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AP Physics 2 Exam Question - Thermodynamic Processes
A gas undergoes a cyclic thermodynamic process as shown in the diagram below:
The gas starts at point A, undergoes a process to point B, then follows another process to point C, and finally returns to point A. The process from A to B is an isobaric process, the process from B to C is an isochoric process, and the process from C to A is an adiabatic process.
- Is the internal energy of the gas at point B greater than, less than, or equal to the internal energy at point A? Justify your answer.
Answer:
The internal energy of an ideal gas is solely dependent on its temperature and not on its pressure or volume. In an isobaric process, the pressure is constant, meaning the change in internal energy is directly proportional to the change in temperature.
Since the temperature at point B is greater than the temperature at point A (as shown by the diagram), the internal energy at point B is greater than the internal energy at point A.
Therefore, the internal energy of the gas at point B is greater than the internal energy at point A.
- Calculate the work done by the gas during the process from point B to point C.
Answer:
During an isochoric process (constant volume), no work is done by or on the gas. Therefore, the work done by the gas during the process from point B to point C is zero.
- Is the heat added to the gas during the process from point C to point A greater than, less than, or equal to the work done by the gas during the process from point B to point C? Justify your answer.
Answer:
In an adiabatic process, no heat is transferred between the gas and its surroundings. Therefore, the heat added to the gas during the process from point C to point A is zero.
As calculated in the previous question, the work done by the gas during the process from point B to point C is also zero.
Therefore, the heat added to the gas during the process from point C to point A is equal to the work done by the gas during the process from point B to point C (both are zero).
This result is in accordance with the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
Since ΔU is zero for the whole cycle, the heat added (Q) equals the work done (W) in the cycle, which in this case is both zero.
- Is the efficiency of this thermodynamic cycle greater than, less than, or equal to 50%? Justify your answer.
Answer:
To calculate the efficiency of a thermodynamic cycle, we use the formula:
Efficiency = (Work done by the system / Heat input to the system) * 100%
From the previous answers, we know that the work done by the system is zero (since both processes B to C and C to A have no work done) and the heat input to the system is also zero (since Q is zero for the process C to A).
Therefore, the efficiency of this thermodynamic cycle is zero percent.
Hence, the efficiency of this thermodynamic cycle is less than 50%.
Note: The efficiency of a thermodynamic cycle is typically less than 100% due to energy loss in the form of heat.