Final answer:
The heat absorbed by the gas during an isobaric expansion, where it performs 25 J of work, is 75 J as per the first law of thermodynamics and assuming constant internal energy.
option a is the correct
Step-by-step explanation:
The question deals with the concept of work, heat, and internal energy in the context of thermodynamics, which is a branch of physics. Given that the average degrees of freedom per molecule for a gas is 6 and the gas performs 25 J of work when it expands at constant pressure, we can use the first law of thermodynamics to find the heat absorbed by the gas.
The first law of thermodynamics states that the change in internal energy of a system (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W): ΔU = Q - W. For an ideal gas with constant pressure, the change in internal energy can be calculated using the degrees of freedom. A diatomic gas has 5 degrees of freedom under normal conditions but can have 6 degrees of freedom if vibrational modes are activated due to high temperatures or other factors. Here, we assume an average value of 6 degrees of freedom, suggestive of a polyatomic gas with both translational, rotational, and vibrational modes.
According to the equipartition theorem, each degree of freedom contributes ½ kT to the internal energy per molecule (with k being the Boltzmann constant and T the temperature), so the molar internal energy (ΔU) would be ΔU = ½ × degrees of freedom × (R × T), where R is the gas constant. Since we are not given temperature or the number of moles, we cannot calculate ΔU directly. However, we are given that the work done by the gas is 25 J, and if the process is isobaric (constant pressure), then ΔU = Q - W. Assuming no change in the internal energy, which is typically the case in an isobaric process of an ideal gas, we get Q = W. Thus, the heat absorbed by the gas would be the same as the work done on the gas, leading to an answer of 75 J.