then the work will be $$W_{3\to1}=\int_{V_3}^{V_1}\frac{C}{V}dV$$In summary, it is possible to solve this problem without assuming that the gas is ideal, by calculating the total
  • #1
grotiare
5
4
EDIT: My bad, this is from my professors lecture, but I assume this question is more suitable for the homework section of this forum

The problem gives that pV = constant, and thus is assumed to be an isothermal process. However, the given U1 and U3 are not equal to each other. Is it because it is not specified the gas is ideal that we ignore that ΔU = 0 ?

unknown.png
 
  • Like
Likes Delta2
Physics news on Phys.org
  • #2
Yes I think that's the catch, it is not specified that the gas is ideal.

You can solve this problem, without assuming that gas is ideal and without using ##PV=nRT## (or ##\Delta U=nC_V\Delta T##). Just calculate the total work done in the cycle and then use 1st law of thermodynamics for the cycle.

For the part of the cycle that ##pV=constant## you need to calculate that constant ##C## first, and then the work will be $$W_{3\to1}=\int_{V_3}^{V_1}\frac{C}{V}dV$$

P.S The solution I see to this problem , doesn't use anywhere the given values of ##U_1,U_3##.
 
Last edited:
  • #3
Delta2 said:
Yes I think that's the catch, it is not specified that the gas is ideal.

You can solve this problem, without assuming that gas is ideal and without using ##PV=nRT## (or ##\Delta U=nC_V\Delta T##). Just calculate the total work done in the cycle (as the total area in a PV diagram that is enclosed by the cycle) and then use 1st law of thermodynamics for the cycle.

For the part of the cycle that ##pV=constant## you need to calculate that constant ##C## first, and then the work will be $$W_{3\to1}=\int_{V_3}^{V_1}\frac{C}{V}dV$$
Okay cool thank you! So it would thus just be this process-> C = constant = PV, and thus you would integrate 1/V dV from V1 to V3 (leaving c=pv outside of integral) to get W3-1 = PV*ln(V1/V3), where PV = P1V1 or P3V3
 
  • #4
grotiare said:
Okay cool thank you! So it would thus just be this process-> C = constant = PV, and thus you would integrate 1/V dV from V1 to V3 (leaving c=pv outside of integral) to get W3-1 = PV*ln(V1/V3), where PV = P1V1 or P3V3
Yes that's correct. And also very easily you can calculate the work from ##1 \to 2## (isochoric process) and ##2\to 3## (isobaric process).
 
  • #5
grotiare said:
The problem gives that pV = constant, and thus is assumed to be an isothermal process.
Wrong assumption.

You have 2 processes - one constant volume, one constant pressure - for which you can calculate the pressure-volume work. (part b)

Then you use the first law to calculate the heat transfer. (part c)
 
  • Like
Likes Delta2

FAQ: Why doesn't ΔU = 0 (U3= U1) in this problem, where pV = constant?

Why does ΔU = 0 when pV = constant?

According to the first law of thermodynamics, the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W). When pV = constant, the work done by the system is equal to zero, since there is no change in volume. Therefore, ΔU = Q - 0 = Q. This means that the change in internal energy is solely determined by the heat added to the system.

How does the constant pressure affect the change in internal energy?

When the pressure is constant, any change in volume will result in a change in temperature. This is because the internal energy of a gas is directly proportional to its temperature. As the volume increases, the gas molecules have more space to move around, thus increasing their kinetic energy and the overall temperature of the system. Similarly, if the volume decreases, the gas molecules have less space to move around, resulting in a decrease in temperature.

Can the change in internal energy be negative when pV = constant?

Yes, the change in internal energy can be negative when pV = constant. This would occur if the heat removed from the system (Q) is greater than the work done by the system (W). In this case, the system would lose internal energy, resulting in a negative ΔU value.

Is pV = constant always true for all thermodynamic processes?

No, pV = constant is not always true for all thermodynamic processes. This equation only applies to isobaric processes, where the pressure remains constant. In other processes such as isochoric (constant volume) or isothermal (constant temperature), the pressure may vary, and therefore pV = constant would not apply.

How does the ideal gas law relate to pV = constant?

The ideal gas law, PV = nRT, describes the relationship between pressure, volume, temperature, and the number of moles of an ideal gas. When pV = constant, it means that the number of moles (n) and the temperature (T) are also constant. Therefore, the ideal gas law can be simplified to P = constant, which is the same as pV = constant.

Back
Top