Final answer:
TVγ-1 is constant for an ideal gas during an adiabatic process because the adiabatic condition pVγ = constant combined with the ideal gas law implies T1V1γ-1 = T2V2γ-1. This relationship holds for all ideal gases, not just monatomic ones, but the constant value depends on the specific gas due to varying gamma (γ) values. Therefore, the correct answer is d) Adiabatic processes do not follow this relationship.
Step-by-step explanation:
To show that TVγ-1 is constant for an ideal gas undergoing an adiabatic process, we must use the relation for an adiabatic process where pVγ is constant. From the ideal gas law, we have pV = nRT. Combining these two relations, we can express pressure p in terms of temperature T and volume V, leading us to p = (nRT)/V.
Substituting this into the adiabatic condition pVγ = constant, we obtain
((nRT)/V)Vγ = nRTVγ-1 = constant, which implies T1V1γ-1 = T2V2γ-1, thus showing TVγ-1 is indeed constant.
This result is not only applicable to monatomic ideal gases but also to diatomic and polyatomic gases, with each having a specific γ (gamma) value, as γ is the heat capacity ratio and depends on the type of gas.
Therefore, the constant in the relation TVγ-1 does depend on the specific gas because the value of γ varies with the gas. Finally, it is incorrect to say that adiabatic processes do not follow this relationship; indeed, it is one of the fundamental characteristics of an adiabatic process for an ideal gas.