Final answer:
The new temperature when the volume of gas is reduced to 1/4 of its initial volume at 27K with gamma of 1.4 is found using the adiabatic process equation. The correct formula after considering the volume reduction is T2 = 27K × (4)^0.4, leading to the answer a) 27 × (4)^0.4 K.
Step-by-step explanation:
When the volume of the gas is reduced to 1/4 of its initial volume at 27K and given that γ (gamma) = 1.4, we use the adiabatic process formula for ideal gases, which is P1V1^γ = P2V2^γ when the pressure is constant, or T1V1^γ-1 = T2V2^γ-1 when dealing with constant volume processes. In this case, we are focusing on temperature and volume changes, so we will use the latter formula. Given that the volume is reduced by 1/4, we can say that V2 = V1/4. Substituting in the values and solving for the new temperature (T2), we'll find:
T1 = 27K (this is the initial temperature)
V1 = initial volume
V2 = V1 / 4
γ = 1.4
Therefore, T2 = T1 × (V1 / V2)^γ-1
This simplifies to T2 = 27K × (1 / (1/4))^0.4, which simplifies to T2 = 27K × (4)^0.4. Thus, the correct answer is a) 27 × (4)^0.4 K.