A quantity of a monatomic ideal gas undergoes a process in which both its pressure and volume are increased by a factor of n = 3 as shown in the figure below. What is the energy absorbed by heat into the gas during this process? Hint: The internal energy of a monatomic ideal gas at pressure P and occupying volume V is given by U = 32PV. (Give your answer as a multiple of P0V0.) Your response differs from the correct answer by more than 10%. Double check your calculations. ✕ P0V0
Question
A quantity of a monatomic ideal gas undergoes a process in which both its pressure and volume are increased by a factor of n = 3 as shown in the figure below. What is the energy absorbed by heat into the gas during this process? Hint: The internal energy of a monatomic ideal gas at pressure P and occupying volume V is given by U = 32PV. (Give your answer as a multiple of P0V0.) Your response differs from the correct answer by more than 10%. Double check your calculations. ✕ P0V0
Solution
The internal energy of a monatomic ideal gas at pressure P and occupying volume V is given by U = 3/2 PV.
Let's denote the initial state of the gas by subscript 0 and the final state by subscript f.
The initial internal energy of the gas is U0 = 3/2 P0V0.
Since both the pressure and volume are increased by a factor of n = 3, the final internal energy of the gas is Uf = 3/2 P0V0 * n^2 = 3/2 P0V0 * 9 = 13.5 P0V0.
The change in internal energy of the gas is ΔU = Uf - U0 = 13.5 P0V0 - 3/2 P0V0 = 12 P0V0.
According to the first law of thermodynamics, the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In this case, the gas does no work (since it is not expanding or contracting), so the heat added to the system is simply Q = ΔU = 12 P0V0.
So, the energy absorbed by heat into the gas during this process is 12 P0V0.
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